From eda19884250b7a33f1b4966b228fb87ddffbe61e Mon Sep 17 00:00:00 2001 From: Manny Gimond Date: Tue, 20 Feb 2024 09:41:39 -0500 Subject: [PATCH] made a few dplyr chapter edits --- docs/dplyr.html | 208 ++++++++--------- docs/ggplot2.html | 18 +- docs/joins.html | 6 +- docs/lattice_plot.html | 6 +- docs/letter_values.html | 6 +- docs/re_express.html | 2 +- docs/read_write_files.html | 14 +- docs/rf.html | 2 +- docs/search.json | 6 +- docs/strings.html | 6 +- docs/subset_base.html | 12 +- docs/theoretical_qq.html | 2 +- docs/tidyr.html | 8 +- docs/univariate_plots.html | 16 +- dplyr.qmd | 69 +++--- ...k-1_e3ba058d8070d51d61096aadebc5455c.RData | Bin 161 -> 0 bytes ...unk-1_e3ba058d8070d51d61096aadebc5455c.rdb | Bin 1370 -> 0 bytes ...unk-1_e3ba058d8070d51d61096aadebc5455c.rdx | Bin 158 -> 0 bytes ...-29_0f202011f488ca11f660ae78566050e9.RData | Bin 299 -> 0 bytes ...-29_91b267ecfb4302c8f57e3ca23fefdf2e.RData | Bin 0 -> 299 bytes ...k-29_91b267ecfb4302c8f57e3ca23fefdf2e.rdb} | Bin 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univariate_plots_files/figure-html/unnamed-chunk-16-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-30-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-31-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-32-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-39-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-40-1.png create mode 100644 univariate_plots_files/figure-html/unnamed-chunk-41-1.png diff --git a/docs/dplyr.html b/docs/dplyr.html index 60859d29..3dec0363 100644 --- a/docs/dplyr.html +++ b/docs/dplyr.html @@ -942,119 +942,97 @@

[1] "1" "2" "2" "apple" "1" -

The output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. apple =1, banana=2, etc…). Likewise, if you wish to replace an erroneous date with a missing value you will get:

-
+

The output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. apple =1, banana=2, etc…). Likewise, if you wish to replace an erroneous date you will get:

+
library(lubridate)
 y <- mdy("1/23/2016", "3/2/2016", "12/1/1901", "11/23/2016")
-ifelse( year(y) != 2016, NA, y)
+ifelse( y == mdy("12/1/1901"), mdy("12/1/2016"), y)
-
[1] 16823 16862    NA 17128
+
[1] 16823 16862 17136 17128

Here, ifelse converts the date object to its internal numeric representation as number of days since 1970.

9.3.2 dplyr’s if_else

-

ifelse does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector’s class. If you want to ensure that the vector’s class is preserved, a safer alternative is to use dplyr’s if_else function.

+

ifelse does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector’s class. A safer alternative is to use dplyr’s if_else function.

Reworking the above examples:

-
-
if_else( year(y) != 2016, NA, y)
+
+
if_else( y == mdy("12/1/1901"), mdy("12/1/2016"), y)
-
[1] "2016-01-23" "2016-03-02" NA           "2016-11-23"
+
[1] "2016-01-23" "2016-03-02" "2016-12-01" "2016-11-23"
+

The date class is preserved. Now let’s check the output of a factor.

if_else(x == "pear", "apple", x)
[1] "apple"  "banana" "banana" "apple"  "apple"
-

Note that when working with factors, however, if_else will strip the factor class of an input factor. dplyr offers a recode function for this purpose however, this function is being superseded according to the documentation as of version 1.1.4. The workaround is to convert the if_else output to a factor.

-
-
as.factor(if_else(x == "pear", "apple", x))
+

Note that when working with factors, however, if_else will strip the factor class of an input factor. But, instead of returning the factor’s underlying integer values, it outputs the associated levels as a character data type.

+

The workaround is to convert the if_else output to a factor.

+
+
factor(if_else(x == "pear", "apple", x))
[1] apple  banana banana apple  apple 
 Levels: apple banana
-
-

9.3.2.1 Note on replacing NA factor levels

-

One operation you cannot perform with if_else is converting an NA level to another factor level. For example, let’s add an NA value to the x factor:

-
-
x[2] <- NA
-x
-
-
[1] apple  <NA>   banana pear   apple 
-Levels: apple banana pear
-
-
-

If we want to replace NA with banana the following will generate an unexpected output.

-
-
if_else( x == NA, "banana", x)
-
-
[1] NA NA NA NA NA
-
-
-

To evaluate if a value is NA, use the is.na() function as in:

-
-
if_else( is.na(x), "banana", x)
-
-
[1] "apple"  "banana" "banana" "pear"   "apple"
-
-
-
+

If you need to explicitly define the levels, add the levels = c("pear",...).

+

NOTE: dplyr offers the recode function that preserves factors however, this function is being superseded according to the documentation as of version 1.1.4.

9.3.3 Changing values based on multiple conditions: case_when

-

ifelse and if_else work great when a single set of conditions is to be satisfied. But if multiple sets of conditions are to be tested, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.

-
-
unit <- c("F","F","C", "K")
-if_else( unit == "C", "Celsius", if_else(unit == "F", "Fahrenheit", "Kelvin"))
+

ifelse and if_else work great when a single set of conditions is to be satisfied. But, if multiple sets of conditions are to be evaluated, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.

+
+
unit <- c("F","F","C", "K")
+if_else( unit == "C", "Celsius", if_else(unit == "F", "Fahrenheit", "Kelvin"))
[1] "Fahrenheit" "Fahrenheit" "Celsius"    "Kelvin"

A simpler solution is to use the case_when function.

-
-
case_when(unit == "C" ~ "Celsius",
-          unit == "F" ~ "Fahrenheit",
-          unit == "K" ~ "Kelvin")
+
+
case_when(unit == "C" ~ "Celsius",
+          unit == "F" ~ "Fahrenheit",
+          unit == "K" ~ "Kelvin")
[1] "Fahrenheit" "Fahrenheit" "Celsius"    "Kelvin"

case_when can aso be used for more complex operations. For example, given two vectors, unit and temp, we would like to convert all temp values to Fahrenheit by applying a temperature conversion dependent on the unit value.

-
-
temp <- c(45.2, 56.3, 11.0, 285)
+
+
temp <- c(45.2, 56.3, 11.0, 285)
-
-
case_when(unit == "F" ~ temp,
-          unit == "C" ~ (temp * 9/5) + 32,
-          TRUE ~ (temp - 273.15) * 9/5 + 32)
+
+
case_when(unit == "F" ~ temp,
+          unit == "C" ~ (temp * 9/5) + 32,
+          TRUE ~ (temp - 273.15) * 9/5 + 32)
[1] 45.20 56.30 51.80 53.33
-

The last parameter, TRUE ~, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return NA values).

+

The last argument, TRUE ~, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return NA values by default). You only need to add a TRUE ~ condition if you know that all previously listed conditions may not cover all possible outcomes. Here, we know that some observations are associated with unit == "K" yet that condition is not explicitly defined in the case_when arguments. We could have, of course, added the unit == "K" condition to the above code chunk thus alleviating the need for the TRUE ~ condition.

