Python有string.find()和string.rfind()来获取字符串中子字符串的索引。
我想知道是否有像string.find_all()这样的东西可以返回所有找到的索引(不仅是从开始的第一个索引，还是从结束的第一个索引)。
例如:
string = "test test test test"
print string.find('test') # 0
print string.rfind('test') # 15
#this is the goal
print string.find_all('test') # [0,5,10,15]
要统计出现次数，请参见计算字符串中子字符串出现的次数。
当前回答
如果您只想使用numpy，这里是一个解决方案
import numpy as np
S= "test test test test"
S2 = 'test'
inds = np.cumsum([len(k)+len(S2) for k in S.split(S2)[:-1]])- len(S2)
print(inds)
其他回答
下面是我想出的一个解决方案，使用赋值表达式(Python 3.8以来的新特性):
string = "test test test test"
phrase = "test"
start = -1
result = [(start := string.find(phrase, start + 1)) for _ in range(string.count(phrase))]
输出:
[0, 5, 10, 15]
>>> help(str.find)
Help on method_descriptor:
find(...)
S.find(sub [,start [,end]]) -> int
因此，我们可以自己构建它:
def find_all(a_str, sub):
start = 0
while True:
start = a_str.find(sub, start)
if start == -1: return
yield start
start += len(sub) # use start += 1 to find overlapping matches
list(find_all('spam spam spam spam', 'spam')) # [0, 5, 10, 15]
不需要临时字符串或正则表达式。
对于非重叠匹配，可以使用re.finditer()。
>>> import re
>>> aString = 'this is a string where the substring "is" is repeated several times'
>>> print [(a.start(), a.end()) for a in list(re.finditer('is', aString))]
[(2, 4), (5, 7), (38, 40), (42, 44)]
但不适用于:
In [1]: aString="ababa"
In [2]: print [(a.start(), a.end()) for a in list(re.finditer('aba', aString))]
Output: [(0, 3)]
请看看下面的代码
#!/usr/bin/env python
# coding:utf-8
'''黄哥Python'''
def get_substring_indices(text, s):
result = [i for i in range(len(text)) if text.startswith(s, i)]
return result
if __name__ == '__main__':
text = "How much wood would a wood chuck chuck if a wood chuck could chuck wood?"
s = 'wood'
print get_substring_indices(text, s)
你可以试试:
import re
str1 = "This dress looks good; you have good taste in clothes."
substr = "good"
result = [_.start() for _ in re.finditer(substr, str1)]
# result = [17, 32]
