Ruby，如何将一个数组混合到另一个数组中
两个arrays：
a1 = ["a", "b", "c", "d", "e", "f"] a2 = [1, 2, 3] 如何将a2插入a1 ，保持a2顺序但是在a1的 随机索引中？
(0..a1.length).to_a.sample(a2.length).sort .zip(a2) .reverse .each{|i, e| a1.insert(i, e)}
这是我更新的答案：
a1 = ["a", "b", "c", "d", "e", "f"] a2 = [1,2,3] # scales to N arrays by just adding to this hash h = { :a1 => a1.dup, :a2 => a2.dup } # => {:a1=>["a", "b", "c", "d", "e", "f"], :a2=>[1, 2, 3]} # Create an array of size a1+a2 with elements representing which array to pull from sources = h.inject([]) { |s,(k,v)| s += [k] * v.size } # => [:a1, :a1, :a1, :a1, :a1, :a1, :a2, :a2, :a2] # Pull from the array indicated by the hash after shuffling the source list sources.shuffle.map { |a| h[a].shift } # => ["a", "b", 1, "c", 2, "d", "e", 3, "f"]
算法归功于我的同事瑞安。
老答案不保留两种顺序
a1.inject(a2) { |s,i| s.insert(rand(s.size), i) }
使用a2作为目标，在a2的随机偏移量处从a1插入a2中的每个值。
通过模拟真实的混洗来维持两个数组的顺序，一旦数组的元素插入到另一个数组中，下一个元素就不能放在它之前。
class Array def shuffle_into(array) n = 0 self.each.with_object(array.dup) do |e, obj| i = rand(n..obj.size) obj.insert(i, e) n = i + 1 end end end
可能能够清理周围漂浮的n = 0 。
示例： a2.shuffle_into(a1) => [1, "a", "b", "c", "d", 2, "e", "f", 3]
这个丑陋的垃圾完成了这项工作（没有搞乱任何数组顺序）：
class Array def shuffle_into(ary) a1 = ary.dup a2 = dup Array.new(a1.size + a2.size) do [true, false].sample ? (a1.shift || a2.shift) : (a2.shift || a1.shift) end end end
a1.zip((a2 + [nil] * (a1.size - a2.size)).shuffle).flatten.compact
BTW，可能的重复： 在随机位置的ruby中压缩2个数组