Applying functions to elements is a pretty common situation in programming. In R, the purrr library allows you to do iterations with consistent syntax plus other benefits.


    While using purrr though, I found it difficult to remember which function was used for columns of a data frame, and which was used for rows of a data frame. I decided to investigate.


# # Use code below if you haven't installed purrr yet
# install.packages("purrr")

library(purrr)

# Create a data frame of values
vals_1 <- data.frame(a_col = c(1, 2, 3),
                     b_col = c(1, 2, 3))

You may have noticed at some point that you can access the elements of a dataframe by using double square brackets like so:

# Access the first element in the vals_1 list
vals_1[[1]]
## [1] 1 2 3

Data frames are actually a type of list where the elements are vectors, and the lengths of these vectors are the same. The data type within a single vector is consistent as well. Using:

as.list(vals_1)
## $a_col
## [1] 1 2 3
## 
## $b_col
## [1] 1 2 3

We can see that a data frame is a list of its columns. In the purrr cheatsheet it mentions that \(\texttt{map}\) “[applies] a function to each element of a list.” Clearly, \(\texttt{map}\) must be used for columns then, as the data frame is the list and the columns are its elements.

# Go through each column in vals_1, and sum the column
map(vals_1, sum)
## $a_col
## [1] 6
## 
## $b_col
## [1] 6

The description for pmap says that it “[applies] functions to groups of elements from list of lists, vectors.” If we look at the example below, we see that every element in a column corresponds to a row in that column.

# First element in the a column
vals_1$a_col[1]
## [1] 1
# Second element in the a column
vals_1$a_col[2]
## [1] 2
# Third element in the a column
vals_1$a_col[3]
## [1] 3

If we have the first element of every column, then we have the first row.

# Vector of first elements
c(vals_1$a_col[1], vals_1$b_col[1])
## [1] 1 1
# First row of the data frame of values
vals_1[1, ]
##   a_col b_col
## 1     1     1

We see that we get the same values. We can think of rows as groups of elements from a vector within a list. Therefore \(\texttt{pmap}\) applies to rows of a dataframe.

pmap(vals_1, sum)
## [[1]]
## [1] 2
## 
## [[2]]
## [1] 4
## 
## [[3]]
## [1] 6

\(\texttt{modify_if}\) is similar to \(\texttt{map}\) in that it works on elements of a list, except it returns the same data type that was given to it, and only acts on the elements that fit the condition. In the code below, I show an example of using \(\texttt{modify_if}\) for cleaning “?” into NA values.

# Small example dataframe
ex1 <- data.frame(id = c(1, 2, 3),
                  value = c(NaN, 2, NA),
                  name = c("a", "b", "?"),
                  name_2 = c("?", "?", "c"))

replace_q_mark <- function(fact_vect, target_char = "?"){
  # Function that takes a factor vector and a character variable such as
  # "not applicable" and changes the factor levels that correspond to the
  # character variable so that it represents NA
  
  # Check if char_vect has a question mark
  if (sum(levels(fact_vect) == target_char) >= 1){
    # Replace it with NA
    levels(fact_vect)[levels(fact_vect) == target_char] <- NA
  }
  return(fact_vect)
}

# This returns a dataframe where the question marks are replaced with NA's
modify_if(ex1, is.factor, ~ replace_q_mark(.x, target_char = "?"))
##   id value name name_2
## 1  1   NaN    a   <NA>
## 2  2     2    b   <NA>
## 3  3    NA <NA>      c

Resources

R cheatsheets 

You can find the cheatsheet for purrr in here, among other R cheatsheets.

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