与每第n个元件正在压缩阵列

Question

是否有这样一种方式，2个阵列将用正元件之间的空间，例如将拉链使用zip一种方式：

a = [1,2,3,4,5,6,7,8,9,10] 
b = ["x","y","z"] 
n = 3 

结果将是

res = [[1,"x"],2,3,[4,"y"],5,6,[7,"z"],8,9,10] # note that 10 is alone and b is not cycled 

Answer 1

我会写：

res = a.each_slice(n).zip(b).flat_map do |xs, y| 
    y ? [[xs.first, y], *xs.drop(1)] : xs 
end 
#=> [[1, "x"], 2, 3, [4, "y"], 5, 6, [7, "z"], 8, 9, 10] 

Answer 2

如何：

a.map.with_index{|x, i| i%n < 1 && b.size > i/n ? [x, b[i/n]] : x} 
#=> [[1, "x"], 2, 3, [4, "y"], 5, 6, [7, "z"], 8, 9, 10] 

Answer 3

遍历b为一种可能性：

# Note this destroys array a;use a dup it if it is needed elsewhere 
res = b.flat_map{|el| [[el].unshift(a.shift), *a.shift(n-1)] }.concat(a)