R data.frame : headers based on existing row containing text and numbers

Question

For some reason specific to my R program, I want to assign column names and row names based on existing column and row in a dataframe in R. That is to say, the first line has to become the column names, and the first column has to become the row names.

I first thought it was easy, using :

colnames(myDataFrame) <- myDataFrame[1,]
rownames(MyDataFrame) <- myDataFrame[,1]

As it is also written in this topic.

But I have a lot of cases to handle in the first row and first column of my data frame : only text, text with numbers, text or numbers... That's why this sometimes does not work. See an example with only text in the first line :

I first load my data frame with no headers at all :

> tab <- read.table(file, header = FALSE, sep = "\t")
> tab
         V1   V2  V3   V4   V5     V6  V7   V8   V9
1      TEST this  is only text hoping  it will work
2         I    4   0    0    0      0   0    0    1
3    really    7   6    6    3     10   6   10   10
4      hope  187 141  140  129    130 157  138  168

Here is my data frame without row and column names. I want "TEST this is only text hoping it will work" to become my column name. This doest not work :

> colnames(tab) <- tab[1,]
> tab
          2   10   9    9   10      8   9    8    9
1      TEST this  is only text hoping  it will work
2         I    4   0    0    0      0   0    0    1
3    really    7   6    6    3     10   6   10   10
4      hope  187 141  140  129    130 157  138  168

Whereas this works :

> colnames(tab) <- as.character(unlist(tab[1,]))
> tab
       TEST this  is only text hoping  it will work
1      TEST this  is only text hoping  it will work
2         I    4   0    0    0      0   0    0    1
3    really    7   6    6    3     10   6   10   10
4      hope  187 141  140  129    130 157  138  168

I thought the problem was because R sometimes considers the first column or row as factor. But as you can see :

> is.factor(tab[1,])
FALSE

It can fail even if it is not converted as factor by R.

I tried to tip "as.character(unlist()))" in my program, but in some other cases that I might encounter, it no longer works !... See an example with text and numbers in the first line :

> otherTab <- read.table(otherFile, header = FALSE, sep = "\t")
> otherTab
               V1      V2     V3    V4  V5  V6    V7     V8    V9
1            TEST this45 is 486text 725 with ca257 some numbers
2        number45       4      0     0   0   0     0      0     1
3        254every       7      6     6   3  10     6     10    10
4           where     187    141   140 129 130   157    138   168

> colnames(otherTab) <- as.character(unlist(otherTab[1,]))
> otherTab
                6      10      9     7 725   8     9      8     9
1            TEST this45 is 486text 725 with ca257 some numbers
2        number45       4      0     0   0   0     0      0     1
3        254every       7      6     6   3  10     6     10    10
4           where     187    141   140 129 130   157    138   168

So how to handle these different cases in a easy way (because this seems to be a so simple problem) ? Many thanks in advance.

Steven Beaupré · Accepted Answer · 2015-01-30T21:40:21.747

This happens because, in your initial data frame, V5 is a column of type "int" , not a factor (so you have two different types in your first row)

#> str(df)
#'data.frame':  4 obs. of  9 variables:
# $ V1: Factor w/ 4 levels "254every","TEST",..: 2 3 1 4
# $ V2: Factor w/ 4 levels "187","4","7",..: 4 2 3 1
# $ V3: Factor w/ 4 levels "0","141","6",..: 4 1 3 2
# $ V4: Factor w/ 4 levels "0","140","486text",..: 3 1 4 2
# $ V5: int  725 0 3 129
# $ V6: Factor w/ 4 levels "0","10","130",..: 4 1 2 3
# $ V7: Factor w/ 4 levels "0","157","6",..: 4 1 3 2
# $ V8: Factor w/ 4 levels "0","10","138",..: 4 1 2 3
# $ V9: Factor w/ 4 levels "1","10","168",..: 4 1 2 3

All elements of a vector must be of the same type. When you try to unlist() and store the value in a vector to pass to colnames(), you actually pass a "int" vector (because R coerces the elements to a common type):

#> str(unlist(df[1,]))
# Named int [1:9] 2 4 4 3 725 4 4 4 4
# - attr(*, "names")= chr [1:9] "V1" "V2" "V3" "V4" ...

If you modify the structure of your data frame to specify that column V5 is a factor, your initial method would work:

df[,5] <- as.factor(df[,5])
colnames(df) <- unlist(df[1,])

You would get:

#> df
#      TEST this45  is 486text 725 with ca257 some numbers
#1     TEST this45  is 486text 725 with ca257 some numbers
#2 number45      4   0       0   0    0     0    0       1
#3 254every      7   6       6   3   10     6   10      10
#4    where    187 141     140 129  130   157  138     168

If you don't want to modify your column types, you could apply as.character() to each element of the first row before coercing to a vector and passing to colnames():

colnames(df) <- lapply(df[1,], as.character)

Which results:

#> df
#      TEST this45  is 486text 725 with ca257 some numbers
#1     TEST this45  is 486text 725 with ca257 some numbers
#2 number45      4   0       0   0    0     0    0       1
#3 254every      7   6       6   3   10     6   10      10
#4    where    187 141     140 129  130   157  138     168

Data

structure(list(V1 = structure(c(2L, 3L, 1L, 4L), .Label = c("254every", 
"TEST", "number45", "where"), class = "factor"), V2 = structure(c(4L, 
2L, 3L, 1L), .Label = c("187", "4", "7", "this45"), class = "factor"), 
    V3 = structure(c(4L, 1L, 3L, 2L), .Label = c("0", "141", 
    "6", "is"), class = "factor"), V4 = structure(c(3L, 1L, 4L, 
    2L), .Label = c("0", "140", "486text", "6"), class = "factor"), 
    V5 = c(725L, 0L, 3L, 129L), V6 = structure(c(4L, 1L, 2L, 
    3L), .Label = c("0", "10", "130", "with"), class = "factor"), 
    V7 = structure(c(4L, 1L, 3L, 2L), .Label = c("0", "157", 
    "6", "ca257"), class = "factor"), V8 = structure(c(4L, 1L, 
    2L, 3L), .Label = c("0", "10", "138", "some"), class = "factor"), 
    V9 = structure(c(4L, 1L, 2L, 3L), .Label = c("1", "10", "168", 
    "numbers"), class = "factor")), .Names = c("V1", "V2", "V3", 
"V4", "V5", "V6", "V7", "V8", "V9"), class = "data.frame", row.names = c("1", 
"2", "3", "4"))

Thank you a lot Steven. This made my day : colnames(df) <- lapply(df[1,], as.character). Thank you for all your explainations ! — Totoro, Feb 02 '15 at 09:27

R data.frame : headers based on existing row containing text and numbers

1 Answers1