比较包括重复的数组元素 [英] Comparing array elements including duplicates

问题描述

我正在尝试查看一个数组是否包含另一个数组的每个元素。另外我想说明重复的事项。例如：

I am trying to see if an array contains each element of another array. Plus I want to account for the duplicates. For example:

array = [1, 2, 3, 3, "abc", "de", "f"]

数组包含[1，2，3，3]，但不包含[2，2， abc] - 太多2的

array contains [1, 2, 3, 3] but does not contain [2, 2, "abc"] - too many 2's

我已经尝试了下面，但显然没有考虑到这些重复。

I have tried the below but obviously doesn't take into account the dupes.

other_arrays.each { |i| array.include? i }

推荐答案

此方法在两个数组上迭代一次。
对于每个数组，它创建一个散列，每个元素的出现次数。

This method iterates once over both arrays. For each array, it creates a hash with the number of occurences of each element.

然后检查子集， superset中至少有一些元素。

class Array
  def count_by
    each_with_object(Hash.new(0)) { |e, h| h[e] += 1 }
  end

  def subset_of?(superset)
    superset_counts = superset.count_by
    count_by.all? { |k, count| superset_counts[k] >= count }
  end
end

[1, 2, 3, 3, "abc", "de", "f"].count_by
#=> {1=>1, 2=>1, 3=>2, "abc"=>1, "de"=>1, "f"=>1}

[1, 2, 3, 3].count_by
#=> {1=>1, 2=>1, 3=>2}

[1, 2, 3, 3].subset_of? [1, 2, 3, 3, "abc", "de", "f"]
#=> true
[2, 2, "abc"].subset_of? [1, 2, 3, 3, "abc", "de", "f"]
#=> false

如果您不想修补 Array class可以定义：

If you don't want to patch the Array class, you could define :

count_by（array）和 subset_of？ （array1，array2）。

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