使用 Ruby，检查哈希中的任何键是否与数组中的任何值匹配的最有效方法是什么

发布于 2024-08-31 11:37:29 字数 292 浏览 5 评论 0原文

我想将参数哈希中的键与元素数组进行比较以进行匹配。

例如：

params          = {"key1", "key2", "key3"}
params_to_match = ["key2","key3"]

我可以这样做，但我确信有一种更优雅的方法可以达到相同的结果

params.each_key{|key|
  if params_to_match.include?(key.to_s)
    return
  end
}

原文

I want to compare the keys in a hash of parameters against an array of elements for a match.

For example:

params          = {"key1", "key2", "key3"}
params_to_match = ["key2","key3"]

I could do this, but I'm sure there is a much more elegant way to acheive the same result

params.each_key{|key|
  if params_to_match.include?(key.to_s)
    return
  end
}

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忆伤 2024-09-07 11:37:30

不一定更高效，但在某种意义上可能更优雅：

return unless (params.keys & params_to_match).empty?

比您的示例更有效的方式（在一般情况下，不一定是这样一个小玩具示例）检查哈希是否包含键，因为从数组中查找它们时查找这些键的时间实际上是恒定的，为 O(n)。所以，你的例子会变成这样：

params_to_match.each { |p| return if params.has_key?(p) }

Not necessarily more efficient but perhaps more elegant in some sense:

return unless (params.keys & params_to_match).empty?

A more efficient way than your example would (in the general case, not necessarily with such a small toy example) be to check whether the hash contains the keys, since the time to look those up is practically constant while looking them up from the array is O(n). So, your example would become something like this:

params_to_match.each { |p| return if params.has_key?(p) }

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默嘫て 2024-09-07 11:37:30

使用 `&`

Set 交集 - 返回一个新数组，其中包含以下元素的公共元素：两个数组，没有重复项。

[ 1, 1, 3, 5 ] & [ 1, 2, 3 ]   #=> [ 1, 3 ]

params.keys & params_to_match  #=> ["key2", "key3"]

Use `&`

Set Intersection—Returns a new array containing elements common to the two arrays, with no duplicates.

[ 1, 1, 3, 5 ] & [ 1, 2, 3 ]   #=> [ 1, 3 ]

params.keys & params_to_match  #=> ["key2", "key3"]

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挽你眉间 2024-09-07 11:37:30

我认为优雅和高效的最佳结合是

return if params_to_match.any? { |p| params.has_key?(p) }

如果你有 ActiveSupport，你可以这样做

return if params.slice(*params_to_match).any?

I think the best combination of elegant and efficient would be

return if params_to_match.any? { |p| params.has_key?(p) }

If you have ActiveSupport, you could do

return if params.slice(*params_to_match).any?

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憧憬巴黎街头的黎明 2024-09-07 11:37:30

这是伪代码中的快速算法。

assert keys, input are sorted
pos = beginning of keys
for i in input
  pos = keys.find(pos, i)  # find() starts search at position pos
  if not_found(pos) then return false
  ++pos
return true

该算法假设键和输入一开始就已排序，执行时间为 O(n+m)，n 和 m 是键和输入的计数。我将让您将其翻译为 Ruby，但请密切注意 find() 函数；它必须从上一次迭代中找到的位置开始搜索，而不是在键的开头。否则你的键数就会减少到 O(n^2+m), n 。

Here's a fast algorithm in psedudocode.

assert keys, input are sorted
pos = beginning of keys
for i in input
  pos = keys.find(pos, i)  # find() starts search at position pos
  if not_found(pos) then return false
  ++pos
return true

This algorithm, assuming the keys and the input are sorted to begin with, performs in O(n+m), n and m being the count of keys and input. I'll leave it for you to translate that to Ruby, but pay close attention to the find() function; it must start its search at the position found in the previous iteration, not at the start of keys. Otherwise you're down to O(n^2+m), n the count of keys.

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