Re: How can I sort my array numerically on part of the string? (updated)

How about sort { ($a=~/(\d+),/)[0] <=> ($b=~/(\d+),/)[0] or $a cmp $b } @list? (see also return values of regular expressions) Though a Schwartzian transform would be much more performant:

@list = map { $$_[0] } sort { $$a[1] <=> $$b[1] or $$a[0] cmp $$b[0] }
    map { /(\d+),/; [$_,$1] } @list;
[download]

Note I added the or cmp so that if the numeric parts are equal (e.g. '1,cat' vs. '001,elk'), the list is still reliably sorted.

Update: The above doesn't handle cases of the regex not matching. In my second piece of code above you could handle that with an error via e.g. map { /(\d+),/ or die $_; [$_,$1] } or a replacement value via e.g. map { [$_, /(\d+),/ ? $1 : 0] }.

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Re^2: How can I sort my array numerically on part of the string? (updated) by LanX (Saint) on Dec 02, 2020 at 02:09 UTC
> `sort { ($a=~/(\d+),/)[0] <=> ($b=~/(\d+),/)[0]` You need that `(...)[0]` for the match to return the captures in list context, ( `<=>` is enforcing scalar context and `m//` only returns captures in list context otherwise boolean ) I'm wondering if there is a prettier solution for that. The Schwartzian transform doesn't have that limitation. `... map { [$_, /(\d+),/] } @list;` should do already. update clarification: any better solution than `(...)[0]` to get list context ? Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l] [select]
Re^3: How can I sort my array numerically on part of the string? (updated) by tybalt89 (Monsignor) on Dec 02, 2020 at 03:17 UTC
Here's a capture in list context. Prettier? `#!/usr/bin/perl use strict; # https://perlmonks.org/?node_id=11124461 use warnings; my @list = ( '1,cat', '2,dog', '22,mouse', '11,eel', '001,elk', '13,mi +nk'); my @n; @n[ /(\d+),/ ] .= "$_\n" for @list; print grep defined, @n;` [download] This is why perl is fun :)	[reply] [d/l]
Re^4: How can I sort my array numerically on part of the string? (updated) by LanX (Saint) on Dec 02, 2020 at 11:06 UTC
> This is why perl is fun :) and so memory efficient ... > Prettier? nope! :) Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply]
Re^5: How can I sort my array numerically on part of the string? (updated) by misterperl (Friar) on Dec 02, 2020 at 13:44 UTC
Re^3: How can I sort my array numerically on part of the string? (updated) by misterperl (Friar) on Dec 02, 2020 at 13:24 UTC
TY I didnt realise I needed the [0] I'm gonna try that. Although I'm unclear as to what "list" is involved with $a and $b which appear to be scalar?	[reply]
Re^4: How can I sort my array numerically on part of the string? (updated) by LanX (Saint) on Dec 02, 2020 at 13:41 UTC
> > You need that (...)[0] for the match to return the captures in list context, ( <=> is enforcing scalar context and m// only returns captures in list context otherwise boolean ) `DB<96> $_= "123,foo" DB<97> p scalar m/(\d+,)/ # boolean value 1 DB<98> p m/(\d+,)/ # list of captures 123, DB<99>` [download] see https://perldoc.perl.org/perlop#Matching-in-list-context If the `/g` option is not used, `m//` in list context returns a list consisting of the subexpressions matched by the parentheses in the pattern, that is, ($1, $2, $3...) (Note that here $1 etc. are also set). When there are no parentheses in the pattern, the return value is the list (1) for success. With or without parentheses, an empty list is returned upon failure. Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l] [select]
Re^4: How can I sort my array numerically on part of the string? (updated) by Anonymous Monk on Dec 02, 2020 at 13:26 UTC
the list is the return values of the regex	[reply]