I have this input:
sdkxyosl 1
safkls 2
asdf--asdfasxy_asd 5
dkd8k jasd 29
sdi44sw 43
asasd afsdfs 10
rklyasd 4
I need this output:
sdi44sw 43
dkd8k jasd 29
asasd afsdfs 10
asdf--asdfasxy_asd 5
rklyasd 4
safkls 2
sdkxyosl 1
So i need to sort the lines by the last column.
I don't know how many columns are in one line.
I just can't figure it out, how to do it. I don't have "perl powers". I just have ~average scripting powers with sed, awk, cut, etc..
Does somebody know how to do it?
The following command line uses awk to prepend the last field of each line of file.txt, does a reverse numerical sort, then uses cut to remove the added field:
awk '{print $NF,$0}' file.txt | sort -nr | cut -f2- -d' '
sort -k. The begfield function in GNU sort just counts down to zero. Your decorate-sort-undecorate (DSU) approach seems to be the best way I think.
– Mikel
Apr 2 '11 at 23:20
$NF,$RS and not $NF,$0? I didn't know $RS did that. (I guess it's equivalent to $NF,$"\n", which does the same, but that's also surprising I think.)
– Mikel
Apr 3 '11 at 0:36
$ can be followed by any expression. “The effect of the field number expression evaluating to anything other than a non-negative integer is unspecified”. GNU awk (on my system) treats a string like "\n" as zero. Others (e.g. the original implementation by A, W and K) abort with an error message. If RS happened to be a digit, you'd get the corresponding field on any implementation. So don't do this, use $0.
– Gilles 'SO- stop being evil'
Apr 3 '11 at 11:51