Note that the order in which these conditions are listed matters since evaluation stops at the first TRUE outcome encountered. So, had the last condition been moved to the top of the stack, all temp values would be assigned the first conversion option.

-
-
# What not to do ...
-case_when(TRUE ~ (temp - 273.15) * 9/5 + 32,
-          unit == "F" ~ temp,
-          unit == "C" ~ (temp * 9/5) + 32)
+
+
# What not to do ...
+case_when(TRUE ~ (temp - 273.15) * 9/5 + 32,
+          unit == "F" ~ temp,
+          unit == "C" ~ (temp * 9/5) + 32)
[1] -378.31 -358.33 -439.87   53.33

Note that case_when can also be used inside of a mutate function. For example, to replace Canada and United States of America in variable Country with CAN and USA respectively and to create a new variable called Type which will take on the values of 1, 2 or 3 depending on the values in variable Source, type the following:

-
-
dat1 <- dat %>% 
-  mutate(Country = case_when(Country == "Canada" ~ "CAN",
-                             Country == "United States of America" ~ "USA"),
-         Type = case_when(Source == "Calculated data" ~ 1,
-                          Source == "Official data" ~ 2,
-                          TRUE ~ 3)) 
-head(dat1)
+
+
dat1 <- dat %>% 
+  mutate(Country = case_when(Country == "Canada" ~ "CAN",
+                             Country == "United States of America" ~ "USA"),
+         Type = case_when(Source == "Calculated data" ~ 1,
+                          Source == "Official data" ~ 2,
+                          TRUE ~ 3)) 
+head(dat1)   
  Country         Crop         Information Year      Value          Source Type
 1     CAN       Barley Area harvested (Ha) 2012 2060000.00   Official data    2
@@ -1070,38 +1048,47 @@ 

 
9.4 Replacing values with NA

-
You can replace a specific set of values (numeric or character) using the na_if() function. For example, to replace -999 with NA:

-

-
val <- c(-999, 6, -1, -999)
-na_if( val , -999 )

+
So far, we’ve used the ifelse or if_else functions to replace certain values with NA. dplyr offers the na_if() function to simplify the syntax. For example, to replace -999 with NA:

+

+
val <- c(-999, 6, -1, -999)
+na_if( val , -999 )

 

 
[1] NA  6 -1 NA

 

 

 
Likewise, to replace all empty character values:

-

-
val <- c("ab", "", "A b", "  ")
-na_if( val , "" )

+

+
val <- c("ab", "", "A b", "  ")
+na_if( val , "" )

 

 
[1] "ab"  NA    "A b" "  " 

 

 

-
Note that na_if() will return the datatype passed to its first argument and not necessarily that of the parent vector. For example, if you use stringr’s str_length() function to identify zero length vectors in a character vector, you get a numeric output.

-

-
library(stringr)
-val <- c("ab", "", "A b", "  ")
-na_if( str_length(val) , 0 )

+
na_if will also preserve the object’s class. For example:

+

+
x <- as.factor( c("apple", "walnut", "banana", "pear", "apple"))
+na_if(x , "walnut")

+

+
[1] apple  <NA>   banana pear   apple 
+Levels: apple banana pear walnut

+

+

+
But, note that it does not automatically drop the level being replaced with NA.

+
na_if also works with dates, but don’t forget to evaluate a date object with a date value. For example, to replace dates of 12/1/1901 with NA, we need to make a date object of that value. Here, we’ll make use of the mdy() function as in mdy("12/1/1901").

+

+
y <- mdy("1/23/2016", "3/2/2016", "12/1/1901", "11/23/2016")
+na_if(y, mdy("12/1/1901"))

 

-
[1]  2 NA  3  2

+
[1] "2016-01-23" "2016-03-02" NA           "2016-11-23"

 

 

 
To use na_if() in a piping operation, it needs to be embedded in a mutate() function. For example, to replace "Calculated data" with NA in the dat dataframe, type:

-

-
dat1 <- dat %>% 
-    mutate(Source = na_if(Source, "Calculated data" ))

+

+
dat1 <- dat %>% 
+    mutate(Source = na_if(Source, "Calculated data" ))

 

-

-
unique(dat1$Source)

+

+
unique(dat1$Source)

 

 
[1] "Official data"                           
 [2] NA                                        
@@ -1112,51 +1099,52 @@ 

 

 
9.5 Outputting a vector instead of a table using pull

-
Piping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a data table:

-

-
oats <- dat %>% 
-  filter(Crop == "Oats",
-         Information == "Yield (Hg/Ha)") %>% 
-  summarise(Oats_sum = sum(Value))
-oats

+
Piping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a dataframe:

+

+
oats <- dat %>% 
+  filter(Crop == "Oats",
+         Information == "Yield (Hg/Ha)") %>% 
+  summarise(Oats_sum = sum(Value))
+oats

 

 
  Oats_sum
 1  2169334

 

-
class(oats)

+
class(oats)

 

 
[1] "data.frame"

 

 

-
There may be times when you want to output as a vector element and not a data table. To output a vector, use the pull() function.

-

-
oats <- dat %>% 
-  filter(Crop == "Oats",
-         Information == "Yield (Hg/Ha)") %>% 
-  summarise(Oats_sum = sum(Value)) %>% 
-  pull()
-oats

+
There may be times when you want to output a vector element and not a dataframe. To output a vector, use the pull() function.

+

+
oats <- dat %>% 
+  filter(Crop == "Oats",
+         Information == "Yield (Hg/Ha)") %>% 
+  summarise(Oats_sum = sum(Value)) %>% 
+  pull()
+oats

 

 
[1] 2169334

 

-
class(oats)

+
class(oats)

 

 
[1] "numeric"

 

 

-
The pull function can also be used explicitly define the column to extract. For example, to extract the Value column type:

-

-
# This outputs a multi-element vector
-yield <- dat %>% 
-  filter(Crop == "Oats",
-         Information == "Yield (Hg/Ha)") %>% 
-  pull(Value)
-
-head(yield)

+
The pull function can also be used to explicitly define the column to extract. For example, to extract the Value column type:

+

+
# This outputs a multi-element vector
+yield <- dat %>% 
+  filter(Crop == "Oats",
+         Information == "Yield (Hg/Ha)") %>% 
+  pull(Value)

+

+

+
head(yield)

 

 
[1] 24954.79 21974.70 29109.36 20492.37 27364.53 23056.62

 

-
class(yield)

+
class(yield)

 

 
[1] "numeric"

 

diff --git a/docs/ggplot2.html b/docs/ggplot2.html
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+++ b/docs/ggplot2.html
@@ -336,7 +336,7 @@
 
           
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+  
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@@ -498,28 +498,28 @@ 
    15  
  
   
-    dplyr 
-    ggplot2 
-    forcats 
-   15   1.1.4 
-    3.4.4 
-    1.0.0 
-   
           
  • 
   
     
-  
+  
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    • 
@@ -468,7 +468,7 @@ 
    12  
  
   
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@@ -471,7 +471,7 @@ 
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-
+
 
 
 
@@ -338,7 +338,7 @@
 
           
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+  
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-      
+      
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diff --git a/docs/re_express.html b/docs/re_express.html
index 10ee8c42..d2430a71 100755
--- a/docs/re_express.html
+++ b/docs/re_express.html
@@ -338,7 +338,7 @@
 
           
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index 05c67f82..0c8e6731 100644
--- a/docs/read_write_files.html
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    • 
@@ -476,14 +476,14 @@ 
    4  
  
   
-    readxl 
-   4   1.4.3 
-   You can save an entire R session (which includes all data objects) using the save function.
    
