Switch the last two fields with awk

Question

I'm looking for a way to switch the last two fields of a tab-separated file. The number of fields vary among rows, so I can't specify the field number.

Most of what I could find online was field-specific scripts. E.g. swap two columns with awk

Thanks!

Answer 1

awk -F"\t" -v OFS="\t" 'NF>1 {t=$(NF-1); $(NF-1)=$NF; $NF=t} {print}' file

Do a standard swap between $NF and $(NF-1) only when we have at least 2 fields. It's fatal error to access field -1 (in case of empty line, NF==0).

See also: GNU awk fields

Answer 2

Using Raku (formerly known as Perl_6)

raku -ne 'join( "\t", .words[0..*-3], .words[*-1] // "", .words[*-2] // "").trim-leading.put ;' 

Basically, the code above creates a list of whitespace-separated elements (words) in the desired order (last two elements switched). The elements are joined with \t tab, any leading whitespace is trimmed away with trim-leading, and the result is output. This code works on tab-separated columns as well as on space-separated columns.

Note, Raku's // defined-OR operator is used to avoid issues with Nil values in String context--an "" empty string is 'inserted' where necessary.

Sample Input (first line is blank):

A
A   B
A   B   C
A   B   C   D
A   B   C   D   E
A   B   C   D   E   F

Sample Output (last two elements on each line are swapped--first line is still blank):

A   
B   A
A   C   B
A   B   D   C
A   B   C   E   D
A   B   C   D   F   E

https://docs.raku.org/routine/words
https://docs.raku.org/routine/join
https://raku.org

Answer 3

You can make use of the built-in variable NF to refer to the last field to print it before the second last field (NF-1):

awk 'NF <= 1; NF > 1 {for (i = 1; i <= NF-2; i++) printf "%s\t", $i; print $NF"\t"$(NF-1)}' file.tsv

Answer 4

One way can be :

awk -F '\t' '
BEGIN {OFS=FS}
f=(NF>1){
  p = length-length($(NF-1)"x"$NF)
  print substr($0,1,p) $NF, $(NF-1)
}!f' file