Find the index of the n'th item in a list


I want to find the index of the n’th occurrence of an item in a list. e.g.,


What is the index of the n’th true? If I wanted the fifth occurrence (4th if zero-indexed), the answer is 10.

I’ve come up with:

indargs = [ i for i,a in enumerate(x) if a ]

Note that x.index returns the first occurrence or the first occurrence after some point, and therefore as far as I can tell is not a solution.

There is also a solution in numpy for cases similar to the above, e.g. using cumsum and where, but I’d like to know if there’s a numpy-free way to solve the problem.

I’m concerned about performance since I first encountered this while implemented a Sieve of Eratosthenes for a Project Euler problem, but this is a more general question that I have encountered in other situations.

EDIT: I’ve gotten a lot of great answers, so I decided to do some performance tests. Below are timeit execution times in seconds for lists with len nelements searching for the 4000’th/1000’th True. The lists are random True/False. Source code linked below; it’s a touch messy. I used short / modified versions of the posters’ names to describe the functions except listcomp, which is the simple list comprehension above.

True Test (100'th True in a list containing True/False)
         nelements      eyquem_occur eyquem_occurrence            graddy            taymon          listcomp       hettinger26         hettinger
             3000:          0.007824          0.031117          0.002144          0.007694          0.026908          0.003563          0.003563
            10000:          0.018424          0.103049          0.002233          0.018063          0.088245          0.003610          0.003769
            50000:          0.078383          0.515265          0.002140          0.078074          0.442630          0.003719          0.003608
           100000:          0.152804          1.054196          0.002129          0.152691          0.903827          0.003741          0.003769
           200000:          0.303084          2.123534          0.002212          0.301918          1.837870          0.003522          0.003601
True Test (1000'th True in a list containing True/False)
         nelements      eyquem_occur eyquem_occurrence            graddy            taymon          listcomp       hettinger26         hettinger
             3000:          0.038461          0.031358          0.024167          0.039277          0.026640          0.035283          0.034482
            10000:          0.049063          0.103241          0.024120          0.049383          0.088688          0.035515          0.034700
            50000:          0.108860          0.516037          0.023956          0.109546          0.442078          0.035269          0.035373
           100000:          0.183568          1.049817          0.024228          0.184406          0.906709          0.035135          0.036027
           200000:          0.333501          2.141629          0.024239          0.333908          1.826397          0.034879          0.036551
True Test (20000'th True in a list containing True/False)
         nelements      eyquem_occur eyquem_occurrence            graddy            taymon          listcomp       hettinger26         hettinger
             3000:          0.004520          0.004439          0.036853          0.004458          0.026900          0.053460          0.053734
            10000:          0.014925          0.014715          0.126084          0.014864          0.088470          0.177792          0.177716
            50000:          0.766154          0.515107          0.499068          0.781289          0.443654          0.707134          0.711072
           100000:          0.837363          1.051426          0.501842          0.862350          0.903189          0.707552          0.706808
           200000:          0.991740          2.124445          0.498408          1.008187          1.839797          0.715844          0.709063
Number Test (750'th 0 in a list containing 0-9)
         nelements      eyquem_occur eyquem_occurrence            graddy            taymon          listcomp       hettinger26         hettinger
             3000:          0.026996          0.026887          0.015494          0.030343          0.022417          0.026557          0.026236
            10000:          0.037887          0.089267          0.015839          0.040519          0.074941          0.026525          0.027057
            50000:          0.097777          0.445236          0.015396          0.101242          0.371496          0.025945          0.026156
           100000:          0.173794          0.905993          0.015409          0.176317          0.762155          0.026215          0.026871
           200000:          0.324930          1.847375          0.015506          0.327957          1.536012          0.027390          0.026657

Hettinger’s itertools solution is almost always the best. taymon’s and graddy’s solutions are next best for most situations, though the list comprehension approach can be better for short arrays when you want the n’th instance such that n is high or lists in which there are fewer than n occurrences. If there’s a chance that there are fewer than n occurrences, the initial count check saves time. Also, graddy’s is more efficient when searching for numbers instead of True/False… not clear why that is. eyquem’s solutions are essentially equivalent to others with slightly more or less overhead; eyquem_occur is approximately the same as taymon’s solution, while eyquem_occurrence is similar to listcomp.

