lapply returns the same value for every data frame element

January 10, 2025

I created a function that splits a string by ":" and takes the first element, which is the information I need from a vcf:

remove_semicolon = function(x){
    newstr = strsplit(x,":")[[1]][1]
    return(newstr)
}

I wish to apply it to every element of a data frame such as the following:

>rubbish
              NS05                   NS113                   NS137
1              0/0:1                  0/0:15                  0/0:25
2              0/0:1                  0/0:15                  0/0:25
3              0/0:1                  0/0:16                  0/0:25
4 1/1:0,1:1:3:39,3,0 1/1:0,16:16:48:621,48,0 1/1:0,26:26:78:969,78,0
5              0/0:1                  0/0:16                  0/0:29

So that for rubbish[1,1] the desired output is "0/0", for rubbish[4,1] "1/1" etc, with the matrix/data frame structure left intact. However,

 rubbish[]=lapply(rubbish,remove_semicolon)

returns:

> rubbish
NS05 NS113 NS137
1  0/0   0/0   0/0
2  0/0   0/0   0/0
3  0/0   0/0   0/0
4  0/0   0/0   0/0
5  0/0   0/0   0/0

even though, in contrast,

sapply(rubbish[,1],remove_semicolon)

returns what I want, i.e. a vector 0/0, 0/0, 0/0, 1/1, 0/0 rather than all 0/0:

         0/0:1              0/0:1              0/0:1 1/1:0,1:1:3:39,3,0 
         "0/0"              "0/0"              "0/0"              "1/1" 
         0/0:1 
         "0/0"

What am I doing incorrectly when implement lapply? Shouldn’t it just apply the remove_semicolon function to every element of rubbish in the same way that sapply does it for every element of a column vector?

>Solution :

Using apply(., MARGIN = 1:2, .) seems to work:

rubbish[] <- apply(rubbish, 1:2, remove_semicolon)

 NS05 NS113 NS137
1  0/0   0/0   0/0
2  0/0   0/0   0/0
3  0/0   0/0   0/0
4  1/1   1/1   1/1
5  0/0   0/0   0/0

If you look at the output of lapply(rubbish, remove_semicolon) (before assigning it back to the data frame) you’ll see that each element of the output is a length-1 vector (which then gets replicated to fill the column). This happens because remove_semicolon isn’t vectorized.

This would work with lapply():

rs2 <- function(x) {
   newstr <- strsplit(x, ":")
   return(sapply(newstr, head, 1))
}
lapply(rubbish, rs2)

Another alternative would be to use gsub() (or stringr::str_extract) with a regular expression, e.g.

rs3 <- function(x) gsub("^([^:]+):.*$", "\\1", x)

(it does admittedly look a little like magic)

example

rubbish <- read.table(header = TRUE, text = "
             NS05                   NS113                   NS137
              0/0:1                  0/0:15                  0/0:25
              0/0:1                  0/0:15                  0/0:25
              0/0:1                  0/0:16                  0/0:25
 1/1:0,1:1:3:39,3,0 1/1:0,16:16:48:621,48,0 1/1:0,26:26:78:969,78,0
              0/0:1                  0/0:16                  0/0:29
")