Count number of occurrences of a substring in a string

Question

How can I count the number of times a given substring is present within a string in Python?

For example:

>>> 'foo bar foo'.numberOfOccurrences('foo')
2

---

_{To get indices of the substrings, see How to find all occurrences of a substring?.}

Answer 1

string.count(substring), like in:

>>> "abcdabcva".count("ab")
2

This is for non overlapping occurrences.
If you need to count overlapping occurrences, you'd better check the answers here, or just check my other answer below.

Answer 2

s = 'arunununghhjj'
sb = 'nun'
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print results

Answer 3

Depending what you really mean, I propose the following solutions:

1. You mean a list of space separated sub-strings and want to know what is the sub-string position number among all sub-strings:

   s = 'sub1 sub2 sub3'
   s.split().index('sub2')
   >>> 1
   

2. You mean the char-position of the sub-string in the string:

   s.find('sub2')
   >>> 5
   

3. You mean the (non-overlapping) counts of appearance of a su-bstring:

   s.count('sub2')
   >>> 1
   s.count('sub')
   >>> 3
   

Answer 4

The best way to find overlapping sub-strings in a given string is to use a regular expression. With lookahead, it will find all the overlapping matches using the regular expression library's findall(). Here, left is the substring and right is the string to match.

>>> len(re.findall(r'(?=aa)', 'caaaab'))
3

Answer 5

To find overlapping occurences of a substring in a string in Python 3, this algorithm will do:

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

I myself checked this algorithm and it worked.

Answer 6

You can count the frequency using two ways:

1. Using the count() in str:

   a.count(b)

2. Or, you can use:

   len(a.split(b))-1

Where a is the string and b is the substring whose frequency is to be calculated.

Answer 7

Scenario 1: Occurrence of a word in a sentence. eg: str1 = "This is an example and is easy". The occurrence of the word "is". lets str2 = "is"

count = str1.count(str2)

Scenario 2 : Occurrence of pattern in a sentence.

string = "ABCDCDC"
substring = "CDC"

def count_substring(string,sub_string):
    len1 = len(string)
    len2 = len(sub_string)
    j =0
    counter = 0
    while(j < len1):
        if(string[j] == sub_string[0]):
            if(string[j:j+len2] == sub_string):
                counter += 1
        j += 1

    return counter

Thanks!

Answer 8

The current best answer involving method count doesn't really count for overlapping occurrences and doesn't care about empty sub-strings as well. For example:

>>> a = 'caatatab'
>>> b = 'ata'
>>> print(a.count(b)) #overlapping
1
>>>print(a.count('')) #empty string
9

The first answer should be 2 not 1, if we consider the overlapping substrings. As for the second answer it's better if an empty sub-string returns 0 as the asnwer.

The following code takes care of these things.

def num_of_patterns(astr,pattern):
    astr, pattern = astr.strip(), pattern.strip()
    if pattern == '': return 0

    ind, count, start_flag = 0,0,0
    while True:
        try:
            if start_flag == 0:
                ind = astr.index(pattern)
                start_flag = 1
            else:
                ind += 1 + astr[ind+1:].index(pattern)
            count += 1
        except:
            break
    return count

Now when we run it:

>>>num_of_patterns('caatatab', 'ata') #overlapping
2
>>>num_of_patterns('caatatab', '') #empty string
0
>>>num_of_patterns('abcdabcva','ab') #normal
2

Answer 9

One way is to use re.subn. For example, to count the number of occurrences of 'hello' in any mix of cases you can do:

import re
_, count = re.subn(r'hello', '', astring, flags=re.I)
print('Found', count, 'occurrences of "hello"')

Answer 10

The question isn't very clear, but I'll answer what you are, on the surface, asking.

A string S, which is L characters long, and where S[1] is the first character of the string and S[L] is the last character, has the following substrings:

• The null string ''. There is one of these.
• For every value A from 1 to L, for every value B from A to L, the string S[A]..S[B] (inclusive). There are L + L-1 + L-2 + ... 1 of these strings, for a total of 0.5*L*(L+1).
• Note that the second item includes S[1]..S[L], i.e. the entire original string S.

So, there are 0.5*L*(L+1) + 1 substrings within a string of length L. Render that expression in Python, and you have the number of substrings present within the string.

Answer 11

I will keep my accepted answer as the "simple and obvious way to do it", however, it does not cover overlapping occurrences. Finding out those can be done naively, with multiple checking of the slices - as in:

sum("GCAAAAAGH"[i:].startswith("AAA") for i in range(len("GCAAAAAGH")))

which yields 3.

