有没有办法以zip
这样的方式使用 2 个数组将在 n 元素之间用空格压缩,例如:
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
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]
怎么样:
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]
迭代 b 是一种可能性:
# 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)