Overlapping count of substring in a string in Python

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I want to find all the counts (overlapping and non-overlapping) of a sub-string in a string. I found two answers one of which is using regex which is not my intention and the other was much more in-efficient than I need. I need something like:

'ababaa'.count('aba') == 2

str.count() just counts simple substrings. What should I do?

def sliding(a, n):
    return (a[i:i+n] for i in xrange(len(a) - n + 1))

def substring_count(a, b):
    return sum(s == b for s in sliding(a, len(b)))

assert list(sliding('abcde', 3)) == ['abc', 'bcd', 'cde']    
assert substring_count('ababaa', 'aba') == 2

Python, Given a string and a sub-string, the task is to get the count of overlapping substring from the given string. Note that in Python, the count() function returns the� Python | Count overlapping substring in a given string Given a string and a sub-string, the task is to get the count of overlapping substring from the given string. Note that in Python, the count() function returns the number of substrings in a given string, but it does not give correct results when two occurrences of the substring overlap.

Does this do the trick?

def count(string, substring):
    n = len(substring)
    cnt = 0
    for i in range(len(string) - n):
        if string[i:i+n] == substring:
            cnt += 1
    return cnt

print count('ababaa', 'aba') # 2

I don't know if there's a more efficient solution, but this should work.

String count with overlapping occurrences, Python's str.count counts non-overlapping substrings: In [3]: "ababa".count("aba") Out[3]: 1. Here are a few ways to count overlapping� While the accepted answer is fastest for shorter strings, if you are searching relatively short substrings within long strings the Counter approach starts to take the edge. Also, if you have need to refactor this to perform multiple substring count queries against the same main string, then the Counter approach starts looking much more attractive

count = len(set([string.find('aba',x) for x in range(len(string)) if string.find('aba',x) >= 0]))

Python : Find occurrence count & all indices of a sub-string in , Python's String class contains a method to count the non overlapping Get a list of strings that matches the given pattern i.e. substring. To get the total number of occurrences of a substring in a string, where the occurrences could be overlapping, follow these steps. Traverse through the string using for loop and range(len(string)). Find the index of the first occurrence of the substring using String. find (substring, start).

Here, using re.finditer() is the best way to achieve what you want.

import re 

def get_substring_count(s, sub_s):
    return sum(1 for m in re.finditer('(?=%s)' % sub_s, s))

get_substring_count('ababaa', 'aba')
# 2 as response

Python Challenge: Find A Substring Count With Overlaps, In this Python Challenge, you will be asked to find a substring and return the total number of Duration: 8:49 Posted: Jan 5, 2019 String count with overlapping occurrences (23 answers) Closed 3 years ago . I have a long sequence, and I would like to know how often some sub-sequences occur in this sequence.

Here's a function you could use:

def count(haystack, needle):
    return len([x for x in [haystack[i:j+1] for i in xrange(len(haystack)) for j in xrange(i,len(haystack))] if x == needle])


>>> count("ababaa", "aba")

Count overlapping substrings in a given string in Python , Python has some inbuilt functions which perform specific tasks. string.count(sub- string) is an inbuilt function that counts the occurrences of a substring in the given � I'm trying to find every 10 digit series of numbers within a larger series of numbers using re in Python 2.6. I'm easily able to grab no overlapping matches, but I want every match in the number series. Eg. in "123456789123456789" I should get the following list:

string count with overlapping occurrences, def function(string, str_to_search_for): count = 0 for x in xrange(len(string) - len( str_to_search_for) + 1): if Python's str.count counts non-overlapping substrings: . Python | Ways to count number of substring in string Given a string and a substring, write a Python program to find how many numbers of substring are there in the string (including overlapping cases). Let’s discuss a few methods below. Method #1: Using re.findall ()

Count overlapping substrings, Python allows you to index from a beginning to an end of a sliceable object. To use this is quite simple. def overlapping_count(string, seek):� To count the number of occurrences of a sub-string in a string, use String.count () method on the main string with sub-string passed as argument. Syntax of String.count () string. count (substring) count () method returns an integer that represents the number of times a specified sub-string appeared in this string.

Non-Overlapping occurrences (Help) : learnpython, Suppose the count function for a string didn't exist. Define a function that returns the number of non-overlapping occurrences of a substring in a string. Currently I � It returns the count & indices of non-overlapping sub-strings only. To find the occurrence count & indices of overlapping sub-strings let’s modify the above create function. Find indices of overlapping sub-string in string using Python '''' Find occurrence count of overlapping substrings and get their count and index positions.

  • What language is this?
  • Here's a Scala solution in case that helps :) "ababaa".sliding(3).count(_ == "aba")
  • sorry, it's python I'm using.
  • Sorry @Chris, I forgot to add I'm using python.
  • what would be time complexity?
  • For fixed |b|, it's O(|a|). As a function of both |a| and |b|, I'd have to think more... probably O(|a| * |b|). Might be able to give a better bound, but I'm guessing you only care about the situation where |b| << |a| ?
  • Well, it's much efficient. Thanks @ChrisMartin
  • It does, but it's efficiency is what matters. If the iteration was all I had to do, I could have done it, in fact, I've done it but I need less time consuming because the problem I'm solving consists of very large strings, where I don't think iteration would work.
  • @lavee_singh This is essentially the same algorithm as mine, just written in a procedural rather than functional style. Same complexity.
  • Sorry, but there is difference, a lot of it. You are using by creating lists and iterators, which is essentially slower than generators. Try to mark the difference in () and [], this is a huge difference.
  • what would be the time complexity?
  • It's good, but I've got a solution with better time complexity. But this is really good as well.