 
    To save all objects, set the list= parameter to ls():
    
 
    
-
    save(list = ls(), file = "ACS_all.Rdata")
    
+
    save(list=ls(), file = "ACS_all.Rdata")
    
 
    
 
    To save only two R session objects–dat and dat.sub–to a file, pass the list of objects to the list= parameter:
    
 
    
-
    save(list = c(dat, dat.sub), file = "ACS_subset.Rdata")
    
+
    save(list=c(dat, dat.sub), file = "ACS_subset.Rdata")
    
 
    
 
    
 
    
diff --git a/docs/rf.html b/docs/rf.html
index e02f6210..db4d5542 100644
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diff --git a/docs/search.json b/docs/search.json
index 4b0351c1..afe4977e 100755
--- a/docs/search.json
+++ b/docs/search.json
@@ -305,21 +305,21 @@
     "href": "dplyr.html#conditional-statements",
     "title": "9  Manipulating data tables with dplyr",
     "section": "9.3 Conditional statements",
-    "text": "9.3 Conditional statements\n  \n\n9.3.1 The base ifelse\nConditional statements are used when you want to create an output value that is conditioned on an evaluation. For example, if you want to output a value of 1 if an input value is less than 23 and a value of 0 otherwise, you can make use of the ifelse function as follows:\n\nx <- c(12,102, 43, 20, 90, 0, 12, 6)\nifelse(x < 23, 1, 0)\n\n[1] 1 0 0 1 0 1 1 1\n\n\nifelse takes three arguments:\n\nThe condition to evaluate (x < 23 in the above example);\nThe value to output if the condition is TRUE (1 in our example);\nThe value to output if the condition is FALSE (0 in our example).\n\nThe base ifelse function works as expected when the input/output values are numeric or character, but it does not work as expected when applied to factors or dates. For example, if you wish to replace one factor level with another, the following example will not return the expected output.\n\nx <- as.factor( c(\"apple\", \"banana\", \"banana\", \"pear\", \"apple\"))\nifelse(x == \"pear\", \"apple\", x)\n\n[1] \"1\"     \"2\"     \"2\"     \"apple\" \"1\"    \n\n\nThe output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. apple =1, banana=2, etc…). Likewise, if you wish to replace an erroneous date with a missing value you will get:\n\nlibrary(lubridate)\ny <- mdy(\"1/23/2016\", \"3/2/2016\", \"12/1/1901\", \"11/23/2016\")\nifelse( year(y) != 2016, NA, y)\n\n[1] 16823 16862    NA 17128\n\n\nHere, ifelse converts the date object to its internal numeric representation as number of days since 1970.\n\n\n9.3.2 dplyr’s if_else\nifelse does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector’s class. If you want to ensure that the vector’s class is preserved, a safer alternative is to use dplyr’s if_else function.\nReworking the above examples:\n\nif_else( year(y) != 2016, NA, y)\n\n[1] \"2016-01-23\" \"2016-03-02\" NA           \"2016-11-23\"\n\n\n\nif_else(x == \"pear\", \"apple\", x)\n\n[1] \"apple\"  \"banana\" \"banana\" \"apple\"  \"apple\" \n\n\nNote that when working with factors, however, if_else will strip the factor class of an input factor. dplyr offers a recode function for this purpose however, this function is being superseded according to the documentation as of version 1.1.4. The workaround is to convert the if_else output to a factor.\n\nas.factor(if_else(x == \"pear\", \"apple\", x))\n\n[1] apple  banana banana apple  apple \nLevels: apple banana\n\n\n\n9.3.2.1 Note on replacing NA factor levels\nOne operation you cannot perform with if_else is converting an NA level to another factor level. For example, let’s add an NA value to the x factor:\n\nx[2] <- NA\nx\n\n[1] apple  <NA>   banana pear   apple \nLevels: apple banana pear\n\n\nIf we want to replace NA with banana the following will generate an unexpected output.\n\nif_else( x == NA, \"banana\", x)\n\n[1] NA NA NA NA NA\n\n\nTo evaluate if a value is NA, use the is.na() function as in:\n\nif_else( is.na(x), \"banana\", x)\n\n[1] \"apple\"  \"banana\" \"banana\" \"pear\"   \"apple\" \n\n\n\n\n\n9.3.3 Changing values based on multiple conditions: case_when\nifelse and if_else work great when a single set of conditions is to be satisfied. But if multiple sets of conditions are to be tested, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.\n\nunit <- c(\"F\",\"F\",\"C\", \"K\")\nif_else( unit == \"C\", \"Celsius\", if_else(unit == \"F\", \"Fahrenheit\", \"Kelvin\"))\n\n[1] \"Fahrenheit\" \"Fahrenheit\" \"Celsius\"    \"Kelvin\"    \n\n\nA simpler solution is to use the case_when function.\n\ncase_when(unit == \"C\" ~ \"Celsius\",\n          unit == \"F\" ~ \"Fahrenheit\",\n          unit == \"K\" ~ \"Kelvin\")\n\n[1] \"Fahrenheit\" \"Fahrenheit\" \"Celsius\"    \"Kelvin\"    \n\n\ncase_when can aso be used for more complex operations. For example, given two vectors, unit and temp, we would like to convert all temp values to Fahrenheit by applying a temperature conversion dependent on the unit value.\n\ntemp <- c(45.2, 56.3, 11.0, 285)\n\n\ncase_when(unit == \"F\" ~ temp,\n          unit == \"C\" ~ (temp * 9/5) + 32,\n          TRUE ~ (temp - 273.15) * 9/5 + 32)\n\n[1] 45.20 56.30 51.80 53.33\n\n\nThe last parameter, TRUE ~, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return NA values).\nNote that the order in which these conditions are listed matters since evaluation stops at the first TRUE outcome encountered. So, had the last condition been moved to the top of the stack, all temp values would be assigned the first conversion option.\n\n# What not to do ...\ncase_when(TRUE ~ (temp - 273.15) * 9/5 + 32,\n          unit == \"F\" ~ temp,\n          unit == \"C\" ~ (temp * 9/5) + 32)\n\n[1] -378.31 -358.33 -439.87   53.33\n\n\nNote that case_when can also be used inside of a mutate function. For example, to replace Canada and United States of America in variable Country with CAN and USA respectively and to create a new variable called Type which will take on the values of 1, 2 or 3 depending on the values in variable Source, type the following:\n\ndat1 <- dat %>% \n  mutate(Country = case_when(Country == \"Canada\" ~ \"CAN\",\n                             Country == \"United States of America\" ~ \"USA\"),\n         Type = case_when(Source == \"Calculated data\" ~ 1,\n                          Source == \"Official data\" ~ 2,\n                          TRUE ~ 3)) \nhead(dat1)   \n\n  Country         Crop         Information Year      Value          Source Type\n1     CAN       Barley Area harvested (Ha) 2012 2060000.00   Official data    2\n2     CAN       Barley       Yield (Hg/Ha) 2012   38894.66 Calculated data    1\n3     CAN    Buckwheat Area harvested (Ha) 2012       0.00    FAO estimate    3\n4     CAN  Canary seed Area harvested (Ha) 2012  101900.00   Official data    2\n5     CAN  Canary seed       Yield (Hg/Ha) 2012   12161.92 Calculated data    1\n6     CAN Grain, mixed Area harvested (Ha) 2012   57900.00   Official data    2"
+    "text": "9.3 Conditional statements\n  \n\n9.3.1 The base ifelse\nConditional statements are used when you want to create an output value that is conditioned on an evaluation. For example, if you want to output a value of 1 if an input value is less than 23 and a value of 0 otherwise, you can make use of the ifelse function as follows:\n\nx <- c(12,102, 43, 20, 90, 0, 12, 6)\nifelse(x < 23, 1, 0)\n\n[1] 1 0 0 1 0 1 1 1\n\n\nifelse takes three arguments:\n\nThe condition to evaluate (x < 23 in the above example);\nThe value to output if the condition is TRUE (1 in our example);\nThe value to output if the condition is FALSE (0 in our example).\n\nThe base ifelse function works as expected when the input/output values are numeric or character, but it does not work as expected when applied to factors or dates. For example, if you wish to replace one factor level with another, the following example will not return the expected output.\n\nx <- as.factor( c(\"apple\", \"banana\", \"banana\", \"pear\", \"apple\"))\nifelse(x == \"pear\", \"apple\", x)\n\n[1] \"1\"     \"2\"     \"2\"     \"apple\" \"1\"    \n\n\nThe output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. apple =1, banana=2, etc…). Likewise, if you wish to replace an erroneous date you will get:\n\nlibrary(lubridate)\ny <- mdy(\"1/23/2016\", \"3/2/2016\", \"12/1/1901\", \"11/23/2016\")\nifelse( y == mdy(\"12/1/1901\"), mdy(\"12/1/2016\"), y)\n\n[1] 16823 16862 17136 17128\n\n\nHere, ifelse converts the date object to its internal numeric representation as number of days since 1970.\n\n\n9.3.2 dplyr’s if_else\nifelse does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector’s class. A safer alternative is to use dplyr’s if_else function.\nReworking the above examples:\n\nif_else( y == mdy(\"12/1/1901\"), mdy(\"12/1/2016\"), y)\n\n[1] \"2016-01-23\" \"2016-03-02\" \"2016-12-01\" \"2016-11-23\"\n\n\nThe date class is preserved. Now let’s check the output of a factor.\n\nif_else(x == \"pear\", \"apple\", x)\n\n[1] \"apple\"  \"banana\" \"banana\" \"apple\"  \"apple\" \n\n\nNote that when working with factors, however, if_else will strip the factor class of an input factor. But, instead of returning the factor’s underlying integer values, it outputs the associated levels as a character data type.\nThe workaround is to convert the if_else output to a factor.