Sort Hash by Size of the Arrays it References

arunhorne has asked for the wisdom of the Perl Monks concerning the following question:

I have a hash in which each mapping references an array. Anyone know how I could sort the hash based on the size of the referenced arrays?

Thus:
key0 -> (f, a)
key1 -> (a, b, c, d, e)
key2 -> (g)
key3 -> (h, i, j)

From this example I want to generate the output

key2 -> 1
key0 -> 2
key3 -> 3
key1 -> 5

I.e. The keys are sorted in ascending order based on the size of their referenced arrays (and I want to print the above keys/sizes to a text file).

Best wishes, Arun

Comment on Sort Hash by Size of the Arrays it References

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Re: Sort Hash by Size of the Arrays it References by broquaint (Abbot) on May 02, 2002 at 10:09 UTC
Make judicious use of dereferencing and the `sort()` function `my %hash = (foo => [1..2], bar => [1..3], baz => [1]); my @keys = sort { @{$hash{$a}} <=> @{$hash{$b}} } keys %hash; print "$_ - @{$hash{$_}}", $/ for @keys; __output__ baz - 1 foo - 1 2 bar - 1 2 3` [download] HTH `_________ broquaint`	[reply] [d/l]
Re:x2 Sort Hash by Size of the Arrays it References (mention scalar context explicitly) by grinder (Bishop) on May 02, 2002 at 13:41 UTC
Yep, no argument with that. The only thing I would add is that my personal taste is to mention explicitly when I am using arrays in scalar context. A newbie^W^WSomeone looking at that code might wonder whether entire arrays were being compared, and if so, how? Saying `sort { scalar @{$hash{$a}} <=> scalar @{$hash{$b}} } keys %hash`provides a more explicit idea of what is going on, the fact that the number of elements in the arrays are being compared. Hmm, of course the person might not realise that an array in scalar context gives the number of elements, but at least the `scalar` keyword gives them some to ponder and hopefully understand. print@_{sort keys %_},$/if%_=split//,'= & *a?b:e\f/h^h!j+n,o@o;r$s-t%t#u'	[reply] [d/l]