Re: Difference between two arrays - is there a better way?

This one-liner (versus answers with hashes and inspired by another answer) is working for me but has not been very rigorously tested (as run on AS Perl 5.16.3):

@foo = (1, 2, 3, 4, 5, 6, 7,  8,  9,  10);
@bar = (0, 1, 2, 5, 7, 9, 11, 12, 13, 14);
my @diff = grep {my $baz = $_; ! grep($_ == $baz, @bar)} @foo;
print join(",", @diff);

Gives:
3,4,6,8,10
[download]

Switch the lists around in the nested greps to go from being "in foo but not also bar" to "in bar but not also foo".

-Simon

Comment on Re: Difference between two arrays - is there a better way? Download Code

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Re^2: Difference between two arrays - is there a better way? by Anonymous Monk on Jan 05, 2015 at 09:36 UTC
wow this is mind-bending! can you please go through it in detail?	[reply]
Re^3: Difference between two arrays - is there a better way? by Anonymous Monk on May 12, 2016 at 18:54 UTC
`@cities = (qw(London Oslo Paris Amsterdam Berlin )); @visited = (qw(Berlin Oslo)); say "Still need to visit:", grep { ! ({ $_, 0 } ~~ @visited) } @cities;` [download]	[reply] [d/l]
Re^4: Difference between two arrays - is there a better way? by SimonSaysCake (Sexton) on Jun 28, 2016 at 17:42 UTC
That's certainly wins for brevity and works a treat but it would seem the smartmatch op is (still) considered deprecated/experimental: https://metacpan.org/pod/release/RJBS/perl-5.18.0/pod/perldelta.pod#The-smartmatch-family-of-features-are-now-experimental -Simon	[reply]
Re^2: Difference between two arrays - is there a better way? by Anonymous Monk on Nov 24, 2016 at 12:01 UTC
What if the 2 lists are array of hashes?	[reply]
Re^3: Difference between two arrays - is there a better way? by Corion (Pope) on Nov 24, 2016 at 12:13 UTC
Then I recommend you adapt the concept in perlfaq4 to your situation.	[reply]


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