Ruby 哈希排列

发布于 2024-12-07 22:06:02 字数 757 浏览 1 评论 0原文

有没有快速的方法来获得给定哈希的（随机）排列？例如，对于数组，我可以使用 sample 方法作为对于

ruby-1.9.2-p180 :031 > a = (1..5).to_a
 => [1, 2, 3, 4, 5] 
ruby-1.9.2-p180 :032 > a.sample(a.length)
 => [3, 5, 1, 2, 4]

哈希，我可以在哈希键上使用相同的方法并构建一个新的哈希，

ruby-1.9.2-p180 :036 > h = { 1 => 'a', 2 => 'b', 3 => 'c' }
 => {1=>"a", 2=>"b", 3=>"c"} 
ruby-1.9.2-p180 :037 > h.keys.sample(h.length).inject({}) { |h2, k| h2[k] = h[k]; h2 }
 => {3=>"c", 2=>"b", 1=>"a"}

但这太丑陋了。是否有任何可以避免所有代码的哈希“样本”方法？

更新正如@Michael Kohl 在评论中指出的，这个问题仅对 ruby 1.9.x 有意义。由于在 1.8.x 中哈希是无序的，因此无法做到这一点。

原文

Is there any quick way to get a (random) permutation of a given hash? For example with arrays I can use the sample method as in

ruby-1.9.2-p180 :031 > a = (1..5).to_a
 => [1, 2, 3, 4, 5] 
ruby-1.9.2-p180 :032 > a.sample(a.length)
 => [3, 5, 1, 2, 4]

For hashes I can use the same method on hash keys and build a new hash with

ruby-1.9.2-p180 :036 > h = { 1 => 'a', 2 => 'b', 3 => 'c' }
 => {1=>"a", 2=>"b", 3=>"c"} 
ruby-1.9.2-p180 :037 > h.keys.sample(h.length).inject({}) { |h2, k| h2[k] = h[k]; h2 }
 => {3=>"c", 2=>"b", 1=>"a"}

but this is so ugly. Is there any 'sample' method for hashes which can avoid all that code?

Update As pointed out by @Michael Kohl in comments, this question is meaningful only for ruby 1.9.x. Since in 1.8.x Hash are unordered there is no way to do that.

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相思故 2024-12-14 22:06:02

mu 的稍微细化太短了答案：

h = Hash[h.to_a.shuffle]

A slight refinement of mu is too short's answer:

h = Hash[h.to_a.shuffle]

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说谎友 2024-12-14 22:06:02

只需添加 to_a和 哈希[] 到你的数组版本以获得哈希版本：

h = Hash[h.to_a.sample(h.length)]

例如：

>> h = { 1 => 'a', 2 => 'b', 3 => 'c' }
=> {1=>"a", 2=>"b", 3=>"c"}
>> h = Hash[h.to_a.sample(h.length)]
=> {2=>"b", 1=>"a", 3=>"c"}

Just add a to_a and Hash[] to your array version to get a Hash version:

h = Hash[h.to_a.sample(h.length)]

For example:

>> h = { 1 => 'a', 2 => 'b', 3 => 'c' }
=> {1=>"a", 2=>"b", 3=>"c"}
>> h = Hash[h.to_a.sample(h.length)]
=> {2=>"b", 1=>"a", 3=>"c"}

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一个人的夜不怕黑 2024-12-14 22:06:02

您真的需要随机播放还是只需要一种访问/迭代随机密钥的方法？

否则，一个可能更便宜的解决方案是对散列键进行洗牌，并根据这些散列键的排列来访问您的项目，

h = your_hash
shuffled_hash_keys = hash.keys.shuffle

shuffled_hash_keys.each do |key|
  # do something with h[key]
end

我相信（但需要带有基准的证明），这避免了构建全新的需要/成本哈希，如果你有大哈希，可能会更有效（你只需要支付数组排列的成本）

Do you really need to shuffle or do you just need a way to access/iterate on a random key ?

Otherwise, a maybe less expensive solution would be to shuffle the hash keys and access your items based on the permutation of those hash keys

h = your_hash
shuffled_hash_keys = hash.keys.shuffle

shuffled_hash_keys.each do |key|
  # do something with h[key]
end

I believe (but would need a proof with a benchmark) that this avoids the need/cost to build a brand new hash and is probably more efficient if you have big hashes (you only need to pay the cost of an array permutation)

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