Or it can be done by trick use of regular expressions, as can be seen at How to use regex to find all overlapping matches - and it can also make for fine code golfing.

This is my "hand made" count for overlapping occurrences of patterns in a string which tries not to be extremely naive (at least it does not create new string objects at each interaction):

def find_matches_overlapping(text, pattern):
    lpat = len(pattern) - 1
    matches = []
    text = array("u", text)
    pattern = array("u", pattern)
    indexes = {}
    for i in range(len(text) - lpat):
        if text[i] == pattern[0]:
            indexes[i] = -1
        for index, counter in list(indexes.items()):
            counter += 1
            if text[i] == pattern[counter]:
                if counter == lpat:
                    matches.append(index)
                    del indexes[index]
                else:
                    indexes[index] = counter
            else:
                del indexes[index]
    return matches
            
def count_matches(text, pattern):
    return len(find_matches_overlapping(text, pattern))

Answer 12

How about a one-liner with a list comprehension? Technically its 93 characters long, spare me PEP-8 purism. The regex.findall answer is the most readable if its a high level piece of code. If you're building something low level and don't want dependencies, this one is pretty lean and mean. I'm giving the overlapping answer. Obviously just use count like the highest score answer if there isn't overlap.

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

Answer 13

If you want to count all the sub-string (including overlapped) then use this method.

import re
def count_substring(string, sub_string):
    regex = '(?='+sub_string+')'
    # print(regex)
    return len(re.findall(regex,string))

Answer 14

For overlapping count we can use use:

def count_substring(string, sub_string):
    count=0
    beg=0
    while(string.find(sub_string,beg)!=-1) :
        count=count+1
        beg=string.find(sub_string,beg)
        beg=beg+1
    return count

For non-overlapping case we can use count() function:

string.count(sub_string)

Answer 15

Overlapping occurences:

def olpcount(string,pattern,case_sensitive=True):
    if case_sensitive != True:
        string  = string.lower()
        pattern = pattern.lower()
    l = len(pattern)
    ct = 0
    for c in range(0,len(string)):
        if string[c:c+l] == pattern:
            ct += 1
    return ct

test = 'my maaather lies over the oceaaan'
print test
print olpcount(test,'a')
print olpcount(test,'aa')
print olpcount(test,'aaa')

Results:

my maaather lies over the oceaaan
6
4
2

Answer 16

You could use the startswith method:

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

Answer 17

Here's a solution that works for both non-overlapping and overlapping occurrences. To clarify: an overlapping substring is one whose last character is identical to its first character.

def substr_count(st, sub):
    # If a non-overlapping substring then just
    # use the standard string `count` method
    # to count the substring occurences
    if sub[0] != sub[-1]:
        return st.count(sub)

    # Otherwise, create a copy of the source string,
    # and starting from the index of the first occurence
    # of the substring, adjust the source string to start
    # from subsequent occurences of the substring and keep
    # keep count of these occurences
    _st = st[::]
    start = _st.index(sub)
    cnt = 0

    while start is not None:
        cnt += 1
        try:
            _st = _st[start + len(sub) - 1:]
            start = _st.index(sub)
        except (ValueError, IndexError):
            return cnt

    return cnt

Answer 18

If you're looking for a power solution that works every case this function should work:

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

Answer 19

If you want to find out the count of substring inside any string; please use below code. The code is easy to understand that's why i skipped the comments. :)

string=raw_input()
sub_string=raw_input()
start=0
answer=0
length=len(string)
index=string.find(sub_string,start,length)
while index<>-1:
    start=index+1
    answer=answer+1
    index=string.find(sub_string,start,length)
print answer

Answer 20

2+ others have already provided this solution, and I even upvoted one of them, but mine is probably the easiest for newbies to understand.

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

Answer 21

Here's the solution in Python 3 and case insensitive:

s = 'foo bar foo'.upper()
sb = 'foo'.upper()
results = 0
sub_len = len(sb)
for i in range(len(s)):
    if s[i:i+sub_len] == sb:
        results += 1
print(results)

Answer 22

s = input('enter the main string: ')
p=input('enter the substring: ')
l=[]
for i in range(len(s)):
    l.append(s[i:i+len(p)])
print(l.count(p))

Answer 23

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

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()

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

Answer 24

def count_substring(string, sub_string):
    k=len(string)
    m=len(sub_string)
    i=0
    l=0
    count=0
    while l<k:
        if string[l:l+m]==sub_string:
            count=count+1
        l=l+1
    return count

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()

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

Answer 25

I'm not sure if this is something looked at already, but I thought of this as a solution for a word that is 'disposable':

for i in xrange(len(word)):
if word[:len(term)] == term:
    count += 1
word = word[1:]

print count

Where word is the word you are searching in and term is the term you are looking for

Answer 26

string="abc"
mainstr="ncnabckjdjkabcxcxccccxcxcabc"
count=0
for i in range(0,len(mainstr)):
    k=0
    while(k<len(string)):
        if(string[k]==mainstr[i+k]):
            k+=1
        else:
            break   
    if(k==len(string)):
        count+=1;   
print(count)

Answer 27

import re
d = [m.start() for m in re.finditer(pattern, string)] 
print(d)

This finds the number of times sub string found in the string and displays index.

