How to Count String Occurrence in String

Count the number of occurrences of a character in a string

str.count(sub[, start[, end]])

Return the number of non-overlapping occurrences of substring sub in the range [start, end]. Optional arguments start and end are interpreted as in slice notation.

>>> sentence = 'Mary had a little lamb'
>>> sentence.count('a')
4

How to count string occurrence in string?

The g in the regular expression (short for global) says to search the whole string rather than just find the first occurrence. This matches is twice:

var temp = "This is a string.";

var count = (temp.match(/is/g) || []).length;

console.log(count);

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 do I count the number of occurrences of a char in a String?

My 'idiomatic one-liner' for this is:

int count = StringUtils.countMatches("a.b.c.d", ".");

Why write it yourself when it's already in commons lang?

Spring Framework's oneliner for this is:

int occurance = StringUtils.countOccurrencesOf("a.b.c.d", ".");

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

Count occurrences of a substring in a list of strings

You can do this by using the sum built-in function. No need to use list.count as well:

>>> data = ["the foo is all fooed", "the bar is all barred", "foo is now a bar"]
>>> sum('foo' in s for s in data)
2
>>>

This code works because booleans can be treated as integers. Each time 'foo' appears in a string element, True is returned. the integer value of True is 1. So it's as if each time 'foo' is in a string, we return 1. Thus, summing the 1's returned will yield the number of times 1 appeared in an element.

A perhaps more explicit but equivalent way to write the above code would be:

>>> sum(1 for s in data if 'foo' in s)
2
>>> 

Count the number of occurrences of a character in a string in Javascript

I have updated this answer. I like the idea of using a match better, but it is slower:

console.log(("str1,str2,str3,str4".match(/,/g) || []).length); //logs 3

console.log(("str1,str2,str3,str4".match(new RegExp("str", "g")) || []).length); //logs 4

How would you count occurrences of a string (actually a char) within a string?

If you're using .NET 3.5 you can do this in a one-liner with LINQ:

int count = source.Count(f => f == '/');

If you don't want to use LINQ you can do it with:

int count = source.Split('/').Length - 1;

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You might be surprised to learn that your original technique seems to be about 30% faster than either of these! I've just done a quick benchmark with "/once/upon/a/time/" and the results are as follows:

Your original = 12s

source.Count = 19s

source.Split = 17s

foreach (from bobwienholt's answer) = 10s

(The times are for 50,000,000 iterations so you're unlikely to notice much difference in the real world.)

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