Simple Hash problem

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

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Re: Simple Hash problem by Juerd (Abbot) on Feb 08, 2002 at 20:45 UTC
You forgot a bunch of commas. Golfing: `$new{$_}=defined$old{$_}?$old{$_}:999 for@mips` [download] If there are no `0` values: `$new{$_}=$old{$_}\|\|999 for@mips` [download] The first, but perl6 (I think): `$new{$_}=$old{$_}//999 for@mips` [download] The first one written in a more readable way: `for (@mips) { $new{$_} = defined($old{$_}) ? $old{$_} : 999; }` [download] Another approach: `%new=map{($_,defined$old{$_}?$old{$_}:999}@mips` [download] If 0 may be 999: `%new=map{($_,$old{$_}\|\|999)}@mips` [download] Another perl6 try: `%new=map{($_,$old{$_}//999)}@mips` [download] Again, the first of this approach written in a readable manner: `%new = map { ($_ => defined($old{$_}) ? $old{$_} : 999) } @mips;` [download] Good luck. `2;0 juerd@ouranos:~$ perl -e'undef christmas' Segmentation fault 2;139 juerd@ouranos:~$` [download]	[reply] [d/l] [select]
Re: Simple Hash problem by VSarkiss (Monsignor) on Feb 08, 2002 at 20:16 UTC
The simplest way is with a single loop: `foreach my $k (@mips) { $hash{$k} = 999 unless exists $hash{$k}; }` [download] This might even touch off a round of golf... HTH	[reply] [d/l]
Re: Re: Simple Hash problem by Vavoom (Scribe) on Feb 08, 2002 at 20:41 UTC
It is simple, but it doesn't quite match the sample results. This approach ignores elements in %hash that are not in @mips, whereas the original example removed them. Here's a small rework that fixes it: `my %newhash = (); for (@mips) { $newhash{$_} = $hash{$_} \|\| 999; } %hash = %newhash;` [download] lestrrat's solution is much more fun to look at, though. Vavoom	[reply] [d/l]
Re: Re: Re: Simple Hash problem by trs80 (Priest) on Feb 08, 2002 at 21:38 UTC
I am not following why you needed to create the temp hash. My output appears to be correct for the example given with the following line of code. `for(@mips){$hash{$_}\|\|=999}` UPDATE: Thanks crazyinsomniac, after reading these posts several times and talking with you in CB I realize that the example above did not have G in it once it was processed. I read the question and did not examine the output. In the text of the original question it does not state that missing keys should be removed, it merely states that the values should be 999 if they have none.	[reply] [d/l]
Re: Re: Re: Re: Simple Hash problem by crazyinsomniac (Prior) on Feb 09, 2002 at 10:13 UTC
Re: Re: Simple Hash problem by particle (Vicar) on Feb 08, 2002 at 20:47 UTC
okay, i'll bite... (41 chars--surely it can be done in less?) `map{exists$hash{$_}or$hash{$_}=999}@mips` update: (39 chars) `exists$hash{$_}or$hash{$_}=999for@mips` [download] update2: vavoom is right. back to the drawing board... (75 chars--results in %h, and it's a little obfuscated) `@$_{@mips}=(999)x@mips for\my%h;exists$hash{$_}and$h{$_}=$hash{$_}for@ +mips;` [download] ~Particle	[reply] [d/l] [select]
Re: Simple Hash problem by lestrrat (Deacon) on Feb 08, 2002 at 20:19 UTC
Will this work? `my %hash = { ... original data ... }; my %tmp = map{ ( $_, exists $hash{ $_ } ? $hash{ $_ } : 999 ) } @mi +ps;` [download]	[reply] [d/l]
Re: Simple Hash problem by Zaxo (Archbishop) on Feb 08, 2002 at 20:48 UTC
39 is probably not golf standard, but one character names would make it 30: `exists$hash{$_}or$hash{$_}=999for@mips; 123456789012345678901234567890123456789` [download] After Compline, Zaxo	[reply] [d/l]
Re: Simple Hash problem by impossiblerobot (Deacon) on Feb 08, 2002 at 20:53 UTC
I did it with a hash slice. Probably not very memory efficient: `@hash{ grep( !defined($hash{$_}), @mips) } = (999) x @mips;` [download] Update: Better hash slice solution: `@new{@mips}=map(($hash{$_}\|\|999),@mips);` [download] I had also missed the requirement to eliminate key/values not in @mips. ++Vavoom. Impossible Robot	[reply] [d/l] [select]