Adding hashes

[Graham Drabble]

Hi,

I have 2 arrays

%a = ('apples' => 2,
'pears' => 1);
%b = ('apples' => 3,
'pears' => 6);

Want I want is to be able to get

%c = ('apples' => 5,
'pears' => 7);

This can be achieved by

my %c;
foreach my $key (keys %a){
$c{$key} = $a{$key} + $b{$key};
}

but I was wondering if there was any way to do it without the loop.

Any ideas?

 

[Paul Lalli]

I have 2 arrays

No you don't [1].

%a = ('apples' => 2,
'pears' => 1);
%b = ('apples' => 3,
'pears' => 6);

You have two hashes.

Want I want is to be able to get

%c = ('apples' => 5,
'pears' => 7);

This can be achieved by

my %c;
foreach my $key (keys %a){
$c{$key} = $a{$key} + $b{$key};
}

but I was wondering if there was any way to do it without the loop.

How about a map?

my %c = map { $_ => $a{$_} + $b{$_} } keys %a;

Of course, this really is just a loop in disguise, as a map is
basically the same as a foreach loop with a push inside....

Paul Lalli

[1] Okay, *technically*, hashes are also known as "associative arrays",
but they're never referred to as simply "arrays". That's a different
structure entirely.

 

[Dr.Ruud]

Graham Drabble schreef:

Hi,

I have 2 arrays

%a = ('apples' => 2,
'pears' => 1);
%b = ('apples' => 3,
'pears' => 6);

Want I want is to be able to get

%c = ('apples' => 5,
'pears' => 7);

This can be achieved by

my %c;
foreach my $key (keys %a){
$c{$key} = $a{$key} + $b{$key};
}

but I was wondering if there was any way to do it without the loop.

ITYM: without the block. There will always be some kind of loop, even if
only hidden like with map, or the Perl6 hyper-operators.

perl -wle '
my %a = (apples => 2, pears => 1 ) ;
my %b = (apples => 3, bears => 6 ) ;

my %c ; $c{$_} = $a{$_} + ($b{$_} or 0) for keys %a ;

print "@{[%c]}"
'

Prints: apples 5 pears 1

 

[Graham Drabble]

I have 2 arrays

No you don't [1].

Sorry, I know it is a hash, see subject. Brain not quite in gear when
writing.

You have two hashes.

How about a map?

my %c = map { $_ => $a{$_} + $b{$_} } keys %a;

Of course, this really is just a loop in disguise, as a map is
basically the same as a foreach loop with a push inside....

Thanks for that and also to Dr Ruud.

I possibly should have also said that in the case I'm currently
working on I can guarantee the hashes have identical keys.

 

[Dr.Ruud]

Graham Drabble schreef:

I possibly should have also said that in the case I'm currently
working on I can guarantee the hashes have identical keys.

Your 'apples pears' communicated the opposite.