Answer 28

my_string = """Strings are amongst the most popular data types in Python. 
               We can create the strings by enclosing characters in quotes.
               Python treats single quotes the same as double quotes."""

Count = my_string.lower().strip("\n").split(" ").count("string")
Count = my_string.lower().strip("\n").split(" ").count("strings")
print("The number of occurance of word String is : " , Count)
print("The number of occurance of word Strings is : " , Count)

Answer 29

Below logic will work for all string & special characters

def cnt_substr(inp_str, sub_str):
    inp_join_str = ''.join(inp_str.split())
    sub_join_str = ''.join(sub_str.split())

    return inp_join_str.count(sub_join_str)

print(cnt_substr("the sky is   $blue and not greenthe sky is   $blue and not green", "the sky"))

Answer 30

For a simple string with space delimitation, using Dict would be quite fast, please see the code as below

def getStringCount(mnstr:str, sbstr:str='')->int:
    """ Assumes two inputs string giving the string and 
        substring to look for number of occurances 
        Returns the number of occurances of a given string
    """
    x = dict()
    x[sbstr] = 0
    sbstr = sbstr.strip()
    for st in mnstr.split(' '):
        if st not in [sbstr]:
            continue
        try:
            x[st]+=1
        except KeyError:
            x[st] = 1
    return x[sbstr]

s = 'foo bar foo test one two three foo bar'
getStringCount(s,'foo')

Answer 31

j = 0
    while i < len(string):
        sub_string_out = string[i:len(sub_string)+j]
        if sub_string == sub_string_out:
            count += 1
        i += 1
        j += 1
    return count

Answer 32

#counting occurence of a substring in another string (overlapping/non overlapping)
s = input('enter the main string: ')# e.g. 'bobazcbobobegbobobgbobobhaklpbobawanbobobobob'
p=input('enter the substring: ')# e.g. 'bob'

counter=0
c=0

for i in range(len(s)-len(p)+1):
    for j in range(len(p)):
        if s[i+j]==p[j]:
            if c<len(p):
                c=c+1
                if c==len(p):
                    counter+=1
                    c=0
                    break
                continue
        else:
            break
print('number of occurences of the substring in the main string is: ',counter)

Answer 33

This makes a list of all the occurrences (also overlapping) in the string and counts them

def num_occ(str1, str2):
    l1, l2 = len(str1), len(str2)
    return len([str1[i:i + l2] for i in range(l1 - l2 + 1) if str1[i:i + l2] == str2])

Example:

str1 ='abcabcd'
str2 = 'bc'

will create this list but save only the BOLD values:

[ab, bc, ca, ab, bc, cd]

that will return:

len([bc, bc])

Answer 34

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

if __name__ == '__main__':
    string = input().strip()
    sub_string = input().strip()

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

Answer 35

Using the assignment operator introduced in Python 3.8, we can write a short function that uses str.find() in a loop to find overlapping instances of a target substring within a string. There are a few other solutions already posted that use the same approach, but this one is shorter and faster.

The assignment expression not only serves to start the next find operation at the character after the last-found instance, it also provides the terminal expression for the while loop. str.find() returns -1 if the substring isn't found, and adding 1 to this yields 0, which is false, thereby exiting the loop when no more matches are found.

# count overlapping occurrences of a substring in a string
def count_overlapping(haystack, needle, start=0, count=0):
    while start := haystack.find(needle, start) + 1:
        count += 1
    return count

print(count_overlapping("moomoooo", "oo"))    # 4

To optimize performance further, we can look up haystack.find once outside the loop and store that in a local variable. This will be faster when there are more than a couple of matches.

# count overlapping occurrences of a substring in a string
def count_overlapping(haystack, needle, start=0, count=0):
    haystack_find = haystack.find
    while start := haystack_find(needle, start) + 1:
        count += 1
    return count

Answer 36

If you're looking to count the whole string this can works.

stri_count="If you're looking to count the whole string this can works"
print(len(stri_count))