Determining how many times a substring occurs in a string in Python

Question

I am trying to figure out how many times a string occurs in a string. For example:

nStr = '000123000123'

Say the string I want to find is 123. Obviously it occurs twice in nStr but I am having trouble implementing this logic into Python. What I have got at the moment:

pattern = '123'
count = a = 0
while pattern in nStr[a:]:
    a = nStr[a:].find(pattern)+1
    count += 1
return count

The answer it should return is 2. I'm stuck in an infinite loop at the moment.

I was just made aware that count is a much better way to do it but out of curiosity, does anyone see a way to do it similar to what I have already got?

Ashwini Chaudhary · Accepted Answer · 2021-01-25 15:39:16Z

130

Use str.count:

>>> nStr = '000123000123'
>>> nStr.count('123')
2

A working version of your code:

nStr = '000123000123'
pattern = '123'
count = 0
flag = True
start = 0

while flag:
    a = nStr.find(pattern, start)  # find() returns -1 if the word is not found, 
    #start i the starting index from the search starts(default value is 0)
    if a == -1:          #if pattern not found set flag to False
        flag = False
    else:               # if word is found increase count and set starting index to a+1
        count += 1        
        start = a + 1
print(count)

edited Jan 25, 2021 at 15:39

answered Jul 13, 2012 at 19:00

Ashwini Chaudhary

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1 Comment

Nasrin Over a year ago

count function doesn't work right in every situation. for example : pattern = "323" and nStr = "10032323000123". As you see 323 appears 2 times in the main string. But count's result is 1. So the second solution is right.

João Pesce · Accepted Answer · 2020-08-17 21:23:49Z

36

The problem with count() and other methods shown here is in the case of overlapping substrings.

For example: "aaaaaa".count("aaa") returns 2

If you want it to return 4 [(aaa)aaa, a(aaa)aa, aa(aaa)a, aaa(aaa)] you might try something like this:

def count_substrings(string, substring):
    string_size = len(string)
    substring_size = len(substring)
    count = 0
    for i in xrange(0,string_size-substring_size+1):
        if string[i:i+substring_size] == substring:
            count+=1
    return count

count_substrings("aaaaaa", "aaa")
# 4

Not sure if there's a more efficient way of doing it, but I hope this clarifies how count() works.

edited Aug 17, 2020 at 21:23

answered Jun 5, 2013 at 3:49

João Pesce

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1 Comment

TawabG Over a year ago

Note that xrange() was renamed to range() in Python 3.

David C · Accepted Answer · 2013-01-30 19:13:35Z

7

import re

pattern = '123'

n =re.findall(pattern, string)

We can say that the substring 'pattern' appears len(n) times in 'string'.

edited Jan 30, 2013 at 19:13

David C

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answered Jul 16, 2012 at 5:18

Prasanna

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2 Comments

B Custer Over a year ago

This computes the count WITHOUT overlaps!

gnoodle Over a year ago

@BCuster good point. See my answer below which uses regex to compute the count WITH overlaps.

TawabG · Accepted Answer · 2019-06-07 17:36:40Z

4

In case you are searching how to solve this problem for overlapping cases.

s = 'azcbobobegghaklbob'
str = 'bob'
results = 0
sub_len = len(str) 
for i in range(len(s)):
    if s[i:i+sub_len] == str: 
        results += 1
print (results)

Will result in 3 because: [azc(bob)obegghaklbob] [azcbo(bob)egghaklbob] [azcbobobegghakl(bob)]

answered Jun 7, 2019 at 17:36

TawabG

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Comments

Harshal Parekh · Accepted Answer · 2019-09-14 03:08:13Z

1

I'm pretty new, but I think this is a good solution? maybe?

def count_substring(str, sub_str):
    count = 0
    for i, c in enumerate(str):
        if sub_str == str[i:i+2]:
            count += 1
    return count

edited Sep 14, 2019 at 3:08

Harshal Parekh

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answered Sep 14, 2019 at 0:47

muramena

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Comments

Gaurav Parashar · Accepted Answer · 2017-02-16 10:56:44Z

0

string.count(substring) is not useful in case of overlapping.

