我想通过删除所有期望的字母字符来分割字符串。
默认情况下,split
仅在单词之间按空格分隔。但是我想按所有期望的字母字符分开。如何为split
添加多个定界符?
例如:
word1 = input().lower().split()
# if you input " has 15 science@and^engineering--departments, affiliated centers, Bandar Abbas&&and Mahshahr."
#the result will be ['has', '15', 'science@and^engineering--departments,', 'affiliated', 'centers,', 'bandar', 'abbas&&and', 'mahshahr.']
但是我正在寻找这种结果:
['has', '15', 'science', 'and', 'engineering', 'departments', 'affiliated', 'centers', 'bandar', 'abbas', 'and', 'mahshahr']
答案 0 :(得分:4)
为了提高性能,您应该根据标记的重复项使用正则表达式。请参阅下面的基准测试。
您可以将itertools.groupby
与str.isalnum
结合使用,以按字母数字字符进行分组。
使用此解决方案,您不必担心会被明确指定的字符分割。
from itertools import groupby
x = " has 15 science@and^engineering--departments, affiliated centers, Bandar Abbas&&and Mahshahr."
res = [''.join(j) for i, j in groupby(x, key=str.isalnum) if i]
print(res)
['has', '15', 'science', 'and', 'engineering', 'departments',
'affiliated', 'centers', 'Bandar', 'Abbas', 'and', 'Mahshahr']
一些性能基准测试与正则表达式解决方案(在Python 3.6.5上测试):
from itertools import groupby
import re
x = " has 15 science@and^engineering--departments, affiliated centers, Bandar Abbas&&and Mahshahr."
z = x*10000
%timeit [''.join(j) for i, j in groupby(z, key=str.isalnum) if i] # 184 ms
%timeit list(filter(None, re.sub(r'\W+', ',', z).split(','))) # 82.1 ms
%timeit list(filter(None, re.split('\W+', z))) # 63.6 ms
%timeit [_ for _ in re.split(r'\W', z) if _] # 62.9 ms
答案 1 :(得分:2)
您可以将所有非字母数字字符替换为一个字符(我使用逗号)
s = 'has15science@and^engineering--departments,affiliatedcenters,bandarabbas&&andmahshahr.'
alphanumeric = re.sub(r'\W+', ',',s)
然后用逗号将其分割:
splitted = alphanumeric.split(',')
编辑:
正如@DeepSpace所建议的,这可以在单个语句中完成:
splitted = re.split('\W+', s)