\n\nfactor(if_else(x == \"pear\", \"apple\", x))\n\n[1] apple  banana banana apple  apple \nLevels: apple banana\n\n\nIf you need to explicitly define the levels, add the levels = c(\"pear\",...).\nNOTE: dplyr offers the recode function that preserves factors however, this function is being superseded according to the documentation as of version 1.1.4.\n\n\n9.3.3 Changing values based on multiple conditions: case_when\nifelse and if_else work great when a single set of conditions is to be satisfied. But, if multiple sets of conditions are to be evaluated, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.\n\nunit <- c(\"F\",\"F\",\"C\", \"K\")\nif_else( unit == \"C\", \"Celsius\", if_else(unit == \"F\", \"Fahrenheit\", \"Kelvin\"))\n\n[1] \"Fahrenheit\" \"Fahrenheit\" \"Celsius\"    \"Kelvin\"    \n\n\nA simpler solution is to use the case_when function.\n\ncase_when(unit == \"C\" ~ \"Celsius\",\n          unit == \"F\" ~ \"Fahrenheit\",\n          unit == \"K\" ~ \"Kelvin\")\n\n[1] \"Fahrenheit\" \"Fahrenheit\" \"Celsius\"    \"Kelvin\"    \n\n\ncase_when can aso be used for more complex operations. For example, given two vectors, unit and temp, we would like to convert all temp values to Fahrenheit by applying a temperature conversion dependent on the unit value.\n\ntemp <- c(45.2, 56.3, 11.0, 285)\n\n\ncase_when(unit == \"F\" ~ temp,\n          unit == \"C\" ~ (temp * 9/5) + 32,\n          TRUE ~ (temp - 273.15) * 9/5 + 32)\n\n[1] 45.20 56.30 51.80 53.33\n\n\nThe last argument, TRUE ~, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return NA values by default). You only need to add a TRUE ~ condition if you know that all previously listed conditions may not cover all possible outcomes. Here, we know that some observations are associated with unit == \"K\" yet that condition is not explicitly defined in the case_when arguments. We could have, of course, added the unit == \"K\" condition to the above code chunk thus alleviating the need for the TRUE ~ condition.\nNote that the order in which these conditions are listed matters since evaluation stops at the first TRUE outcome encountered. So, had the last condition been moved to the top of the stack, all temp values would be assigned the first conversion option.\n\n# What not to do ...\ncase_when(TRUE ~ (temp - 273.15) * 9/5 + 32,\n          unit == \"F\" ~ temp,\n          unit == \"C\" ~ (temp * 9/5) + 32)\n\n[1] -378.31 -358.33 -439.87   53.33\n\n\nNote that case_when can also be used inside of a mutate function. For example, to replace Canada and United States of America in variable Country with CAN and USA respectively and to create a new variable called Type which will take on the values of 1, 2 or 3 depending on the values in variable Source, type the following:\n\ndat1 <- dat %>% \n  mutate(Country = case_when(Country == \"Canada\" ~ \"CAN\",\n                             Country == \"United States of America\" ~ \"USA\"),\n         Type = case_when(Source == \"Calculated data\" ~ 1,\n                          Source == \"Official data\" ~ 2,\n                          TRUE ~ 3)) \nhead(dat1)   \n\n  Country         Crop         Information Year      Value          Source Type\n1     CAN       Barley Area harvested (Ha) 2012 2060000.00   Official data    2\n2     CAN       Barley       Yield (Hg/Ha) 2012   38894.66 Calculated data    1\n3     CAN    Buckwheat Area harvested (Ha) 2012       0.00    FAO estimate    3\n4     CAN  Canary seed Area harvested (Ha) 2012  101900.00   Official data    2\n5     CAN  Canary seed       Yield (Hg/Ha) 2012   12161.92 Calculated data    1\n6     CAN Grain, mixed Area harvested (Ha) 2012   57900.00   Official data    2"
   },
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     "objectID": "dplyr.html#replacing-values-with-na",
     "href": "dplyr.html#replacing-values-with-na",
     "title": "9  Manipulating data tables with dplyr",
     "section": "9.4 Replacing values with NA",
-    "text": "9.4 Replacing values with NA\nYou can replace a specific set of values (numeric or character) using the na_if() function. For example, to replace -999 with NA:\n\nval <- c(-999, 6, -1, -999)\nna_if( val , -999 )\n\n[1] NA  6 -1 NA\n\n\nLikewise, to replace all empty character values:\n\nval <- c(\"ab\", \"\", \"A b\", \"  \")\nna_if( val , \"\" )\n\n[1] \"ab\"  NA    \"A b\" \"  \" \n\n\nNote that na_if() will return the datatype passed to its first argument and not necessarily that of the parent vector. For example, if you use stringr’s str_length() function to identify zero length vectors in a character vector, you get a numeric output.\n\nlibrary(stringr)\nval <- c(\"ab\", \"\", \"A b\", \"  \")\nna_if( str_length(val) , 0 )\n\n[1]  2 NA  3  2\n\n\nTo use na_if() in a piping operation, it needs to be embedded in a mutate() function. For example, to replace \"Calculated data\" with NA in the dat dataframe, type:\n\ndat1 <- dat %>% \n    mutate(Source = na_if(Source, \"Calculated data\" ))\n\n\nunique(dat1$Source)\n\n[1] \"Official data\"                           \n[2] NA                                        \n[3] \"FAO estimate\"                            \n[4] \"FAO data based on imputation methodology\""
+    "text": "9.4 Replacing values with NA\nSo far, we’ve used the ifelse or if_else functions to replace certain values with NA. dplyr offers the na_if() function to simplify the syntax. For example, to replace -999 with NA:\n\nval <- c(-999, 6, -1, -999)\nna_if( val , -999 )\n\n[1] NA  6 -1 NA\n\n\nLikewise, to replace all empty character values:\n\nval <- c(\"ab\", \"\", \"A b\", \"  \")\nna_if( val , \"\" )\n\n[1] \"ab\"  NA    \"A b\" \"  \" \n\n\nna_if will also preserve the object’s class. For example:\n\nx <- as.factor( c(\"apple\", \"walnut\", \"banana\", \"pear\", \"apple\"))\nna_if(x , \"walnut\")\n\n[1] apple  <NA>   banana pear   apple \nLevels: apple banana pear walnut\n\n\nBut, note that it does not automatically drop the level being replaced with NA.\nna_if also works with dates, but don’t forget to evaluate a date object with a date value. For example, to replace dates of 12/1/1901 with NA, we need to make a date object of that value. Here, we’ll make use of the mdy() function as in mdy(\"12/1/1901\").\n\ny <- mdy(\"1/23/2016\", \"3/2/2016\", \"12/1/1901\", \"11/23/2016\")\nna_if(y, mdy(\"12/1/1901\"))\n\n[1] \"2016-01-23\" \"2016-03-02\" NA           \"2016-11-23\"\n\n\nTo use na_if() in a piping operation, it needs to be embedded in a mutate() function. For example, to replace \"Calculated data\" with NA in the dat dataframe, type:\n\ndat1 <- dat %>% \n    mutate(Source = na_if(Source, \"Calculated data\" ))\n\n\nunique(dat1$Source)\n\n[1] \"Official data\"                           \n[2] NA                                        \n[3] \"FAO estimate\"                            \n[4] \"FAO data based on imputation methodology\""
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     "href": "dplyr.html#outputting-a-vector-instead-of-a-table-using-pull",
     "title": "9  Manipulating data tables with dplyr",
     "section": "9.5 Outputting a vector instead of a table using pull",
-    "text": "9.5 Outputting a vector instead of a table using pull\nPiping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a data table:\n\noats <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  summarise(Oats_sum = sum(Value))\noats\n\n  Oats_sum\n1  2169334\n\nclass(oats)\n\n[1] \"data.frame\"\n\n\nThere may be times when you want to output as a vector element and not a data table. To output a vector, use the pull() function.\n\noats <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  summarise(Oats_sum = sum(Value)) %>% \n  pull()\noats\n\n[1] 2169334\n\nclass(oats)\n\n[1] \"numeric\"\n\n\nThe pull function can also be used explicitly define the column to extract. For example, to extract the Value column type:\n\n# This outputs a multi-element vector\nyield <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  pull(Value)\n\nhead(yield)\n\n[1] 24954.79 21974.70 29109.36 20492.37 27364.53 23056.62\n\nclass(yield)\n\n[1] \"numeric\""
+    "text": "9.5 Outputting a vector instead of a table using pull\nPiping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a dataframe:\n\noats <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  summarise(Oats_sum = sum(Value))\noats\n\n  Oats_sum\n1  2169334\n\nclass(oats)\n\n[1] \"data.frame\"\n\n\nThere may be times when you want to output a vector element and not a dataframe. To output a vector, use the pull() function.\n\noats <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  summarise(Oats_sum = sum(Value)) %>% \n  pull()\noats\n\n[1] 2169334\n\nclass(oats)\n\n[1] \"numeric\"\n\n\nThe pull function can also be used to explicitly define the column to extract. For example, to extract the Value column type:\n\n# This outputs a multi-element vector\nyield <- dat %>% \n  filter(Crop == \"Oats\",\n         Information == \"Yield (Hg/Ha)\") %>% \n  pull(Value)\n\n\nhead(yield)\n\n[1] 24954.79 21974.70 29109.36 20492.37 27364.53 23056.62\n\nclass(yield)\n\n[1] \"numeric\""
   },
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     "objectID": "group_by.html#summarizing-data-by-group",
diff --git a/docs/strings.html b/docs/strings.html
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--- a/docs/strings.html
+++ b/docs/strings.html
@@ -336,7 +336,7 @@
 
