2011-01-29 45 views

回答

18

嘗試使用Range.step

> (1..19).step(3).to_a 
=> [1, 4, 7, 10, 13, 16, 19] 
4

在Ruby 1.9:

1.step(12).to_a #=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] 
1.step(12,3).to_a #=> [1, 4, 7, 10] 

或者你可以圖示的to_a代替:

a = *1.step(12,3) #=> [1, 4, 7, 10] 
+0

我真希望我能選擇多個正確答案 – NullVoxPopuli 2011-01-30 02:48:05