Ruby每周一测 - 发牌趣题擂台

虎炮你的代码有bug，每次shuffle出来的数组都不一样的

(1..4).map{|i| (1..52).to_a.shuffle.indexes(i..i + 13)}

这个给力，给你加了3票

Recipe 4.10. Shuffling an Array
Problem

You want to put the elements of an array in random order.
Solution

The simplest way to shuffle an array (in Ruby 1.8 and above) is to sort it randomly:

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


This is not the fastest way, though.
Discussion

It's hard to beat a random sort for brevity of code, but it does a lot of extra work. Like any general sort, a random sort will do about n log n variable swaps. But to shuffle a list, it suffices to put a randomly selected element in each position of the list. This can be done with only n variable swaps.

class Array
  def shuffle! 
    each_index do |i|
      j = rand(length-i) + i
      self[j], self[i] = self[i], self[j] 
    end
  end

  def shuffle 
    dup.shuffle! 
  end
end


If you're shuffling a very large list, either Array#shuffle or Array#shuffle! will be significantly faster than a random sort. Here's a real-world example of shuffling using Array#shuffle:

class Card
  def initialize(suit, rank)
    @suit = suit 
    @rank = rank 
  end

  def to_s
    "#{@suit} of #{@rank}"
  end
end

class Deck < Array
  attr_reader :cards 
  @@suits = %w{Spades Hearts Clubs Diamonds}
  @@ranks = %w{Ace 2 3 4 5 6 7 8 9 10 Jack Queen King}

  def initialize
    @@suits.each { |suit| @@ranks.each { |rank| self << Card.new(rank, suit) } }
  end
end

deck = Deck.new
deck.collect { |card| card.to_s }
# => ["Ace of Spades", "2 of Spades", "3 of Spades", "4 of Spades",…]
deck.shuffle!
deck.collect { |card| card.to_s }
# => ["6 of Clubs", "8 of Diamonds", "2 of Hearts", "5 of Clubs",…]


See Also

    *

      Recipe 2.5, "Generating Random Numbers"
    *

      The Facets Core library provides implementations of Array#shuffle and Array#shuffle!

(1..52).to_a.shuffle.each_slice(13).map

引用的回复已经非常短了. 我蛋疼的提供另一种思路...

(1..52).to_a.shuffle.each_slice(4).inject{|r,x| r.zip x}.each{|s| s.flatten!}

从一次分好到. 分好后再zip..    蛋疼无极限.

如果没要求格式的话. 可以不需要最后面的 each .. zip一次其实也分好了. 嗯..

 
(1..52).to_a.shuffle.each_slice(13).map