Find all occurrences of a substring in Python

Question

Python has string.find() and string.rfind() to get the index of a substring in string.

I wonder, maybe there is something like string.find_all() which can return all founded indexes (not only first from beginning or first from end)?

For example:

string = "test test test test"

print string.find('test') # 0
print string.rfind('test') # 15

#that's the goal
print string.find_all('test') # [0,5,10,15]

Answer 1

This does the trick for me using re.finditer

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()))

Answer 2

This thread is a little old but this worked for me:

numberString = "onetwothreefourfivesixseveneightninefiveten"
testString = "five"

marker = 0
while marker < len(numberString):
    try:
        print(numberString.index("five",marker))
        marker = numberString.index("five", marker) + 1
    except ValueError:
        print("String not found")
        marker = len(numberString)

Answer 3

>>> help(str.find)
Help on method_descriptor:

find(...)
    S.find(sub [,start [,end]]) -> int

Thus, we can build it ourselves:

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]

No temporary strings or regexes required.

Answer 4

There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:

import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]

If you want to find overlapping matches, lookahead will do that:

[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]

If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:

search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]

re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.