Question

现在，这是数组，

[1,2,3,4,5,6,7,8,9]

我想，

[1,2],[2,3],[3,4] upto [8,9]

当我这样做时，我得到each_slice（2），

[[1,2],[3,4]..[8,9]]

我目前正在这样做，

arr.each_with_index do |i,j|
  p [i,arr[j+1]].compact #During your arr.size is a odd number, remove nil.
end

有更好的方法吗？

Answer 1

Ruby读你的想法。您想要缺点执行元素吗？

[1, 2, 3, 4, 5, 6, 7, 8, 9].each_cons(2).to_a
# => [[1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9]]

Answer 2

.each_cons完全符合您的要求。

[1] pry(main)> a = [1,2,3,4,5,6,7,8,9]
=> [1, 2, 3, 4, 5, 6, 7, 8, 9]
[2] pry(main)> a.each_cons(2).to_a
=> [[1, 2], [2, 3], [3, 4], [4, 5], [5, 6], [6, 7], [7, 8], [8, 9]]

Answer 3

你几乎做对了：）

arr = [1,2,3,4,5,6,7,8,9]
arr.each_cons(2) do |chunk|
  p chunk
end
# >> [1, 2]
# >> [2, 3]
# >> [3, 4]
# >> [4, 5]
# >> [5, 6]
# >> [6, 7]
# >> [7, 8]
# >> [8, 9]

Answer 4

如果你想实现自己的each_cons：

arr = [1, 2, 3, 4, 5, 6, 7, 8, 9]
cons = 2

0.upto(arr.size - cons) do |i|
  p arr[i, cons]
end

输出：

[1, 2]
[2, 3]
[3, 4]
[4, 5]
[5, 6]
[6, 7]
[7, 8]
[8, 9]

Ruby数组一次访问2个连续（链接）元素

4 个答案: