Find the nth occurrence of substring in a string

Here's a more Pythonic version of the straightforward iterative solution:

def find_nth(haystack, needle, n):
    start = haystack.find(needle)
    while start >= 0 and n > 1:
        start = haystack.find(needle, start+len(needle))
        n -= 1
    return start

Example:

>>> find_nth("foofoofoofoo", "foofoo", 2)
6

If you want to find the nth overlapping occurrence of needle, you can increment by 1 instead of len(needle), like this:

def find_nth_overlapping(haystack, needle, n):
    start = haystack.find(needle)
    while start >= 0 and n > 1:
        start = haystack.find(needle, start+1)
        n -= 1
    return start

Example:

>>> find_nth_overlapping("foofoofoofoo", "foofoo", 2)
3

This is easier to read than Mark's version, and it doesn't require the extra memory of the splitting version or importing regular expression module. It also adheres to a few of the rules in the Zen of python, unlike the various re approaches:

1. Simple is better than complex.
2. Flat is better than nested.
3. Readability counts.

Mark's iterative approach would be the usual way, I think.

Here's an alternative with string-splitting, which can often be useful for finding-related processes:

def findnth(haystack, needle, n):
    parts= haystack.split(needle, n+1)
    if len(parts)<=n+1:
        return -1
    return len(haystack)-len(parts[-1])-len(needle)

And here's a quick (and somewhat dirty, in that you have to choose some chaff that can't match the needle) one-liner:

'foo bar bar bar'.replace('bar', 'XXX', 1).find('bar')

Understanding that regex is not always the best solution, I'd probably use one here:

>>> import re
>>> s = "ababdfegtduab"
>>> [m.start() for m in re.finditer(r"ab",s)]
[0, 2, 11]
>>> [m.start() for m in re.finditer(r"ab",s)][2] #index 2 is third occurrence 
11

I'd probably do something like this, using the find function that takes an index parameter:

def find_nth(s, x, n):
    i = -1
    for _ in range(n):
        i = s.find(x, i + len(x))
        if i == -1:
            break
    return i

print find_nth('bananabanana', 'an', 3)

It's not particularly Pythonic I guess, but it's simple. You could do it using recursion instead:

def find_nth(s, x, n, i = 0):
    i = s.find(x, i)
    if n == 1 or i == -1:
        return i 
    else:
        return find_nth(s, x, n - 1, i + len(x))

print find_nth('bananabanana', 'an', 3)

It's a functional way to solve it, but I don't know if that makes it more Pythonic.

Here is another approach using re.finditer.
The difference is that this only looks into the haystack as far as necessary

from re import finditer
from itertools import dropwhile
needle='an'
haystack='bananabanana'
n=2
next(dropwhile(lambda x: x[0]<n, enumerate(re.finditer(needle,haystack))))[1].start() 

Here's another re + itertools version that should work when searching for either a str or a RegexpObject. I will freely admit that this is likely over-engineered, but for some reason it entertained me.

import itertools
import re

def find_nth(haystack, needle, n = 1):
    """
    Find the starting index of the nth occurrence of ``needle`` in \
    ``haystack``.

    If ``needle`` is a ``str``, this will perform an exact substring
    match; if it is a ``RegexpObject``, this will perform a regex
    search.

    If ``needle`` doesn't appear in ``haystack``, return ``-1``. If
    ``needle`` doesn't appear in ``haystack`` ``n`` times,
    return ``-1``.

    Arguments
    ---------
    * ``needle`` the substring (or a ``RegexpObject``) to find
    * ``haystack`` is a ``str``
    * an ``int`` indicating which occurrence to find; defaults to ``1``

    >>> find_nth("foo", "o", 1)
    1
    >>> find_nth("foo", "o", 2)
    2
    >>> find_nth("foo", "o", 3)
    -1
    >>> find_nth("foo", "b")
    -1
    >>> import re
    >>> either_o = re.compile("[oO]")
    >>> find_nth("foo", either_o, 1)
    1
    >>> find_nth("FOO", either_o, 1)
    1
    """
    if (hasattr(needle, 'finditer')):
        matches = needle.finditer(haystack)
    else:
        matches = re.finditer(re.escape(needle), haystack)
    start_here = itertools.dropwhile(lambda x: x[0] < n, enumerate(matches, 1))
    try:
        return next(start_here)[1].start()
    except StopIteration:
        return -1

For the special case where you search for the n'th occurence of a character (i.e. substring of length 1), the following function works by building a list of all positions of occurences of the given character:

def find_char_nth(string, char, n):
    """Find the n'th occurence of a character within a string."""
    return [i for i, c in enumerate(string) if c == char][n-1]

If there are fewer than n occurences of the given character, it will give IndexError: list index out of range.