           
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-    4.3.2 
+             border-radius: 5px;width:5px;"> 4.3.1 
   
 
 
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diff --git a/dplyr.qmd b/dplyr.qmd
index ea48a525..53a14e7c 100755
--- a/dplyr.qmd
+++ b/dplyr.qmd
@@ -276,12 +276,12 @@ x <- as.factor( c("apple", "banana", "banana", "pear", "apple"))
 ifelse(x == "pear", "apple", x)
 ```
 
-The output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. `apple` =1, `banana`=2, etc...). Likewise, if you wish to replace an erroneous date with a missing value you will get:
+The output is a character representation of the level number (recall that factors encode level values as numbers behind the scenes, i.e. `apple` =1, `banana`=2, etc...). Likewise, if you wish to replace an erroneous date you will get:
 
 ```{r}
 library(lubridate)
 y <- mdy("1/23/2016", "3/2/2016", "12/1/1901", "11/23/2016")
-ifelse( year(y) != 2016, NA, y)
+ifelse( y == mdy("12/1/1901"), mdy("12/1/2016"), y)
 ```
 
 Here, `ifelse` converts the date object to its internal numeric representation as number of days since 1970. 
@@ -289,49 +289,36 @@ Here, `ifelse` converts the date object to its internal numeric representation a
 
 ### `dplyr`'s `if_else`
 
-`ifelse` does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector's `class`. If you want to ensure that the vector's `class` is preserved, a safer alternative is to use `dplyr`'s `if_else` function.
+`ifelse` does not preserve the attributes that might be present in a vector. In other words, it will strip away the vector's `class`. A safer alternative is to use `dplyr`'s `if_else` function.
 
 Reworking the above examples: 
 
 ```{r}
-if_else( year(y) != 2016, NA, y)
+if_else( y == mdy("12/1/1901"), mdy("12/1/2016"), y)
 ```
 
-```{r error=TRUE}
-if_else(x == "pear", "apple", x)
-```
-
-Note that when working with factors, however, `if_else` will strip the `factor` class of an input factor. `dplyr` offers a `recode` function for this purpose however, this function is being superseded according to the documentation as of version `1.1.4`. The workaround is to convert the `if_else` output to a factor.
+The date class is preserved. Now let's check the output of a factor.
 
