什么是Python的Ruby中itertools.product相当于？

问题描述

我在寻找一个具有如Python的 itertools.product 在Ruby中同样效果的方法。看看下面的Python code：

I'm looking for a method that has the same effect as Python's itertools.product in Ruby. Take the following Python code:

from itertools import product

chars = []
for i in range(97,123):
    chars.append(chr(i))

for a in range(1,3):
    for i in product(chars,repeat=a):
        s = ''.join(i)
        print s

这输出是这样的：

a, b, c... x, y, z, aa, ab, ac... ax, ay, az, ba, bb, bc.. etc.

我试着翻译成红宝石：

I tried to translate that into Ruby:

(1..2).each do |n|
  ('a'..'z').to_a.combination(n).each do |c|
    s = c.join
    puts s
  end
end

但输出是不一样的。单字符那些做工精细（A-Z），但是当它进入两个字符的人，如我所料不工作：

But the output isn't the same. The one-character ones work fine (a-z), but when it goes into two-character ones, it doesn't work as I expected:

ab, ac, ad.. ax, ay, az, bc, bd, be

它不产生 AA ， BA 或 BB - 因此它看起来它生成所有组合，而无需重复字符或什么

It isn't generating aa, ba or bb - so it appears it's generating all the combinations without repeating characters or something?

所以，我应该用什么方法来生成的所有的如 itertools.product 组合确实在Python？

So what method should I use to generate all the combinations like itertools.product does in Python?

推荐答案

我会写（简化为3个元素，Ruby 1.9的要求）：

I would write (simplified for 3 elements, Ruby 1.9 required):

xs = ["a", "b", "c"]
strings = 1.upto(xs.size).flat_map do |n|
  xs.repeated_permutation(n).map(&:join)
end
#=> ["a", "b", "c", "aa", "ab", "ac", ...,  "cca", "ccb", "ccc"]

一个懒惰的解决方案：你可以很容易地将其与写每个的S代替地图 S，但让我们检查懒在Ruby 2.0：

A lazy solution: you could easily write it with eachs instead of maps, but let's check "lazy" from Ruby 2.0:

xs = ("a".."z").to_a
strings = 1.upto(xs.size).lazy.flat_map do |n|
  xs.repeated_permutation(n).lazy.map(&:join)
end

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