荏苒2阵列中的红宝石在随机位置 [英] Zipping 2 arrays in ruby in random locations

问题描述

有没有一种简单的方法来压缩在随机位置2阵列和保持其原有的秩序

例如：

  A = [0,1,2,3,4,5,6,7,8,9,10]
B = [一，二，三，四]
 

和从0到5的随机数与兰特（5）

 压缩= [0，一，1,2,3，两节; 4，三，5,6,7,8，四个一，9 10]
 

和随机系列将 1,3,1,4 的位置在哪里ZIPB的每个元素成

我能做到的最好的是

  I = 0
合并=一
b.each做| X |
RND =兰特（5）
merged.insert（RND我+ X）
I = I + RND
结束
 

解决方案

这个版本将提供一个平衡的洗牌，以插入不偏向阵列的两端。

  DEF ordered_random_merge（A，B）
  A，B = a.dup，b.dup
  。a.map {兰特（b.size + 1）} sort.reverse.each办|首页|
    b.insert（索引，a.pop）
  结束
  b
结束
 

Is there an easy way to zip 2 arrays in random locations and keep their original order?

for example

a=[0,1,2,3,4,5,6,7,8,9,10]


b=["one","two","three","four"]

and a random number from 0 to 5 with rand(5)

zipped = [0,"one",1,2,3,"two",4,"three",5,6,7,8,"four",9,10]

and the random series would be 1,3,1,4 as location to where to "zip" each element of b into a

The best I could do is

i=0
merged=a
b.each do |x|
rnd = rand(5)
merged.insert(i+rnd,x)
i=i+rnd
end

解决方案

This version will give a balanced shuffling, with insertions not biased to either end of the array.

def ordered_random_merge(a,b)
  a, b = a.dup, b.dup
  a.map{rand(b.size+1)}.sort.reverse.each do |index|
    b.insert(index, a.pop)
  end
  b
end

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