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

当在一份文件中寻找大量的关键词时,使用flash文本

from flashtext import KeywordProcessor
words = ['test', 'exam', 'quiz']
txt = 'this is a test'
kwp = KeywordProcessor()
kwp.add_keywords_from_list(words)
result = kwp.extract_keywords(txt, span_info=True)

在大量搜索词列表上,Flashtext比正则表达式运行得更快。

其他回答

你可以试试:

>>> string = "test test test test"
>>> for index,value in enumerate(string):
    if string[index:index+(len("test"))] == "test":
        print index

0
5
10
15

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”)

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]

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()))