Using lapply to label the values of specific variables

Question

I would like to use lapply to label the values of specific variables. I have found an example that gets me close ( here ), but I can't get it to work for only certain variables in the data set.

Working example:

df1 <- tribble(
 ~var1, ~var2, ~var3, ~var4,
 "1",   "1",   "1", "a",
 "2",   "2",   "2", "b",
 "3",   "3",   "3", "c"
)

Here is the code that seems like it should work, but doesn't:

df1["var1", "var2"] <- lapply(df1["var1", "var2"], factor,
                          levels=c(1, 
                                   2, 
                                   3), 
                          labels = c("Agree", 
                                     "Neither Agree/Disagree", 
                                     "Disagree"))

The code runs, but give the following output:

# A tibble: 4 x 4
  var1  var2  var3  var4
* <chr> <chr> <chr> <chr>
1     1     1     1     a
2     2     2     2     b
3     3     3     3     c
4  <NA>  <NA>  <NA>  <NA>

If I try with just one variable, it works:

df1["var1"] <- lapply(df1["var1"], factor,
                          levels=c(1, 
                                2, 
                                3), 
                          labels = c("Agree", 
                                  "Neither Agree/Disagree", 
                                  "Disagree"))

It gives the following output (which is correct):

# A tibble: 3 x 4
                    var1  var2  var3  var4
                  <fctr> <chr> <chr> <chr>
1                  Agree     1     1     a
2 Neither Agree/Disagree     2     2     b
3               Disagree     3     3     c

I have tried a lot of different ways to change the code to get it to work, but I just can't figure it out.

Answer 1

You were close. We need df1[c("var1", "var2")] to specify columns.

df1[c("var1", "var2")] <- lapply(df1[c("var1", "var2")], factor,
                              levels=c("1", 
                                       "2", 
                                       "3"), 
                              labels = c("Agree", 
                                         "Neither Agree/Disagree", 
                                         "Disagree"))
df1
# # A tibble: 3 x 4
#                     var1                   var2  var3  var4
#                   <fctr>                 <fctr> <chr> <chr>
# 1                  Agree                  Agree     1     a
# 2 Neither Agree/Disagree Neither Agree/Disagree     2     b
# 3               Disagree               Disagree     3     c

Answer 2

Your problem is arising because you're trying to subset your data.frame incorrectly.

In a data.frame or tbl , extracting using [ works in a couple of ways.

• Since the data is in a matrix -like rectangular form, you can use a [row, column] approach to get specific values. For example to get a single value, you can do something like df1[2, 1] .
• Since a tbl / data.frame is a special type of list , if you don't supply a comma, it assumes you want the entire list element.

Thus, when you did ["var1", "var2"] , it went into matrix subsetting mode and was looking for a row named "var1", which it couldn't find, so it inserted a row of NA values in your dataset.

Here's a small set of examples for you to experiment with.

• Get rows 1:4 and columns 1:4

   df <- mtcars[1:4, 1:4] df # mpg cyl disp hp # Mazda RX4 21.0 6 160 110 # Mazda RX4 Wag 21.0 6 160 110 # Datsun 710 22.8 4 108 93 # Hornet 4 Drive 21.4 6 258 110 

• Extract a single value using a [row, column] approach

   df["Mazda RX4", "mpg"] # [row, column] # [1] 21 

• Check whether a data.frame is a list

   is.list(df) # [1] TRUE 

• Convert a data.frame to a list and try to extract using [row, column] .

   L <- unclass(df) L["Mazda RX4", "mpg"] # A list doesn't have `dim`s. # Error in L["Mazda RX4", "mpg"] : incorrect number of dimensions 

• Providing just one value to extract from a data.frame or a list

   df["mpg"] # Treats it as asking for a single value from a list # mpg # Mazda RX4 21.0 # Mazda RX4 Wag 21.0 # Datsun 710 22.8 # Hornet 4 Drive 21.4 L["mpg"] # $mpg # [1] 21.0 21.0 22.8 21.4 

• Providing a vector of values to extract

   df[c("mpg", "hp")] # mpg hp # Mazda RX4 21.0 110 # Mazda RX4 Wag 21.0 110 # Datsun 710 22.8 93 # Hornet 4 Drive 21.4 110 L[c("mpg", "hp")] # $mpg # [1] 21.0 21.0 22.8 21.4 # # $hp # [1] 110 110 93 110 

• Since a data.frame is a special type of list with dim s, using an empty [, vals] would work

   df[, c("mpg", "hp")] # mpg hp # Mazda RX4 21.0 110 # Mazda RX4 Wag 21.0 110 # Datsun 710 22.8 93 # Hornet 4 Drive 21.4 110 

• Looking for a row that is not there would return NA s

   df["not here", ] # mpg cyl disp hp # NA NA NA NA NA 

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Keeping those details in mind, your best approach is to just use (as suggested in @www's answer :

df1[c("var1", "var2")]