@subscriptions = {}
[:free, :lite, :brisk, :jammin].each { |sub_type| subscriptions[sub_type] = SubscriptionType.make(sub_type) }
Refactorings
No refactoring yet !
danielharan
August 3, 2009, August 03, 2009 20:32, permalink
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Use inject({}). Technically you're still initializing an empty hash, but it's more idiomatic.
chipcastle.myopenid.com
August 4, 2009, August 04, 2009 01:42, permalink
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Of course! inject is definitely the way to go. Here's what I ended up with:
@subscriptions = [:free, :lite, :brisk, :jammin].inject({}) {|result, ele| result[ele] = SubscriptionType.make(ele); result }
danielharan
August 4, 2009, August 04, 2009 03:55, permalink
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Hmm... that's a bit long. I prefer splitting it up for readability
@subscriptions = [:free, :lite, :brisk, :jammin].inject({}) do |result, ele|
result[ele] = SubscriptionType.make(ele)
result
end
Chip Castle
August 4, 2009, August 04, 2009 13:56, permalink
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I agree. Thanks for the suggestions!
Vidar Hokstad
August 5, 2009, August 05, 2009 20:01, permalink
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Here's an alternative using Hash[]
@subscriptions = Hash[*[:free, :lite, :brisk, :jammin].collect {|ele| [ele, SubscriptionType.make(ele)] }.flatten]
ReinH
September 9, 2009, September 09, 2009 18:35, permalink
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One option is to make a factory Hash. More a design alternative than a strict refactoring.
@subscriptions = Hash.new {|h,k| h[k] = SubscriptionType.make(k)}
Pharmd189
November 3, 2009, November 03, 2009 16:14, permalink
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Very nice site!
Very nice site!
OvXjWl
November 18, 2009, November 18, 2009 22:00, permalink
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Hi! UZWEDf
Hi! UZWEDf
sNjhqbc
November 19, 2009, November 19, 2009 19:36, permalink
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Hi! ZnEHzrwV
Hi! ZnEHzrwV
Here's the long way of writing what I want:
subscriptions[:free] = SubscriptionType.make(:free)
subscriptions[:lite] = SubscriptionType.make(:lite)
subscriptions[:brisk] = SubscriptionType.make(:brisk)
subscriptions[:jammin] = SubscriptionType.make(: jammin)
Please see how I currently do this below, but I'd rather not initialize an empty hash first. Any suggestions? Thanks!