Question

我怎样才能拍摄像'aaaaaaaaaaaaaaaaaaaaaaa'这样的字符串并将其拆分为4个长度元组，如（aaaa，aaaa，aaaa）

Answer 1

使用textwrap.wrap：

>>> import textwrap
>>> s = 'aaaaaaaaaaaaaaaaaaaaaaa'
>>> textwrap.wrap(s, 4)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa']

Answer 2

使用列表理解，生成器表达式：

>>> s = 'aaaaaaaaaaaaaaaaaaaaaaa'
>>> [s[i:i+4] for i in range(0, len(s), 4)]
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa']

>>> tuple(s[i:i+4] for i in range(0, len(s), 4))
('aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaa')

>>> s = 'a bcdefghi j'
>>> tuple(s[i:i+4] for i in range(0, len(s), 4))
('a bc', 'defg', 'hi j')

Answer 3

使用正则表达式的另一种解决方案：

>>> s = 'aaaaaaaaaaaaaaaaaaaaaaa'
>>> import re
>>> re.findall('[a-z]{4}', s)
['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']
>>>

Answer 4

您可以使用grouper recipe，zip(*[iter(s)]*4)：

In [113]: s = 'aaaaaaaaaaaaaaaaaaaaaaa'

In [114]: [''.join(item) for item in zip(*[iter(s)]*4)]
Out[114]: ['aaaa', 'aaaa', 'aaaa', 'aaaa', 'aaaa']

请注意，如果字符串包含空格，textwrap.wrap可能不会将s拆分为长度为4的字符串：

In [43]: textwrap.wrap('I am a hat', 4)
Out[43]: ['I am', 'a', 'hat']

石斑鱼配方比使用textwrap更快：

In [115]: import textwrap

In [116]: %timeit [''.join(item) for item in zip(*[iter(s)]*4)]
100000 loops, best of 3: 2.41 µs per loop

In [117]: %timeit textwrap.wrap(s, 4)
10000 loops, best of 3: 32.5 µs per loop

石斑鱼配方可以使用任何迭代器，而textwrap仅适用于字符串。

Answer 5

s = 'abcdef'

我们需要分成两部分

[s[pos:pos+2] for pos,i in enumerate(list(s)) if pos%2 == 0]

答案：

['ab', 'cd', 'ef']

Answer 6

s = 'abcdefghi'

k-字符串的任何部分

k = 3

parts-列表以存储字符串的一部分

parts = [s[i:i+k] for i in range(0, len(s), k)]

parts-> ['abc'，'def'，'ghi']

Answer 7

我认为这种方法更简单。但是消息长度必须用split_size分割。或必须在邮件中添加字母。例如：message =“ lorem ipsum_”，则可以删除添加的字母。

message = "lorem ipsum"

array = []

temp = ""

split_size = 3

for i in range(1, len(message) + 1):
    temp += message[i - 1]

    if i % split_size == 0:
        array.append(temp)
        temp = ""

print(array)

输出： ['lor'，'em'，'ips']

Answer 8

这是给定问题的另一种可能的解决方案：

def split_by_length(text, width):
    width = max(1, width)
    chunk = ""
    for v in text:
        chunk += v
        if len(chunk) == width:
            yield chunk
            chunk = ""

    if chunk:
        yield chunk

if __name__ == '__main__':
    x = "123456789"
    for i in range(20):
        print(i, list(split_by_length(x, i)))

输出：

0 ['1', '2', '3', '4', '5', '6', '7', '8', '9']
1 ['1', '2', '3', '4', '5', '6', '7', '8', '9']
2 ['12', '34', '56', '78', '9']
3 ['123', '456', '789']
4 ['1234', '5678', '9']
5 ['12345', '6789']
6 ['123456', '789']
7 ['1234567', '89']
8 ['12345678', '9']
9 ['123456789']
10 ['123456789']
11 ['123456789']
12 ['123456789']
13 ['123456789']
14 ['123456789']
15 ['123456789']
16 ['123456789']
17 ['123456789']
18 ['123456789']
19 ['123456789']

Answer 9

孩子的方式

def wrap(string, max_width):
    i=0
    strings = []
    s = ""
    for x in string:
        i+=1
        if i == max_width:
            s = s + x
            strings.append(s)
            s = ""
            i = 0
        else:
            s = s + x
    strings.append(s)
    return strings

wrap('ABCDEFGHIJKLIMNOQRSTUVWXYZ',4)
# output: ['ABCD', 'EFGH', 'IJKL', 'IMNO', 'QRST', 'UVWX', 'YZ']

将字符串拆分为偶数大小的块

9 个答案: