紅寶石數組訪問連續2個（鏈接）元素在時間

Question

現在，這是一個數組，

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

我想，

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

當我這樣做，each_slice（2）我得到的，

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

即時通訊目前這樣做，

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

有沒有更好的方法？

Answer 1

紅寶石讀你的心。你想要缺點行政元素？

[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確實如你所願。

[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

你幾乎得到它吧:)

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

如果你想實現自己的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 

輸出：

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