 ```{r error=TRUE}
-as.factor(if_else(x == "pear", "apple", x))
+if_else(x == "pear", "apple", x)
 ```
 
-#### Note on replacing NA factor levels
+Note that when working with factors, however, `if_else` will strip the `factor` class of an input factor. But, instead of returning the factor's underlying integer values, it outputs the associated levels as a `character` data type.
 
-One operation you cannot perform with `if_else` is converting an `NA` level to another factor level. For example, let's add an `NA` value to the `x` factor:
+The workaround is to convert the `if_else` output to a factor.
 
-```{r}
-x[2] <- NA
-x
+```{r error=TRUE}
+factor(if_else(x == "pear", "apple", x))
 ```
 
-If we want to replace `NA` with `banana` the following will generate an unexpected output.
-
-
-```{r}
-if_else( x == NA, "banana", x)
-```
+If you need to explicitly define the levels, add the `levels = c("pear",...)`.
 
-To evaluate if a value is `NA`, use the `is.na()` function as in:
+NOTE: `dplyr` offers the `recode` function that preserves factors however, this function is being superseded according to the documentation as of version `1.1.4`.
 
-```{r}
-if_else( is.na(x), "banana", x)
-```
 
 ### Changing values based on multiple conditions: `case_when`
 
-`ifelse` and `if_else` work great when a single set of conditions is to be satisfied. But if multiple sets of conditions are to be tested, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.
+`ifelse` and `if_else` work great when a single set of conditions is to be satisfied. But, if multiple sets of conditions are to be evaluated, nested if/else statements become cumbersome and are prone to clerical error. The following code highlights an example of nested if/else statements.
 
 ```{r}
 unit <- c("F","F","C", "K")
@@ -359,7 +346,7 @@ case_when(unit == "F" ~ temp,
           TRUE ~ (temp - 273.15) * 9/5 + 32)
 ```
 
-The last parameter, `TRUE ~ `, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return `NA` values).
+The last argument, `TRUE ~ `, applies to all conditions not satisfied by the previous two conditions (otherwise, not doing so would return `NA` values by default). You only need to add a `TRUE ~` condition if you know that all previously listed conditions may not cover all possible outcomes. Here, we know that some observations are associated with `unit == "K"` yet that condition is not explicitly defined in the `case_when` arguments. We could have, of course, added the `unit == "K"` condition to the above code chunk thus alleviating the need for the `TRUE ~` condition.
 
 Note that the order in which these conditions are listed matters since evaluation stops at the first `TRUE` outcome encountered. So, had the last condition been moved to the top of the stack, all `temp` values would be assigned the first conversion option.
 
@@ -387,7 +374,7 @@ head(dat1)
 
 ## Replacing values with `NA`
 
-You can replace a specific set of values (numeric or character) using the `na_if()` function. For example, to replace `-999` with `NA`:
+So far, we've used the `ifelse` or `if_else` functions to replace certain values with `NA`. `dplyr` offers the `na_if()` function to simplify the syntax. For example, to replace `-999` with `NA`:
 
 ```{r}
 val <- c(-999, 6, -1, -999)
@@ -401,12 +388,20 @@ val <- c("ab", "", "A b", "  ")
 na_if( val , "" )
 ```
 
-Note that `na_if()` will return the datatype passed to its first argument and not necessarily that of the parent vector. For example, if you use `stringr`'s  `str_length()` function to identify zero length vectors in a character vector, you get a numeric output.
+`na_if` will also preserve the object's class. For example:
 
 ```{r}
-library(stringr)
-val <- c("ab", "", "A b", "  ")
-na_if( str_length(val) , 0 )
+x <- as.factor( c("apple", "walnut", "banana", "pear", "apple"))
+na_if(x , "walnut")
+```
+
+But, note that it does not automatically drop the level being replaced with `NA`.
+
+`na_if` also works with dates, but don't forget to evaluate a date object with a date value. For example, to replace dates of `12/1/1901` with `NA`, we need to make a date object of that value. Here, we'll make use of the `mdy()` function as in `mdy("12/1/1901")`.
+
+```{r}
+y <- mdy("1/23/2016", "3/2/2016", "12/1/1901", "11/23/2016")
+na_if(y, mdy("12/1/1901"))
 ```
 
 To use `na_if()` in a piping operation, it needs to be embedded in a `mutate()` function. For example, to replace `"Calculated data"` with `NA` in the `dat` dataframe, type:
@@ -422,7 +417,7 @@ unique(dat1$Source)
 
 ## Outputting a vector instead of a table using `pull`
 
-Piping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a data table:
+Piping operations will output a table, even if a single value is returned. For example, the following summarization operation returns the total oats yield as a dataframe:
 
 ```{r}
 oats <- dat %>% 
@@ -433,7 +428,7 @@ oats
 class(oats)
 ```
 
-There may be times when you want to output as a vector element and not a data table. To output a vector, use the `pull()` function.
+There may be times when you want to output a vector element and not a dataframe. To output a vector, use the `pull()` function.
 
 ```{r}
 oats <- dat %>% 
@@ -446,7 +441,7 @@ oats
 class(oats)
 ```
 
-The `pull` function can also be used explicitly define the column to extract. For example, to extract the `Value` column type:
+The `pull` function can also be used to explicitly define the column to extract. For example, to extract the `Value` column type:
 
 
 ```{r}
@@ -455,8 +450,10 @@ yield <- dat %>%
   filter(Crop == "Oats",
          Information == "Yield (Hg/Ha)") %>% 
   pull(Value)
+```
 
+```{r}
 head(yield)
-
 class(yield)
 ```
+
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diff --git a/sl_plot.knit.md b/sl_plot.knit.md
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--- a/sl_plot.knit.md
+++ /dev/null
@@ -1,218 +0,0 @@
----
-output: html_document
-editor_options: 
-  chunk_output_type: console
----
-
-# Spread-level plots
-
-
-
-::: {.cell}
-
-:::
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-2_a9e0c5c25fa201d342586fad77e0b216'}
-::: {.cell-output-display}
-`````{=html}
-
- 
-  
-   
-   
-  
- 
-
-  
-   
-   
-  
-
-
     dplyr  ggplot2 
     1.1.4  3.4.4 
    
