Sort lexicographically in bash

bashsorting

I want data to be sorted in the way Python sorts it, comparing ASCII values. But sort command seems too clever for that. Take a look. Since '.' < '9':

$ sort
.
9
^D
.
9

and :

$ sort
1.
19
^D
1.
19

These two are fine. But for some reason, if I just add characters to the ends:

$ sort
1.c
19z
^D
19z
1.c

Probably it tries to read that as a number or something. I don't want that, I want it to sort stuff comparing ASCII values of each character. Couldn't find such an option in man, any ideas?

Best Answer

The described behaviour is probably an effect of locale. Turn off locale settings for sort:

$ echo '1.c
19z ' | LC_ALL=C sort
1.c
19z

tl;dr

sort -k1.5 file | uniq -s 6 -w 5

Explanation

My sort is GNU coreutils 8.22. The manpage for my sort shows:

KEYDEF is F[.C][OPTS][,F[.C][OPTS]] for start and stop position, where F is a field number and  C
       a  character  position  in  the  field;  both are origin 1, and the stop position defaults to the
       line's end.

So with your current sort command, sort -k1,1 file uses the first word to the first word as the sort.

What you want is (for the sort command anyway):

sort -k1.5 file | uniq -s 6 -w 5

This will use the fifth character of the first word, which is what you wanted.

Command Line – Sort Packs of Lines Alphabetically

This works in my Debian:

sed '1 ! s/^\[/\x00\[/g' | sort -z | tr -d "\0"

To work with file(s) use redirection(s), e.g. { sed … ; } <input.txt >output.txt, where sed … is the whole command.

The procedure is as follows:

sed inserts a null character before every [ that is in the beginning of a line, unless the line is the first one. This way null characters separate profiles.
sort -z uses these null characters as separators, so it sorts whole profiles, not separate lines.
tr deletes null characters.

Best Answer

Related Solutions

Bash – sort by not first character

tl;dr

Explanation

Command Line – Sort Packs of Lines Alphabetically

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