从字符串获取n个字符

Question

我想找到方法遍历字符串并获取每个n个字符。结果应该类似于each_cons方法中的数组。

你现在关于一些现有的方法或如何实现自己的方法？

Answer 1

使用String#each_char：

"abcdef".each_char.each_cons(3) { |a| p a } 
["a", "b", "c"] 
["b", "c", "d"] 
["c", "d", "e"] 
["d", "e", "f"] 
=> nil 

Answer 2

补充对方的回答。

"abcdef".each_char.to_a.in_groups_of(3) 

Answer 3

如果你想长的3 “连续串”：

r =/
    (?=  # begin a positive lookahead 
     ((.{3}) # match any three characters in capture group 1 
    )   # close the positive lookahead 
    /x  # free-spacing regex definition mode 

"abcdef".scan(r).flatten 
    #=> ["abc", "bcd", "cde", "def"] 

写在传统的方式，这正则表达式是：

r = /(?=(.{3}))/ 

如果你想的数组的数组三个字母，这样做：

"abcdef".scan(/(?=(.{3}))/).flatten.map { |s| s.split('') } 
    #=> [["a", "b", "c"], ["b", "c", "d"], ["c", "d", "e"], ["d", "e", "f"]]