-
-`````
-:::
-:::
-
-
-
-## Introduction
-
-Some batches of data may show a systematic change in spread vs. location. In other words, the variability in each batch may be dependent on that batches median value. Such dependency is often undesirable (e.g. in an ANOVA for instance) and preferably removed in an analysis. A plot well suited for visualizing this dependency is the **spread-level** plot, **s-l** (or **spread-location** plot as Cleveland calls it).
-
-## Constructing the s-l plot
-
-The s-l plot compares a measure of the spread's residual to the location (usually the median) for each batch of data. The spread is usually distilled down to its residual (what remains after subtracting each batch value by the batch median) then it's transformed by taking the square root of its absolute value. The following block walks you through the steps needed to create an s-l plot.
-
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-3_9187cec471b10956d63ffc9a30e6f07a'}
-
-```{.r .cell-code}
-library(dplyr)
-library(ggplot2)
-
-singer <- lattice::singer
-res.sq <-  singer %>% group_by(voice.part) %>% 
-                      mutate(Median   = median(height),
-                             Residual = sqrt(abs(height - Median)))
-
-ggplot(res.sq, aes(x=Median, y=Residual)) + 
-  geom_jitter(alpha=0.4,width=0.2) +
-  stat_summary(fun = median, geom = "line", col = "red") +
-  ylab(expression(sqrt(abs(" Residuals ")))) +
-  geom_text(aes(x = Median, y = 3.3, label = voice.part))
-```
-
-::: {.cell-output-display}
-![](sl_plot_files/figure-html/unnamed-chunk-3-1.png){width=480}
-:::
-:::
-
-
-The red line in the plot helps identify the type of relationship between spread and location. If the line increases monotonically upward, there is an increasing spread as a function of increasing location; if the line decreases monotonically downward, there is a decreasing spread as a function of increasing location; and if line is neither increasing nor decreasing monotonically, there is no change in spread as a function of location.
-
-> Note that if you are to rescale the y-axis when using the `stat_summary()` function, you should use the `coord_cartesian(ylim = c( .. , .. ))` function instead of the `ylim()` function. The latter will mask the values above its maximum range from the `stat_summary()` function, the former will not.
-
-The singer dataset does not seem to exhibit any dependence between a voice part's spread and its median value.
-
-Next, we'll look at an example of a dataset that *does* exhibit a dependence between spread and fitted values.
-
-## Example: the `iris` dataset
-
-R has a built-in dataset called `iris` that provide measurements of sepal and petal dimensions for three different species of the iris family. In this next example, we will plot the spreads of the `Petal.Length` residuals (after removing their group median values) to their group medians.
-
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-4_c31935faa3125f3844ba51b516baafae'}
-
-```{.r .cell-code}
-# Create two new columns: group median and group residuals
- df1 <- iris %>%
-   group_by(Species)  %>%
-   mutate( Median = median(Petal.Length),
-           Residuals = sqrt(abs( Petal.Length - Median)))  
-
-# Generate the s-l plot 
- ggplot(df1, aes(x = Median, y = Residuals)) + 
-   geom_jitter(alpha = 0.4, width = 0.05, height = 0) +
-   stat_summary(fun = median, geom = "line", col = "red") +
-   ylab(expression(sqrt( abs(" Residuals ")))) +
-   geom_text(aes(x = Median, y = 1.3, label = Species))
-```
-
-::: {.cell-output-display}
-![](sl_plot_files/figure-html/unnamed-chunk-4-1.png){width=336}
-:::
-:::
-
-
-A monotonic spread is apparent in this dataset too, i.e. as the median length of the Petal increases, so does the spread.
-
-## How can we stabilize spreads in a dataset?
-
-A technique used to help reduce or eliminate monotonic variations in the spreads as a function of fitted values is to **re-express** the original values. Re-expression, which involves transforming values via a pwoer transformation, will be covered in the next chapter. However, in the next section, we learn of a variation of the s-l plot that can identify a power transformation that can help stabilize spread. 
-
-
-### Variations of the S-L plot
-
-Another version of the S-L plot (and one that seems to be more mainstream) pits the log of the inter-quartile spread vs the log of the median. This approach only works for positive values (this may require that values be adjusted so that the minimum value be no less than or equal to 0).  
-
-This approach is appealing in that the slope of the best fit line can be used to come up with a power transformation (a topic covered in next week's lecture) via **power = 1 - slope**.
-
-This variant of the s-l plot can be computed in R as follows (we will use the food web data as an example).
-
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-5_e930bbfdb7e1af934707bcd46bff54b4'}
-
-```{.r .cell-code}
-sl <- iris %>%
-  group_by(Species)  %>%
-  summarise (level  = log(median(Petal.Length)),
-                IQR = IQR(Petal.Length),  # Computes the interquartile range
-             spread = log(IQR))
-
-ggplot(sl, aes(x = level, y = spread)) + geom_point() + 
-  stat_smooth(method = MASS::rlm, se = FALSE) +
-  xlab("Median (log)") + ylab("Spread (log)") +
-  geom_text(aes(x = level, y = spread, label = Species), cex=2.5)
-```
-
-::: {.cell-output-display}
-![](sl_plot_files/figure-html/unnamed-chunk-5-1.png){width=268.8}
-:::
-:::
-
-
-Note how this plot differs from our earlier s-l plot in that we are only displaying each batch's median spread value and we are *fitting* a straight line to the medians instead of *connecting* them.
-
-The slope suggests a monotonic increase in spread vs location. We can extract the slope value from a regression model. Here, we'll adopt a robust bivariate model (bivariate analysis is covered later in this course).
-
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-6_6d3fd12613d9bfbd45d59b658e228128'}
-
-```{.r .cell-code}
-coefficients(MASS::rlm(spread ~ level, sl))
-```
-
-::: {.cell-output .cell-output-stdout}
-```
-(Intercept)       level 
-  -2.204242    1.143365 
-```
-:::
-:::
-
-
-The slope is the second coefficient in the above output. The computed slope value is `1.14`. This suggests a power transformation of `1 - 1.14` (or about `-0.14`). You will learn in the next chapter that as a power transformation value approaches `0`, when can opt to use the `log` transformation instead. We'll try this next.
-
-
-::: {.cell hash='sl_plot_cache/html/unnamed-chunk-7_2eb9368a33f85e5ca3bb37b49231d052'}
-
-```{.r .cell-code}
-# Create two new columns: group median and group residuals
- df1 <- iris %>%
-   group_by(Species)  %>%
-   mutate( log.length = log(Petal.Length),
-           Median = median(log.length),
-           Residuals = sqrt(abs( log.length - Median)))  
-
-# Generate the s-l plot 
- ggplot(df1, aes(x = Median, y = Residuals)) + 
-   geom_jitter(alpha = 0.4, width = 0.05, height = 0) +
-   stat_summary(fun = median, geom = "line", col = "red") +
-   ylab(expression(sqrt( abs(" Residuals ")))) +
-   geom_text(aes(x = Median, y = 1.3, label = Species))
-```
-
-::: {.cell-output-display}
-![](sl_plot_files/figure-html/unnamed-chunk-7-1.png){width=336}
-:::
-:::
-
-
-Applying a log transformation to the petal length values seems to have helped stabilize the spread given that the connected lines are close to flat. However, there is a very slight upward slope--though probably insignificant. Would applying a power transformation of `-0.14` have helped? We will revisit this dataset in the next chapter.
-
-
diff --git a/sl_plot_files/execute-results/html.json b/sl_plot_files/execute-results/html.json
deleted file mode 100644
index 41ec2c23..00000000
--- a/sl_plot_files/execute-results/html.json
+++ /dev/null
@@ -1,18 +0,0 @@
-{
-  "hash": "47c94f236970986146f503c5c234e361",
-  "result": {