My approach:

def count_substring(string, sub_string):

    length = len(string)
    counter = 0
    for i in range(length):
        for j in range(length):
            if string[i:j+1] == sub_string:
                counter +=1
    return counter

answered Feb 16, 2017 at 10:56

Gaurav Parashar

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Comments

e_i_pi · Accepted Answer · 2018-01-04 05:37:07Z

0

You are not changing a with each loop. You should put:

a += nStr[a:].find(pattern)+1

...instead of:

a = nStr[a:].find(pattern)+1

edited Jan 4, 2018 at 5:37

e_i_pi

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answered Jan 4, 2018 at 4:07

N Prad

1

Comments

Priyanka Kumari · Accepted Answer · 2018-10-23 15:13:22Z

0

def count_substring(string, substring):
         c=0
         l=len(sub_string)
         for i in range(len(string)):
                 if string [i:i+l]==sub_string:
                          c=c+1
         return c
string=input().strip()
sub_string=input().strip()

count= count_substring(string,sub_string)
print(count)

answered Oct 23, 2018 at 15:13

Priyanka Kumari

1

Comments

Aditya Patnaik · Accepted Answer · 2019-02-11 13:10:10Z

0

As mentioned by @João Pesce and @gaurav, count() is not useful in the case of overlapping substrings, try this out...

def count_substring(string, sub_string):
    c=0
    for i in range(len(string)):
        if(string[i:i+len(sub_string)]==sub_string):
            c = c+1
    return c

edited Feb 11, 2019 at 13:10

answered Feb 11, 2019 at 12:44

Aditya Patnaik

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Comments

Shir · Accepted Answer · 2019-04-22 09:52:35Z

0

def countOccurance(str,pat):
    count=0
    wordList=str.split()
    for word in wordList:
        if pat in word:
            count+=1
    return count

edited Apr 22, 2019 at 9:52

Shir

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answered Apr 22, 2019 at 9:13

Bhabani Sharma

11

Comments

ruddy simonpour · Accepted Answer · 2020-04-23 02:25:45Z

0

Usually i'm using enumerate for this kind of problems:

def count_substring(string, sub_string):
        count = 0
        for i, j in enumerate(string):
            if sub_string in string[i:i+3]:
                count = count + 1
        return count

answered Apr 23, 2020 at 2:25

ruddy simonpour

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1 Comment

NaN Over a year ago

@ruddy_simonpour you might have added that count_substring(nStr, pattern) with nStr = '000123000123' and pattern = '123' yields 2, which is correct.

gnoodle · Accepted Answer · 2023-04-11 14:08:33Z

0

Only one approach here uses regex, and that approach doesn't work for overlaps.

Here is how to use regex with "lookaheads" to find overlapping matches also:

import re

nStr = '00012312310001231'
regex_pattern = '(?=(1231))'

matches = re.findall(regex_pattern, nStr)
print(len(matches))

This returns 3, as it found two matches of 1231 in 1231231, despite the overlap.

answered Apr 11, 2023 at 14:08

gnoodle

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Comments

Prince_Israel · Accepted Answer · 2020-06-09 07:02:53Z

-1

def count(sub_string,string):

count = 0
ind = string.find(sub_string)

while True:
    if ind > -1:
        count += 1
        ind = string.find(sub_string,ind + 1)
    else:
        break
return count

answered Jun 9, 2020 at 7:02

Prince_Israel

1

Comments

Avancha Bhargava · Accepted Answer · 2021-12-19 22:53:01Z

-1

def count_substring(string, sub_string):
    count = 0
    len_sub = len(sub_string)
    for i in range(0,len(string)):
        if(string[i:i+len_sub] == sub_string):
            count+=1
    return count

answered Dec 19, 2021 at 22:53

Avancha Bhargava

1

1 Comment

General Grievance Over a year ago

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