Sort – Sort by Contiguous Digits as a Single Number

awksedsort

A command outputs this:

file_0
file_1
file_10
file_11
file_12
file_13
file_14
file_15
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9

How can I use awk or some other posix tool to actually sort it by the contiguous digits as a single number:

file_0
file_1
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9
file_10
file_11
file_12
file_13
file_14
file_15

In general it should also work in case the digits is inside the file name, e.g.:

file_0.txt
file_1.txt
file_10.txt
file_11.txt
file_12.txt
file_13.txt
file_14.txt
file_15.txt
file_2.txt
file_3.txt
file_4.txt
file_5.txt
file_6.txt
file_7.txt
file_8.txt
file_9.txt

Best Answer

sort -nt '_' -k2

Output:

file_0
file_1
file_2
file_3
file_4
file_5
file_6
file_7
file_8
file_9
file_10
file_11
file_12
file_13
file_14
file_15

or:

file_0.txt
file_1.txt
file_2.txt
file_3.txt
file_4.txt
file_5.txt
file_6.txt
file_7.txt
file_8.txt
file_9.txt
file_10.txt
file_11.txt
file_12.txt
file_13.txt
file_14.txt
file_15.txt

Tested with FreeBSD and GNU coreutils implementations of sort but would not work with busybox implementation. All options used are specified by POSIX.

Related Solutions

Sed Command – Extracting a Regex Match Without Printing Surrounding Characters

When a regexp contains groups, there may be more than one way to match a string against it: regexps with groups are ambiguous. For example, consider the regexp ^.*$[0-9][0-9]*$$ and the string a12. There are two possibilities:

Match a against .* and 2 against [0-9]*; 1 is matched by [0-9].
Match a1 against .* and the empty string against [0-9]*; 2 is matched by [0-9].

Sed, like all other regexp tools out there, applies the earliest longest match rule: it first tries to match the first variable-length portion against a string that's as long as possible. If it finds a way to match the rest of the string against the rest of the regexp, fine. Otherwise, sed tries the next longest match for the first variable-length portion and tries again.

Here, the match with the longest string first is a1 against .*, so the group only matches 2. If you want the group to start earlier, some regexp engines let you make the .* less greedy, but sed doesn't have such a feature. So you need to remove the ambiguity with some additional anchor. Specify that the leading .* cannot end with a digit, so that the first digit of the group is the first possible match.

If the group of digits cannot be at the beginning of the line:
```
sed -n 's/^.*[^0-9]$[0-9][0-9]*$.*/\1/p'
```
If the group of digits can be at the beginning of the line, and your sed supports the \? operator for optional parts:
```
sed -n 's/^$.*[^0-9]$\?$[0-9][0-9]*$.*/\1/p'
```
If the group of digits can be at the beginning of the line, sticking to standard regexp constructs:
```
sed -n -e 's/^.*[^0-9]$[0-9][0-9]*$.*/\1/p' -e t -e 's/^$[0-9][0-9]*$.*/\1/p'
```

By the way, it's that same earliest longest match rule that makes [0-9]* match the digits after the first one, rather than the subsequent .*.

Note that if there are multiple sequences of digits on a line, your program will always extract the last sequence of digits, again because of the earliest longest match rule applied to the initial .*. If you want to extract the first sequence of digits, you need to specify that what comes before is a sequence of non-digits.

sed -n 's/^[^0-9]*\([0-9][0-9]*\).*$/\1/p'

More generally, to extract the first match of a regexp, you need to compute the negation of that regexp. While this is always theoretically possible, the size of the negation grows exponentially with the size of the regexp you're negating, so this is often impractical.

Consider your other example:

sed -n 's/.*\(CONFIG_[a-zA-Z0-9_]*\).*/\1/p'

This example actually exhibits the same issue, but you don't see it on typical inputs. If you feed it hello CONFIG_FOO_CONFIG_BAR, then the command above prints out CONFIG_BAR, not CONFIG_FOO_CONFIG_BAR.

There's a way to print the first match with sed, but it's a little tricky:

sed -n -e 's/\(CONFIG_[a-zA-Z0-9_]*\).*/\n\1/' -e T -e 's/^.*\n//' -e p

(Assuming your sed supports \n to mean a newline in the s replacement text.) This works because sed looks for the earliest match of the regexp, and we don't try to match what precedes the CONFIG_… bit. Since there is no newline inside the line, we can use it as a temporary marker. The T command says to give up if the preceding s command didn't match.

When you can't figure out how to do something in sed, turn to awk. The following command prints the earliest longest match of a regexp:

awk 'match($0, /[0-9]+/) {print substr($0, RSTART, RLENGTH)}'

And if you feel like keeping it simple, use Perl.

perl -l -ne '/[0-9]+/ && print $&'       # first match
perl -l -ne '/^.*([0-9]+)/ && print $1'  # last match

Sort by regular expressions

The usual sort command doesn't provide an included way to specify your specific "dictionary", and while the grep command allows you to provide a file of regular expressions, it won't change the order of the output. But you can put both together in a simple foreach loop -- here's an example that works in the bash shell:

for i in `cat fileofregexp`; do grep "$i" myinputfile; done

This takes each regexp line from your file of regular expressions one by one, and outputs any match from your inputfile, so the resulting output will be sorted by your regexp order. Note that any lines in your inputfile that don't match at all will not make it to the output.

Addendum: As requested, here's a version using a while loop:

while IFS= read -r i; do grep "$i" myinputfile; done  < fileofregexp

Best Answer

Related Solutions

Sed Command – Extracting a Regex Match Without Printing Surrounding Characters

Sort by regular expressions

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