Finding a string multiple times in another String - Python

Question

I\'m trying to see if a string exists in another string with out using Python\'s predefined functions such as find and index..

Right now what my function takes 2 strings

Answer 1

@Jacob, I hope you'll find this one very short yet still easy to understand.

def multi_find(s, r):
    return [pos for pos in range(len(s)) if s.startswith(r,pos)]

Answer 2

You can do:

>>> haystack = "abcdefabc. asdli! ndsf acba saa abe?"
>>> needle = "abc"
>>> for i, _ in enumerate(haystack):
...     if haystack[i:i + len(needle)] == needle:
...         print (i)
...
0
6

Answer 3

Note: I think this answer here is still a good "teaching answer", I have submitted a better solution elsewhere in this thread, without recursion.

def multi_find(s, r, start=0):
    if start >= len(s): 
        return []
    if s.startswith(r, start):
        return [start] + multi_find(s, r, start+1)
    else:
        return multi_find(s, r, start+1)

This allows you to pass an optional start position to begin the search in s.

This solution is recursive, which may or may not be the fastest implementation, but it is correct and I believe it makes the code easy to identify each of the three possibilities at each position of s:

1. end of s
2. found another r
3. didn't find another r

Answer 4

def multi_find (s, r):
    s_len = len(s)
    r_len = len(r)
    n = [] # assume r is not yet found in s

    if s_len >= r_len:
        m = s_len - r_len
        i = 0

        while i < m:
            # search for r in s until not enough characters are left
            if s[i:i + r_len] == r:
                n.append(i)
            i = i + 1
    print (n)

multi_find("abcdefabc. asdli! ndsf acba saa abe?", "abc")

Pretty much just replace n with a list so you can keep adding values to it as you find them. You also need to be incrementing i even when a match is found, it would have been stuck in a loop forever except that you had the while n == -1 constraint that made it stop as soon as a match was found.

Answer 5

def multi_find(s, r):

    s_len = len(s)
    r_len = len(r)

    _complete = []

    if s_len < r_len:
        n = -1
    else:

        for i in xrange(s_len):
            # search for r in s until not enough characters are left
            if s[i:i + r_len] == r:
                _complete.append(i)
            else:
                i = i + 1
    print(_complete)

multi_find("abcdefabc. asdli! ndsf abc saa abe?", "abc")

Answer 6

Another alternative using regex:

>>> import re
>>> haystack = "abcdefabc. asdli! ndsf acba saa abe?"
>>> needle = "abc"
>>> [m.start() for m in re.finditer(r'{}'.format(re.escape(needle)), haystack)]
[0, 6]

The above solution will not work for overlapping sub-strings, like there are 3 'aa' in 'aaaa'. So, if you want to find overlapping matches as well, then:

>>> haystack = "bobob"
>>> needle = "bob"
>>> [m.start() for m in re.finditer(r'(?={})'.format(re.escape(needle)), haystack)]
[0, 2]