2

I have a list of data frames with column names that I want to substitute some strings out, but I can't do it right.

list:
[1]
cpg value.TCGA.06.5415.01A     value.TCGA.02.0003.01A   value.TCGA.16.1062.01A
cg02726808  0.934641544 NA  NA
cg04243127  0.8828403   NA  NA
[2]
cpg value.TCGA.QH.A6CV.01A  value.TCGA.E1.A7Z4.01A  value.TCGA.E1.5303.01A
 cg02726808 0.938556343 0.92163563  0.959269597
 cg04243127 0.886928811 0.842963126 0.937700666
[N]
.....

Desired output:
    list:           
[1]         
cpg 06.5415.01A 02.0003.01A 16.1062.01A
cg02726808  0.934641544 NA  NA
cg04243127  0.8828403   NA  NA
[2]         
cpg QH.A6CV.01A E1.A7Z4.01A E1.5303.01A
cg02726808  0.938556343 0.92163563  0.959269597
cg04243127  0.886928811 0.842963126 0.937700666
[N]         
.....

I tried to write the following:

lapply(lst, function(x) { gsub("value.TCGA.", "", colnames(lst[[x]]))})

Error in R Studio:

Error in llis1[[xy]] : invalid subscript type 'list'
Called from: is.data.frame(x)
Browse[1]>

And I don't understand what is it about. Thanks for your help.

Peter Chung
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2 Answers2

4

We can use setNames to replace the column names with the new column names from the sub.

lapply(lst, function(x) setNames(x, sub("value.TCGA.", "", names(x))))

Or another option is substring if the 'value.TCGA.' position is the same in all the column names except the first one.

lapply(lst, function(x) setNames(x, c("cpg", substring(names(x)[-1], 12)))
akrun
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3

Try

lapply(lst, function(x) {
  colnames(x) <- gsub("value.TCGA.", "", colnames(x))
  return(x)
})

x in this case is an element of lst rather than the index of the element of lst. So in your code, you are trying to do lst[[ lst[[1]] ]]. From here, the error message should make sense.

Chrisss
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