sapply not applying a function created to all rows in R dataframe - r

Question

I have the following dataframe in R and am trying to use a stringsplit function to the same to yield a different dataframe
DF
A B C
"1,2,3" "1,2"
"2" "1"
The cells of the dataframe are filled with characters. The empty spaces are blank values. I have created the following function
sepfunc<-function(x){strsplit(as.character(x, split= ","))[[1]][1]}
The function works neatly when i use it on a single column
sapply(DF$A, sepfunc)
[1] "1" "2"
However, the following command yields only a single row
sapply(DF, sepfunc)
A B C
"1" NA "1"
The second row is not displayed. I know I must be missing something rudimentary. I request someone to help.
The expected output is
A B C
"1" NA "1"
"2" "1" "NA"

Answer 1

When we do the strsplit, the output is a list of vectors. If we just subset the first list element with [[1]], then the rest of the elements are skipped. Here the first element corresponds to the first row. But, when we do the same on a single column, it is looping through each element and then do the strsplit. It will not hurt by taking the first element [[1]] because the list is of length 1. Here, the case is different. The number of list elements are the same as the number of rows for each of the columns. So, we need to loop through the list (either with sapply/lapply - former gives a vector depends on the case, while latter always return list)
sapply(DF, function(x) sapply(strsplit(as.character(x), ","), `[`, 1))
# A B C
#[1,] "1" NA "1"
#[2,] "2" "1" NA
Let's look this more closely by splitting the codes into chunks. On each column, we can find the output as list of splitted vectors
lapply(DF, function(x) strsplit(as.character(x), ","))
#$A
#$A[[1]]
#[1] "1" "2" "3"
#$A[[2]]
#[1] "2"
#$B
#$B[[1]]
#[1] NA
#$B[[2]]
#[1] "1"
#$C
#$C[[1]]
#[1] "1" "2"
#$C[[2]]
#character(0)
When we do [[1]], the first element is extracted i.e. the first row of 'A', 'B', 'C'
lapply(DF, function(x) strsplit(as.character(x), ",")[[1]])
#$A
#[1] "1" "2" "3"
#$B
#[1] NA
#$C
#[1] "1" "2"
If we again subset on the above, i.e. the first element, the output will be 1 NA 1.
Instead we want to loop through the list and get the first element of each list

Answer 2

As you only want to extract the first part before the , you can also do
sapply(DF, function(x) gsub("^([^,]*),.*$", "\\1", x))
# A B C
# [1,] "1" NA "1"
# [2,] "2" NA "1"
This extracts the the first group (\\1) which is here marked with brackets. ([^,]*)
Or with stringr:
library(stringr)
sapply(DF, function(x) str_extract(x, "^([^,]*)"))

Answer 3

Here is another version of this
lapply(X = df, FUN = function(x) sapply(strsplit(x = as.character(x), split = ","), FUN = head, n=1))

Answer 4

First of all, notice that your sepfun should always give an error:
sepfunc<-function(x){strsplit(as.character(x, split= ","))[[1]][1]}
split should go with strsplit, not as.character, so what you meant is probably:
sepfunc<-function(x){strsplit(as.character(x), split= ",")[[1]][1]}
Second, the question of data sanity. You have character variables stored as factors, and missing data stored as empty strings. I would recommend dealing with these issues before trying to do anything else. (Why do I say NA is more sensible here than an empty string? Because you told me so. You want NA's in the output, so I guess this means that if there are no numbers in the string, it means that something is missing. Missing = NA. There is also a technical reason which would take a bit longer to explain.)
So in the following, I'm just using an altered version of your DF:
DF <- data.frame(A=c("1,2,3", "2"), B=c(NA, "1"), C=c("1,2", NA), stringsAsFactors=FALSE)
(If DF comes from a file, then you could use read.csv("file", as.is=TRUE). And then DF[DF==""] <- NA.)
The output of strsplit is a list so you'll need sapply to get something useful out from it. And another sapply to apply it to all columns in a data frame.
sapply(DF, function(x) sapply(strsplit(x, ","), head, 1))
# A B C
# [1,] "1" NA "1"
# [2,] "2" "1" NA
Or step by step. Before you can sapply a function over all columns of a data frame, you need it to give meaningful results for all the columns. Let's try:
sf <- function(x) sapply(strsplit(x, ","), head, 1)
# and sepfunc as defined above:
sepfunc<-function(x){strsplit(as.character(x), split= ",")[[1]][1]}
sf(DF$A)
# [1] "1" "2"
# as expected
sepfunc(DF$A)
# [1] "1"
Notice that sepfunc uses only the first element (as you told it to!) of each column, and the rest is discarded. You need sapply or something similar to use all elements. So as a consequence, you get this:
sapply(DF, sepfunc)
# A B C
# "1" NA "1"
(It works, because we've redefined empty strings as NA. But you get the results only for the first row of each variable.)
sapply(DF, sf)
# A B C
# [1,] "1" NA "1"
# [2,] "2" "1" NA