3

我们如何在 Ruby 中为重复元素的集合获取交集和并集。

# given the sets
a = ["A", "B", "B", "C", "D", "D"]
b = ["B", "C", "D", "D", "D", "E"]

# A union function that adds repetitions
union(a, b)
=> ["A", "B", "B", "C", "D", "D", "D", "E"]

# An intersection function that adds repetitions
intersection(a, b)
=> ["B", "C", "D", "D"]

和运算&符|似乎忽略了重复和重复，如文档中所述。

# union without duplicates
a | b
=> ["A", "B", "C", "D", "E"]

# intersections without duplicates
a & b
=> ["B", "C", "D"]
4

2 回答 2

2 
def union(a,b)
  (a|b).flat_map { |s| [s]*[a.count(s), b.count(s)].max }
end

union(a,b)
  # => ["A", "B", "B", "C", "D", "D", "D", "E"] 

def intersection(a,b)
  (a|b).flat_map { |s| [s]*[a.count(s), b.count(s)].min }
end

intersection(a,b)
  #=> ["B", "C", "D", "D"]
于 2016-06-14T02:31:06.970 回答
1

基于Cary Swoveland 的回答，您可以创建一个临时哈希来计算每个数组中每个成员的出现次数：（我已经概括了参数的数量）

def multiplicities(*arrays)
  m = Hash.new { |h, k| h[k] = Array.new(arrays.size, 0) }
  arrays.each_with_index { |ary, idx| ary.each { |x| m[x][idx] += 1 } }
  m
end

multiplicities(a, b)
#=> {"A"=>[1, 0], "B"=>[2, 1], "C"=>[1, 1], "D"=>[2, 3], "E"=>[0, 1]}

实施union并且intersection是直截了当的：

def union(*arrays)
  multiplicities(*arrays).flat_map { |x, m| Array.new(m.max, x) }
end

def intersection(*arrays)
  multiplicities(*arrays).flat_map { |x, m| Array.new(m.min, x) }
end

union(a, b)        #=> ["A", "B", "B", "C", "D", "D", "D", "E"]
intersection(a, b) #=> ["B", "C", "D", "D"]

使用这种方法，每个数组只需要遍历一次。

于 2016-06-15T13:32:28.590 回答