问个问题 'abcdefghigklmnopqrstuvwxyz'怎么切成 ['abc','def','ghi'…]这样的形式啊

[whnzy]

如题！

[seki]

http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-sized-chunks-in-python

[SErHo]

["abcdefghigklmnopqrstuvwxyz"[i: i+3] for i in range(0, len("abcdefghigklmnopqrstuvwxyz"), 3)]

[aec4d]

re.findall('.{3}','abcdefghigklmnopqrstuvwxyz')

[billion]

2 楼正解。

[hcwhan]

2 楼这样 如果不是 3 的倍数会报索引超出 错误吧

[wittyfox]

str = ('a'..'z').to_a.join
# => "abcdefghijklmnopqrstuvwxyz"

str.split('').each_slice(3).to_a.map(&:join)
=> ["abc",
"def",
"ghi",
"jkl",
"mno",
"pqr",
"stu",
"vwx",
"yz"]

[wittyfox]

如果核心库更完善的话，像这样：

```ruby
String.class_eval do
include Enumerable
alias :each :each_char
end
```

那就可以这样：

```ruby
str.each_slice(3).to_a
```

[dant]

@wittyfox 然而这里是 Python 版（

[seki]

@hcwhan 不会

[sjtlqy]

http://php.net/manual/zh/function.str-split.php
http://php.net/manual/zh/function.chunk-split.php

贴一个 PHP 版本的

[xuboying]

能用正则就用正则啊
#!python
import re
x="abcdefghijklmnopqrstuvwxyz"
print re.findall(r".{1,3}",x)

[xcodebuild]

JavaScript 版：

'abcdefghigklmnopqrstuvwxyz'.match(/.{3}/g) // => ["abc", "def", "ghi", "gkl", "mno", "pqr", "stu", "vwx"]

[ty0716]

>>> import textwrap
>>> end.join(textwrap.wrap(body, chunklen))

https://docs.python.org/2/library/textwrap.html

[SakuraSa]

以前用过一个奇怪的方法实现：

```python
s,n='abcdefghigklmnopqrstuvwxyz',3
g=iter(s)
print([''.join(i) for i in zip(*[g]*n)])
```

>>> ['abc', 'def', 'ghi', 'gkl', 'mno', 'pqr', 'stu', 'vwx']


要保留最后的 yz 的话，可以：

```python
import itertools
s,n='abcdefghigklmnopqrstuvwxyz',3
g=iter(s)
print([''.join(j for j in i if j) for i in itertools.izip_longest(*[g]*n)])
```

>>>['abc', 'def', 'ghi', 'gkl', 'mno', 'pqr', 'stu', 'vwx', 'yz']

[leavic]

from more_itertools import chunked


搞定了

[msg7086]

@wittyfox 就知道有人会贴 ruby 版 w

[araraloren]

~~ perl6 版

```perl6
#!/usr/bin/env perl6


# 3 个一组合
say (comb /\w ** 3/, ['a' ... 'z'].join);

# 输出 (abc def ghi jkl mno pqr stu vwx)


# 保留 yz
say (comb /\w ** 3 | \w ** 2/, ['a' ... 'z'].join);

# 输出 (abc def ghi jkl mno pqr stu vwx yz)
```

[liyj144]

参考 2 楼，加上末尾的处理： re.findall('.{3}|.{2}$','abcdefghigklmnopqrstuvwxyz')

[liyj144]

@liyj144 刚写错了，是三楼

[HustLiu]

@codefalling 可以优化一下 /.{1,3}/g ，这样避免字符串长度不是 3 的倍数出现遗漏

[jackal]

Javascript+正则表达式版本
<script type="text/javascript">
var str = "abcdefghigklmnopqrstuvwxyz";
var exp = /[a-z]{3}/g;
var arr = [];
var index = 0;

while ((arr = exp.exec(str)) !== null) {
console.log(arr[0]);
index = exp.lastIndex;
}
console.log(str.substring(index));
</script>

[irenicus]

perl 版
@out = ("abcdef" =~ /(...)/);

[irenicus]

那个不行
@out = split /(...)/, "abcdefghijklmnopqrstuvwxyz";
不是三的倍数也行

[xcodebuild]

@HustLiu 有道理

[extreme]

C 版：
int x = 0, y = 0;
char *string ="abcdefghigklmnopqrstuvwxyz" , three_char[9][4] = {0};
while (*string != '\0') {
three_char[x][y % 3] = *string;
y++;
if (y == 3) {
y = 0;
x++;
}
string++;
}

[saxon]

s = "abcdefghijklmnopqrstuvwxyz"
print zip(s[::3]+s[1::3]+s[2::3])
:)

[saxon]

更正，刚手滑回车
s = "abcdefghijklmnopqrstuvwxyz"
def func(s):
l = list(s)
return ''.join(l)
print map(func,zip(s[::3],s[1::3],s[2::3]))

[dofine]

def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)

[whnzy]

我想是不是应该再来点 C++,Golang,C#,OC,erlang,Java,Haskell.........