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
试试这个，对我有用!
x=input('enter the string')
y=input('enter the substring')
z,r=x.find(y),x.rfind(y)
while z!=r:
print(z,r,end=' ')
z=z+len(y)
r=r-len(y)
z,r=x.find(y,z,r),x.rfind(y,z,r)
其他回答
查找给定字符串中某个字符的所有出现次数，并作为字典返回 例如:你好 结果: {'h':1， 'e':1， 'l':2， 'o':1}
def count(string):
result = {}
if(string):
for i in string:
result[i] = string.count(i)
return result
return {}
否则你就像这样
from collections import Counter
def count(string):
return Counter(string)
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]
>>>
如果你想使用没有re(regex)，那么:
find_all = lambda _str,_w : [ i for i in range(len(_str)) if _str.startswith(_w,i) ]
string = "test test test test"
print( find_all(string, 'test') ) # >>> [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)]
