现在,这是数组,
[1,2,3,4,5,6,7,8,9]
我想要,
[1,2],[2,3],[3,4] upto [8,9]
当我这样做时,each_slice(2)我得到,
[[1,2],[3,4]..[8,9]]
我现在正在做这个,
arr.each_with_index do |i,j|
p [i,arr[j+1]].compact #During your arr.size is a odd number, remove nil.
end
有没有更好的办法?
发布于 2013-03-28 20:49:18
.each_cons
做的正是您想要的。
[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]]
发布于 2013-03-28 20:49:48
您几乎做对了:)
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]
发布于 2018-02-16 20:09:22
如果你想实现你自己的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]
https://stackoverflow.com/questions/15682223
复制相似问题