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

python的方法是:

mystring = 'Hello World, this should work!'
find_all = lambda c,s: [x for x in range(c.find(s), len(c)) if c[x] == s]

# s represents the search string
# c represents the character string

find_all(mystring,'o')    # will return all positions of 'o'

[4, 7, 20, 26] 
>>> 

其他回答

这是我使用re.finditer的技巧

import re

text = 'This is sample text to test if this pythonic '\
       'program can serve as an indexing platform for '\
       'finding words in a paragraph. It can give '\
       'values as to where the word is located with the '\
       'different examples as stated'

#  find all occurances of the word 'as' in the above text

find_the_word = re.finditer('as', text)

for match in find_the_word:
    print('start {}, end {}, search string \'{}\''.
          format(match.start(), match.end(), match.group()))
def count_substring(string, sub_string):
    c=0
    for i in range(0,len(string)-2):
        if string[i:i+len(sub_string)] == sub_string:
            c+=1
    return c

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()
    
    count = count_substring(string, sub_string)
    print(count)
>>> 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)]

其他人提供的解决方案完全基于可用的find()方法或任何可用的方法。

找出a的所有出现点的核心基本算法是什么 字符串中的子字符串?

def find_all(string,substring):
    """
    Function: Returning all the index of substring in a string
    Arguments: String and the search string
    Return:Returning a list
    """
    length = len(substring)
    c=0
    indexes = []
    while c < len(string):
        if string[c:c+length] == substring:
            indexes.append(c)
        c=c+1
    return indexes

你也可以继承str类到新的类,并可以使用这个函数 在下面。

class newstr(str):
def find_all(string,substring):
    """
    Function: Returning all the index of substring in a string
    Arguments: String and the search string
    Return:Returning a list
    """
    length = len(substring)
    c=0
    indexes = []
    while c < len(string):
        if string[c:c+length] == substring:
            indexes.append(c)
        c=c+1
    return indexes

调用方法

newstr。find_all('你觉得这个答案有用吗?然后upvote 这个!”、“这”)