python - 用于从 Python 到 Ruby 查找集合的所有分区的翻译函数

Question

我有以下 python 函数来递归查找集合的所有分区:

def partitions(set_):
    if not set_:
        yield []
        return
    for i in xrange(2**len(set_)/2):
        parts = [set(), set()]
        for item in set_:
            parts[i&1].add(item)
            i >>= 1
        for b in partitions(parts[1]):
            yield [parts[0]]+b

for p in partitions(["a", "b", "c", "d"]):
print(p)

有人可以帮我把它翻译成 ruby​​ 吗？这是我目前所拥有的:

def partitions(set)
  if not set
    yield []
    return
  end
  (0...2**set.size/2).each { |i|
    parts = [Set.new, Set.new]
    set.each { |item|
      parts[i&1] << item
      i >>= 1
    }
    partitions(parts[1]).each { |b|
      yield [parts[0]] << b
    }
  }
end

p partitions([1, 2, 3, 4].to_set)

我收到错误“LocalJumpError:没有给出 block ”。我猜这是因为 yield 函数在 Python 和 Ruby 中的工作方式不同。

Answer 1

#!/usr/bin/ruby1.8

def partitions(set)
  yield [] if set.empty?
  (0 ... 2 ** set.size / 2).each do |i|
    parts = [[], []]
    set.each do |item|
      parts[i & 1] << item
      i >>= 1
    end
    partitions(parts[1]) do |b|
      result = [parts[0]] + b
      result = result.reject do |e|
        e.empty?
      end
      yield result
    end
  end
end

partitions([1, 2, 3, 4]) do |e|
  p e
end

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

有什么不同:

• 守卫调用 set.empty？代替(隐含地)测试 set.nil？
• 调用时省略 .each分区
• 使用数组代替集合
• 从 yielded 中过滤空集结果