awk: print everything but not the last nth columns

Question

I have an input file in this form:

foo bar 08 320984 2384 bla foo baz 23 32425 32532 [...]

there are always three tokens in the end, but an unknown number of tokens in the front. I want to rewrite the file as CSV so it can be automatically parsed by other applications. my current awk command is:

awk '{ print $(NF-2)";"$(NF-1)";"$NF}'

the output should be

foo bar;08;320984;2384 bla foo baz;23;32425;32532 [...]

That's not CSV but that's not the point. You are trying to split your file into four output fields? Where the last three fields are the last three fields of the input and the first field is everything else on the line? — Etan Reisner
– Etan Reisner, Commented Nov 4, 2014 at 21:46
How do you know what is token? Any text with no numbers? You have two words in one line and three words in another. — Jotne
– Jotne, Commented Nov 4, 2014 at 22:15
@Jotne from the sentence "there are always three tokens in the end, but an unknown number of tokens in the front" I understand it is always like tk1 tk2 tk3 ... tkn TK1 TK2 TK3 and has to become tk1 tk2 tk3 ... tkn;TK1;TK2;TK3. — fedorqui
– fedorqui, Commented Nov 5, 2014 at 9:00

glenn jackman · Accepted Answer · 2014-11-04 22:00:07Z

This is unfortunately something that awk just isn't the greatest at (and cut's ability to do field ranges doesn't help here either.

Something like this should work though:

awk '{nfff=$(NF-2); nff=$(NF-1); nf=$NF; NF-=3; printf "%s;%s;%s;%s\n", $0, nfff, nff, nf}' file

A small variation on your approach: $ awk '{ last=";"$(NF-2)";"$(NF-1)";"$NF; NF-=3; print $0 last}' file

Community · Accepted Answer · 2017-05-23 11:56:52Z

If I understand you and fedorqui properly:

awk '{for (i=1;i<NF;i++) printf "%s%s",$i,(i+4>NF?";":FS);print $NF}' file foo bar;08;320984;2384 bla foo baz;23;32425;32532

This will add ; in front of the last three fields.

John's comment may be better way to do it.

chaos · Accepted Answer · 2014-11-04 22:05:17Z

sed could also work:

sed 's/\ \([^\ ]\+\)\ \([^\ ]\+\)\ \([^\ ]\+\)$/;\1;\2;\3/' file

or if your sed supports -r:

sed -r 's/\ ([^\ ]+)\ ([^\ ]+)\ ([^\ ]+)$/;\1;\2;\3/' file

It replaces the last 3 newlines with ;.

Or a bit easier:

rev file | sed 's/\ /;/g; s/;/\ /g4' | rev

glenn jackman · Accepted Answer · 2014-11-04 22:05:37Z

A fancy GNU awk method:

gawk ' function replace(what) { return gensub(/[[:blank:]]+([^[:blank:]]+)$/, ";\\1", 1, what) } {$0 = replace(replace(replace($0))); print} ' file

foo bar;08;320984;2384 bla foo baz;23;32425;32532

Miguel · Accepted Answer · 2014-11-04 22:06:40Z

This should do it for an arbitrary number of fields before the last three:

awk '{for (i=1; i <= NF - 3; i++) if (i == 1) printf $i; else printf " "$i} {print ";"$(NF-2)";"$(NF-1)";"$NF}' input

Arjun Mathew Dan · Accepted Answer · 2014-11-05 08:54:39Z

I am new to awk, but how about this (this will not remove the blank spaces though.):

awk '{for (i=0; i<3; i++) {$(NF-i)=";" $(NF-i)} print $0} ' file

Example:

sdlcb@Goofy-Gen:~/AMD$ cat file foo bar 08 320984 2384 bla foo baz 23 32425 32532 sdlcb@Goofy-Gen:~/AMD$ awk '{for (i=0; i<3; i++) {$(NF-i)=";" $(NF-i)} print $0} ' file foo bar ;08 ;320984 ;2384 bla foo baz ;23 ;32425 ;32532

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