From: dblack@...
Date: 2006-01-05T05:15:36+09:00
Subject: Re: Question on iterating a hash

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Hi --

On Thu, 5 Jan 2006, Simon Kr=F6ger wrote:

> dblack@wobblini.net wrote:
>
>> Hi --
>>=20
>> On Thu, 5 Jan 2006, James Edward Gray II wrote:
>>=20
>>> On Jan 4, 2006, at 12:46 PM, phil swenson wrote:
>>>=20
>>>> I have a hash that looks like this : {"1"=3D>"0, "3"=3D"1", "45"=3D>"1=
",
>>>> "101"=3D>"0"}
>>>>=20
>>>> What I want is an array of all the keys that have hash values equal to
>>>> "1" ["3","45"] in this case.
>>>>=20
>>>> Yeah, i can just iterate the keys... but it seems like there is a "rub=
y"
>>>> way to do this.  Any thoughts?
>>>=20
>>>=20
>>>>> h =3D Hash[*%w{1 0 3 1 45 1 101 0}]
>>>=20
>>> =3D> {"45"=3D>"1", "1"=3D>"0", "101"=3D>"0", "3"=3D>"1"}
>>>=20
>>>>> h.select { |key, value| value =3D=3D "1" }.map { |key, value| key }
>>>=20
>>> =3D> ["45", "3"]
>>=20
>>=20
>> Wouldn't it be easier to iterate through the keys?  I'm not sure why
>> Phil didn't want to.  (Phil?)
>>
>>   h.keys.select {|k| h[k] =3D=3D 1}
>>=20
>>=20
>> David
>
> Don't know, but here is another one:
>
> h.select{|k, v| v =3D=3D "1" }.transpose.first

OK... but I'm still not getting why just "select"ing the ones you want
isn't the best way.


David

--=20
David A. Black
dblack@wobblini.net

"Ruby for Rails", from Manning Publications, coming April 2006!
http://www.manning.com/books/black
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