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This topic created in 3211 days ago, the information mentioned may be changed or developed.

‘ 7w876543217w'怎么拆分成[7w,87,65,43,21,7w]

list['7w876543217w']貌似不行

12 replies • 2017-09-13 00:41:14 +08:00

Cooky

Sep 11, 2017 via Android

[ list[i,I+2] for i in range(len(list)) ]

Cooky

Sep 11, 2017 via Android

@Cooky 第一个逗号改冒号，切片

MrgHOST

Sep 11, 2017

@Cooky 谢谢

qlbr

Sep 11, 2017

>>> import re
>>> string='7w876543217w'
>>> re.findall('.{2}',string)
['7w', '87', '65', '43', '21', '7w']

Cooky

Sep 11, 2017 via Android

@MrgHOST 直接用 str 切片就行，转 list 多余了…

nauynah

Sep 11, 2017 via iPhone

>>> [str[i: i+2] for i in range(0, 10, 2)]

acheapskate

Sep 11, 2017

s = '7w876543217w'
l = [m + n for m,n in zip(s[0::2],s[1::2])]

acheapskate

Sep 11, 2017

@nauynah range(0, 11, 2) ，不然截取不到最后 2 个字符。感觉用 range 分隔还是最明晰方便的

nauynah

Sep 11, 2017

@acheapskate 是的，我把字符串长度看成 10 了

40huo

Sep 11, 2017

[string[x : x + width] for x in range(0, len(string), width)]

yucongo

Sep 13, 2017

In [538]: [elm + s[1::2][idx] for idx, elm in enumerate(s[::2])]
Out[538]: ['7w', '87', '65', '43', '21', '7w']

yucongo

Sep 13, 2017

In [544]: s
Out[544]: '7w876543217w'

In [545]: [ s[2*idx: 2*idx+2] for idx in range(len(s)//2)]
Out[545]: ['7w', '87', '65', '43', '21', '7w']