python: Finding a substring within a string

Question

I'm trying to write a program which counts how many times a substring appears within a string.

word = "wejmfoiwstreetstreetskkjoih"
streets = "streets"
count = 0

if streets in word:
    count += 1

print(count)

as you can see "streets" appears twice but the last s of streets is also the beginning of streets. I can't think of a way to loop this.

Thanks!

Just so you know, a proper substr function typically only searches for one occurrence of the sub-string, and usually returns the index of it's position in the string. In your example, you are only searching for one occurrence as well, if streets in word well only increment count once, no matter how many times the sub string 'streets' shows up in your string. There are already some good answers here, I just wanted to provide some insight into why your function wasn't working. — samrap
– samrap, Commented Jun 21, 2014 at 14:53

Tim · Accepted Answer · 2014-06-21 14:49:56Z

4

Can be done using a regex

>>> import re
>>> text = 'streetstreets'
>>> len(re.findall('(?=streets)', text))
2

From the docs:

(?=...)

Matches if ... matches next, but doesn’t consume any of the string. This is called a lookahead assertion. For example, Isaac (?=Asimov) will match 'Isaac ' only if it’s followed by 'Asimov'.

edited Jun 21, 2014 at 14:49

answered Jun 21, 2014 at 14:43

Tim

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2 Comments

Eric Wilson Over a year ago

That works, but this answer would be better if the regex was explained somewhat. In particular, what does the ?= do, as it doesn't work with that omitted.

Tim Over a year ago

@EricWilson agreed, added the part from the docs where it's explained

UltraInstinct · Accepted Answer · 2014-06-21 14:42:59Z

2

Quick and dirty:

>>> word = "wejmfoiwstreetstreetskkjoih"
>>> streets = "streets"
>>> sum(word[start:].startswith(streets) for start in range(len(word)))
2

answered Jun 21, 2014 at 14:42

UltraInstinct

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Comments

bereal · Accepted Answer · 2014-06-21 14:59:52Z

0

A generic (though not as elegant) way would be a loop like this:

def count_substrings(stack, needle):
    idx = 0
    count = 0
    while True:
        idx = stack.find(needle, idx) + 1 # next time look after this idx
        if idx <= 0:
            break
        count += 1
    return count

My measurement shows that it's ~8.5 times faster than the solution with startswith for every substring.

edited Jun 21, 2014 at 14:59

answered Jun 21, 2014 at 14:54

bereal

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