Ruby array access 2 consecutive(chained) elements

问题:

Now, This is the array,

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

I want,

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

When I do, each_slice(2) I get,

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

Im currently doing this,

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

Is there a better way??

回答1:

Ruby reads your mind. You want cons ecutive elements?

[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]]

回答2:

.each_cons does exactly what you want.

[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]]

回答3:

You almost got it right :)

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]

回答4:

And if you wanted to implement your own 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

Output:

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