string.count(substring), like in:
>>> "abcdabcva".count("ab")
2
As pointed up in the comments, this is the way to do it for non overlapping occurrences. If you need to count overlapping occurrences, you'd better check the answers at: "Python regex find all overlapping matches?", or just check my other answer below.
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))
Risking a downvote because 2+ others have already provided this solution. 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
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
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))
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.
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)
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
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
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)
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
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
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)
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)
Depending what you really mean, I propose the following solutions:
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
You mean the char-position of the sub-string in the string:
s.find('sub2')
>>> 5
You mean the (non-overlapping) counts of appearance of a su-bstring:
s.count('sub2')
>>> 1
s.count('sub')
>>> 3
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])
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"')
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])
I will keep my accepted answer as the "simple and obvious way to do it" - however, that 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) - it can be done by trick use of regular expressions, as can be seen at Python regex find all overlapping matches? - and it can also make for fine code golfing - This is my "hand made" count for overlappingocurrences 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))
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))
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
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)
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)
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
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!
The best way to find overlapping sub-string in a given string is to use the python regular expression it will find all the overlapping matching using the regular expression library. Here is how to do it left is the substring and in right you will provide the string to match
print len(re.findall('(?=aa)','caaaab'))
3
import re
d = [m.start() for m in re.finditer(seaching, string)]
print (d)
This finds the number of times sub string found in the string and displays index.
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:
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.
You can count the frequency using two ways:
Using the count() in str:
a.count(b)
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.
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"))
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)
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
#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)
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')
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
Source: Stackoverflow.com