Ruby 数组减法而不多次删除项目 [英] Ruby array subtraction without removing items more than once

问题描述

Ruby 中的规范数组差异示例是:

The canonical Array difference example in Ruby is:

[ 1, 1, 2, 2, 3, 3, 4, 5 ] - [ 1, 2, 4 ]  #=>  [ 3, 3, 5 ]

获得以下行为的最佳方法是什么?

What's the best way to get the following behavior instead?

[ 1, 1, 2, 2, 3, 3, 4, 5 ].subtract_once([ 1, 2, 4 ])  #=>  [ 1, 2, 3, 3, 5 ]

也就是说，只从第一个数组中删除第二个数组中每个匹配项的第一个实例.

That is, only the first instance of each matching item in the second array is removed from the first array.

推荐答案

减去在另一个数组中出现的次数，或者任何 可枚举:

Subtract values as many times as they appear in the other array, or any Enumerable:

class Array
  # Subtract each passed value once:
  #   %w(1 2 3 1).subtract_once %w(1 1 2) # => ["3"]
  #   [ 1, 1, 2, 2, 3, 3, 4, 5 ].subtract_once([ 1, 2, 4 ]) => [1, 2, 3, 3, 5]
  # Time complexity of O(n + m)
  def subtract_once(values)
    counts = values.inject(Hash.new(0)) { |h, v| h[v] += 1; h }
    reject { |e| counts[e] -= 1 unless counts[e].zero? }
  end

将每个唯一值减去一次:

require 'set'
class Array
  # Subtract each unique value once:
  #   %w(1 2 2).subtract_once_uniq %w(1 2 2) # => [2]
  # Time complexity of O((n + m) * log m)
  def subtract_once_uniq(values)
    # note that set is implemented 
    values_set = Set.new values.to_a 
    reject { |e| values_set.delete(e) if values_set.include?(e) }
  end
end

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