-    "markdown": "---\noutput: html_document\neditor_options: \n  chunk_output_type: console\n---\n\n# Spread-level plots\n\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-1_9d27599af8b18542bf589d5de9a52edf'}\n\n:::\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-2_ddef71e58918eb94c9077c76bf467040'}\n::: {.cell-output-display}\n`````{=html}\n\n \n  \n   \n  \n \n\n  \n   \n  \n\n
     dplyr 
     1.1.4 
    \n\n`````\n:::\n:::\n\n\n\n## Introduction\n\nSome batches of data may show a systematic change in spread vs. location. In other words, the variability in each batch may be dependent on that batches median value. Such dependency is often undesirable (e.g. in an ANOVA for instance) and preferably removed in an analysis. A plot well suited for visualizing this dependency is the **spread-level** plot, **s-l** (or **spread-location** plot as Cleveland calls it).\n\n## Constructing the s-l plot\n\nThe s-l plot compares a measure of the spread's residual to the location (usually the median) for each batch of data. The spread is usually distilled down to its residual (what remains after subtracting each batch value by the batch median) then it's transformed by taking the square root of its absolute value. The following block walks you through the steps needed to create an s-l plot.\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-3_9187cec471b10956d63ffc9a30e6f07a'}\n\n```{.r .cell-code}\nlibrary(dplyr)\nlibrary(ggplot2)\n\nsinger <- lattice::singer\nres.sq <-  singer %>% group_by(voice.part) %>% \n                      mutate(Median   = median(height),\n                             Residual = sqrt(abs(height - Median)))\n\nggplot(res.sq, aes(x=Median, y=Residual)) + \n  geom_jitter(alpha=0.4,width=0.2) +\n  stat_summary(fun = median, geom = \"line\", col = \"red\") +\n  ylab(expression(sqrt(abs(\" Residuals \")))) +\n  geom_text(aes(x = Median, y = 3.3, label = voice.part))\n```\n\n::: {.cell-output-display}\n![](sl_plot_files/figure-html/unnamed-chunk-3-1.png){width=480}\n:::\n:::\n\n\nThe red line in the plot helps identify the type of relationship between spread and location. If the line increases monotonically upward, there is an increasing spread as a function of increasing location; if the line decreases monotonically downward, there is a decreasing spread as a function of increasing location; and if line is neither increasing nor decreasing monotonically, there is no change in spread as a function of location.\n\n> Note that if you are to rescale the y-axis when using the `stat_summary()` function, you should use the `coord_cartesian(ylim = c( .. , .. ))` function instead of the `ylim()` function. The latter will mask the values above its maximum range from the `stat_summary()` function, the former will not.\n\nThe singer dataset does not seem to exhibit any dependence between a voice part's spread and its median value.\n\nNext, we'll look at an example of a dataset that *does* exhibit a dependence between spread and fitted values.\n\n## R's built in `iris` dataset\n\nR has a built-in dataset called `iris` that provide measurements of sepal and petal dimensions for three different species of the iris family. In this next example, we will plot the spreads of the `Petal.Length` residuals (after removing their group median values) to their group medians.\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-4_c31935faa3125f3844ba51b516baafae'}\n\n```{.r .cell-code}\n# Create two new columns: group median and group residuals\n df1 <- iris %>%\n   group_by(Species)  %>%\n   mutate( Median = median(Petal.Length),\n           Residuals = sqrt(abs( Petal.Length - Median)))  \n\n# Generate the s-l plot \n ggplot(df1, aes(x = Median, y = Residuals)) + \n   geom_jitter(alpha = 0.4, width = 0.05, height = 0) +\n   stat_summary(fun = median, geom = \"line\", col = \"red\") +\n   ylab(expression(sqrt( abs(\" Residuals \")))) +\n   geom_text(aes(x = Median, y = 1.3, label = Species))\n```\n\n::: {.cell-output-display}\n![](sl_plot_files/figure-html/unnamed-chunk-4-1.png){width=336}\n:::\n:::\n\n\nA monotonic spread is apparent in this dataset too, i.e. as the median length of the Petal increases, so does the spread.\n\n## How can we stabilize spreads in a dataset?\n\nA technique used to help reduce or eliminate monotonic variations in the spreads as a function of fitted values is to **re-express** the original values. Re-expression, which involves transforming values via a pwoer transformation, will be covered in the next chapter. However, in the next section, we learn of a variation of the s-l plot that can identify a power transformation that can help stabilize spread. \n\n\n### Variations of the S-L plot\n\nAnother version of the S-L plot (and one that seems to be more mainstream) pits the log of the inter-quartile spread vs the log of the median. This approach only works for positive values (this may require that values be adjusted so that the minimum value be no less than or equal to 0).  \n\nThis approach is appealing in that the slope of the best fit line can be used to come up with a power transformation (a topic covered in next week's lecture) via **power = 1 - slope**.\n\nThis variant of the s-l plot can be computed in R as follows (we will use the food web data as an example).\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-5_e930bbfdb7e1af934707bcd46bff54b4'}\n\n```{.r .cell-code}\nsl <- iris %>%\n  group_by(Species)  %>%\n  summarise (level  = log(median(Petal.Length)),\n                IQR = IQR(Petal.Length),  # Computes the interquartile range\n             spread = log(IQR))\n\nggplot(sl, aes(x = level, y = spread)) + geom_point() + \n  stat_smooth(method = MASS::rlm, se = FALSE) +\n  xlab(\"Median (log)\") + ylab(\"Spread (log)\") +\n  geom_text(aes(x = level, y = spread, label = Species), cex=2.5)\n```\n\n::: {.cell-output-display}\n![](sl_plot_files/figure-html/unnamed-chunk-5-1.png){width=268.8}\n:::\n:::\n\n\nNote how this plot differs from our earlier s-l plot in that we are only displaying each batch's median spread value and we are *fitting* a straight line to the medians instead of *connecting* them.\n\nThe slope suggests a monotonic increase in spread vs location. We can extract the slope value from a regression model. Here, we'll adopt a robust bivariate model (bivariate analysis is covered later in this course).\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-6_6d3fd12613d9bfbd45d59b658e228128'}\n\n```{.r .cell-code}\ncoefficients(MASS::rlm(spread ~ level, sl))\n```\n\n::: {.cell-output .cell-output-stdout}\n```\n(Intercept)       level \n  -2.204242    1.143365 \n```\n:::\n:::\n\n\nThe slope is the second coefficient in the above output. The computed slope value is `1.14`. This suggests a power transformation of `1 - 1.14` (or about `-0.14`). You will learn in the next chapter that as a power transformation value approaches `0`, when can opt to use the `log` transformation instead. We'll try this next.\n\n\n::: {.cell hash='sl_plot_cache/html/unnamed-chunk-7_2eb9368a33f85e5ca3bb37b49231d052'}\n\n```{.r .cell-code}\n# Create two new columns: group median and group residuals\n df1 <- iris %>%\n   group_by(Species)  %>%\n   mutate( log.length = log(Petal.Length),\n           Median = median(log.length),\n           Residuals = sqrt(abs( log.length - Median)))  \n\n# Generate the s-l plot \n ggplot(df1, aes(x = Median, y = Residuals)) + \n   geom_jitter(alpha = 0.4, width = 0.05, height = 0) +\n   stat_summary(fun = median, geom = \"line\", col = \"red\") +\n   ylab(expression(sqrt( abs(\" Residuals \")))) +\n   geom_text(aes(x = Median, y = 1.3, label = Species))\n```\n\n::: {.cell-output-display}\n![](sl_plot_files/figure-html/unnamed-chunk-7-1.png){width=336}\n:::\n:::\n\n\nApplying a log transformation to the petal length values seems to have helped stabilize the spread given that the connected lines are close to flat. However, there is a very slight upward slope--though probably insignificant. Would applying a power transformation of `-0.14` have helped? We will revisit this dataset in the next chapter.\n\n",
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