Re: hash keys compare

i'm not sure i understand the question, but if i do, here's a possible answer:

perl -wMstrict -e
"my %X = qw(a 1  b 2  c 3  z 9);  my %Y = qw(d 4  e 5  f 6  z 8);
 my %X_not_in_Y = %X;  my %Y_not_in_X = %Y;
 delete @X_not_in_Y{ keys %Y };  delete @Y_not_in_X{ keys %X };
 print qq(X not in Y: @{[ keys %X_not_in_Y ]} \n);
 print qq(Y not in X: @{[ keys %Y_not_in_X ]})"
X not in Y: c a b
Y not in X: e d f
[download]

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Re^2: hash keys compare by Anonymous Monk on Dec 18, 2007 at 19:26 UTC
hi Super doc, What I am trying to explain is that the list of keys in $c in hash1 should be exactly similar to list of keys in $f in hash2. I just want a list of keys that are not present in either of them. For ex: html $a = 'a, b , c , d, e, f'; $f = 'a, 1, c, d, 2, , f'; SO I want a report saying : $a is missing keys "1, 2". $f is missing keys "b, e". /html The hash that you used to explain works fine for simple hash tables with one key and one value. I am not sure how to build a data structure, for a hash like this one: html $hash1->{$a}{$b}{$c} $hash2->{$d}{$e}{$f} /html Can you help me out?	[reply]
Re^3: hash keys compare by Anonymous Monk on Dec 19, 2007 at 07:50 UTC
oops... created a little too soon -- it's late. I meant to say, See: perlref perlreftut perldsc to get you started. (by the way -- the output of the `print` statement in the code above is `fum`.)	[reply] [d/l] [select]
Re^3: hash keys compare by Anonymous Monk on Dec 19, 2007 at 07:36 UTC
I believe you are laboring under a misapprehension (or else I am). You write of a "... list of keys in $f" and then give an example. Your example is not of a list, but of a string literal assigned to a scalar. A string literal can be used as a single key in a hash. Likewise, a scalar holding a string can be used as a single key in a hash. The two string literals given in your example are not the same, so they can be used as two keys in the same hash. It is possible to ask questions about how similar or different two strings are and get useful answers, but you must first be very clear about just what similar and different mean in the context of your problem. On The Other Hand... If you just want to construct a nested data structure using hashes, you are almost there with the other example you give. `my $hash_reference = {}; # empty hash ref for now my $a = 'fee'; my $b = 'fie'; my $c = 'foe'; $hash_reference->{$a}{$b}{$c} = 'fum'; # same as... $hash_reference->{'fee'}{'fie'}{'foe'} = 'fum'; # same as... $hash_reference->{fee}{fie}{foe} = 'fum'; print $hash_reference->{fee}{fie}{foe};` [download] See:	[reply] [d/l]