This is derived from @Zv_oDD's answer and simplified for the case of a single character.

>>> s="abcdefabcdefababcdef"
>>> j=0
>>> for n,i in enumerate(s):
...   if s[n:n+2] =="ab":
...     print n,i
...     j=j+1
...     if j==2: print "2nd occurence at index position: ",n
...
0 a
6 a
2nd occurence at index position:  6
12 a
14 a

# return -1 if nth substr (0-indexed) d.n.e, else return index
def find_nth(s, substr, n):
    i = 0
    while n >= 0:
        n -= 1
        i = s.find(substr, i + 1)
    return i

Solution without using loops and recursion.

Use the required pattern in compile method and enter the desired occurrence in variable 'n' and the last statement will print the starting index of the nth occurrence of the pattern in the given string. Here the result of finditer i.e. iterator is being converted to list and directly accessing the nth index.

import re
n=2
sampleString="this is history"
pattern=re.compile("is")
matches=pattern.finditer(sampleString)
print(list(matches)[n].span()[0])

Here is my solution for finding nth occurrance of b in string a:

from functools import reduce


def findNth(a, b, n):
    return reduce(lambda x, y: -1 if y > x + 1 else a.find(b, x + 1), range(n), -1)

It is pure Python and iterative. For 0 or n that is too large, it returns -1. It is one-liner and can be used directly. Here is an example:

>>> reduce(lambda x, y: -1 if y > x + 1 else 'bibarbobaobaotang'.find('b', x + 1), range(4), -1)
7

I used findnth() function and ran into some issues, so I rewrote a faster version of the function (no list splitting):

def findnth(haystack, needle, n):
    if not needle in haystack or haystack.count(needle) < n:
        return -1

    last_index = 0
    cumulative_last_index = 0
    for i in range(0, n):
        last_index = haystack[cumulative_last_index:].find(needle)
        cumulative_last_index += last_index
        
        # if not last element, then jump over it
        if i < n-1:
            cumulative_last_index += len(needle)

    return cumulative_last_index

Def:

def get_first_N_words(mytext, mylen = 3):
    mylist = list(mytext.split())
    if len(mylist)>=mylen: return ' '.join(mylist[:mylen])

To use:

get_first_N_words('  One Two Three Four ' , 3)

Output:

'One Two Three'

Avoid a failure or incorrect output when the input value for occurrence provided is higher than the actual count of occurrence. For example, in a string 'overflow' if you would check the 3rd occurrence of 'o' ( it has only 2 occurrences ) then below code will return a warning or message indicating that the occurrence value has exceeded.

Input Occurrence entered has exceeded the actual count of Occurrence.

def check_nth_occurrence (string, substr, n):

## Count the Occurrence of a substr
    cnt = 0
    for i in string:
        if i ==substr:
            cnt = cnt + 1
        else:
            pass

## Check if the Occurrence input has exceeded the actual count of Occurrence

    if n > cnt:
        print (f' Input Occurrence entered has exceeded the actual count of Occurrence')
        return

## Get the Index value for first Occurrence of the substr

   index = string.find(substr)

## Get the Index value for nth Occurrence of Index
    while index >= 0 and n > 1:
        index = string.find(substr, index+ 1)
        n -= 1
  return index

Here's a simple and fun way to do it:

def index_of_nth(text, substring, n) -> int:
    index = 0
    for _ in range(n):
        index = text.index(substring, index) + 1
    return index - 1

Just in-case anyone wants to find n-th from the back:

def find_nth_reverse(haystack: str, needle: str, n: int) -> int:
    end = haystack.rfind(needle)

    while end >= 0 and n > 1:
        end = haystack.rfind(needle, 0, end - len(needle))
        n -= 1

    return end

I solved it like this.

def second_index(text: str, symbol: str) -> [int, None]:
"""
    returns the second index of a symbol in a given text
"""
first = text.find(symbol)
result = text.find(symbol,first+1)
if result > 0: return result 

A simple solution for those with basic programming knowledge:

# Function to find the nth occurrence of a substring in a text
def findnth(text, substring, n):

# variable to store current index in loop
count = -1

# n count
occurance = 0

# loop through string
for letter in text:
    
    # increment count
    count += 1
    
    # if current letter in loop matches substring target
    if letter == substring:
        
        # increment occurance
        occurance += 1
        
        # if this is the nth time the substring is found
        if occurance == n:
            
            # return its index
            return count
        
# otherwise indicate there is no match
return "No match"

# example of how to call function
print(findnth('C$100$150xx', "$", 2))