Occurrences of Substring in a String

Occurrences of substring in a string

The last line was creating a problem. lastIndex would never be at -1, so there would be an infinite loop. This can be fixed by moving the last line of code into the if block.

String str = "helloslkhellodjladfjhello";
String findStr = "hello";
int lastIndex = 0;
int count = 0;

while(lastIndex != -1){

    lastIndex = str.indexOf(findStr,lastIndex);

    if(lastIndex != -1){
        count ++;
        lastIndex += findStr.length();
    }
}
System.out.println(count);

How to find the count of substring in java

What about:

String temp = s.replace(sub, "");
int occ = (s.length() - temp.length()) / sub.length();

Just remove all the substring, then check the difference on string length before and after removal. Divide the temp string with number of characters from the substring gives you the occurrences.

How to count number of occurrences of a substring inside a string in Python?

you can use count

print("hellohel".count("hel"))

2

If you want to count overlapping occurrences... maybe this can help

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

print(countOverlapping("ehehe", "ehe"))

output should be...

2

How does that work?

as @SomeDude mentioned it uses what he calls a sliding window approach

we take the length of the substring and check if its in that "window" of the string each iteration:

is ehe in [ehe]he? yes, count += 1
is ehe in e[heh]e? no, pass
is ehe in eh[ehe]? yes, count += 1

How to count of sub-string occurrences?

Regex.Matches(input, "OU=").Count

Finding number of times a substring exists in a string - Python

You seem to want overlapping counts. str.count will not get you there, unfortunately, because it does not overlap substring searches. Try slicing and counting.

Here's a solution with a collections.Counter though it can be done just about any other way as long as you slice it right.

from collections import Counter

text = 'bobbisbobobugbobobbobbobo'
term = 'bob'
c = Counter([text[i : i + len(term)] for i in range(len(text))])
print(c[term])

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7

How to find all occurrences of a substring?

There is no simple built-in string function that does what you're looking for, but you could use the more powerful regular expressions:

import re
[m.start() for m in re.finditer('test', 'test test test test')]
#[0, 5, 10, 15]

If you want to find overlapping matches, lookahead will do that:

[m.start() for m in re.finditer('(?=tt)', 'ttt')]
#[0, 1]

If you want a reverse find-all without overlaps, you can combine positive and negative lookahead into an expression like this:

search = 'tt'
[m.start() for m in re.finditer('(?=%s)(?!.{1,%d}%s)' % (search, len(search)-1, search), 'ttt')]
#[1]

re.finditer returns a generator, so you could change the [] in the above to () to get a generator instead of a list which will be more efficient if you're only iterating through the results once.

Find all occurrences of substring in string in Java

You can use capturing inside a positive look-ahead to get all overlapping matches and use Matcher#start to get the indices of the captured substrings.

As for the regex, it will look like

(?=(aa))

In Java code:

String s = "aaaaaa";
Matcher m = Pattern.compile("(?=(aa))").matcher(s);
List<Integer> pos = new ArrayList<Integer>();
while (m.find())
{
    pos.add(m.start());
}
System.out.println(pos);

Result:

[0, 1, 2, 3, 4]

See IDEONE demo

Number of occurrences of substring in string in Swift

A simple approach would be to split on "Swift", and subtract 1 from the number of parts:

let s = "hello Swift Swift and Swift"
let tok =  s.components(separatedBy:"Swift")
print(tok.count-1)

This code prints 3.

Edit: Before Swift 3 syntax the code looked like this:

let tok =  s.componentsSeparatedByString("Swift")

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