Asked By: keflavich



I can’t say for certain that this is the fastest way, but I imagine it’d be pretty good:

i = -1
for j in xrange(n):
    i = x.index(True, i + 1)

The answer is i.

Answered By: Taymon

if efficiency is a concern i think its better to iterate the normally ( O(N) ) instead of list comprehension which takes O(L) where L is length of list

Example : Consider a very huge list and you want to find the first occurence N=1 it is obviously better to stop as soon as you find the first occurence

count = 0
for index,i in enumerate(L):
    if i:
        count = count + 1
        if count==N:
            return index
Answered By: Graddy

If you’re concerned with performance, you are best off seeing if there are algorithmic optimizations you can make. For example, if you are calling this function many times on the same values, you may wish to cache previous computations (e.g. once you find the 50th occurrence of an element, you can find any previous occurrence in O(1) time).

Otherwise, you want to make sure your technique works on (lazy) iterators.

The most *in*elegant and performance-happy way I can think of implementing it is as:

def indexOfNthOccurrence(N, element, stream):
    """for N>0, returns index or None"""
    seen = 0
    for i,x in enumerate(stream):
        if x==element:
            seen += 1
            if seen==N:
                return i

(if you really care about the performance difference between enumerate and other techniques, you will need to resort to profiling, especially with the numpy functions, which may resort to C)

To preprocess the entire stream and support O(1) queries:

from collections import *
cache = defaultdict(list)
for i,elem in enumerate(YOUR_LIST):
    cache[elem] += [i]

# e.g. [3,2,3,2,5,5,1]
#       0 1 2 3 4 5 6
# cache: {3:[0,2], 1:[6], 2:[1,3], 5:[4,5]}
Answered By: ninjagecko

The answer from @Taymon using list.index was great.

FWIW, here’s a functional approach using the itertools module. It works with any iterable input, not just lists:

>>> from itertools import compress, count, imap, islice
>>> from functools import partial
>>> from operator import eq

>>> def nth_item(n, item, iterable):
        indicies = compress(count(), imap(partial(eq, item), iterable))
        return next(islice(indicies, n, None), -1)

The example is nice because it shows off how to effectively combine Python’s functional toolset. Note, that once the pipeline is set-up, there are no trips around Python’s eval loop — everything gets done at C speed, with a tiny memory footprint, with lazy evaluation, with no variable assignments, and with separately testable components. IOW, it is everything functional programmers dream about 🙂

Sample run:

>>> x = [False,True,True,False,True,False,True,False,False,False,True,False,True]
>>> nth_item(50, True, x)
>>> nth_item(0, True, x)
>>> nth_item(1, True, x)
>>> nth_item(2, True, x)
>>> nth_item(3, True, x)
Answered By: Raymond Hettinger
[y for y in enumerate(x) if y[1]==True][z][0]

Note: Here Z is the n’th occurance,

Answered By: avasal

A solution that first creates a list object and returns the nth-1 element of this list : function occurence()

And a solution that fulfill functional programmers’dreams too, I think, using generators, because I love them : function occur()

S = ' is a fantastic amazing site'
print 'object S is string %r' % S
print "indexes of 'a' in S :",[indx for indx,elem in enumerate(S) if elem=='a']

def occurence(itrbl,x,nth):
    return [indx for indx,elem in enumerate(itrbl)
            if elem==x ][nth-1] if x in itrbl 
           else None

def occur(itrbl,x,nth):
    return (i for pos,i in enumerate(indx for indx,elem in enumerate(itrbl)
                                     if elem==x)
            if pos==nth-1).next() if x in itrbl
            else   None

print "noccurence(S,'a',4th) ==",occurence(S,'a',4)
print "noccur(S,'a',4th) ==",occur(S,'a',4)


object S is string ' is a fantastic amazing site'
indexes of 'a' in S : [2, 21, 24, 27, 33, 35]

occur(S,'a',4th) == 27

occurence(S,'a',4th) == 27

The second solution seems complex but it isn’t really. It doesn’t need to run completely through the iterable: the process stops as soon as the wanted occurence is found.

Answered By: eyquem

Here is another way to find the nth occurrence of x in a list itrbl:

def nthoccur(nth,x,itrbl):
    count,index = 0,0
    while count < nth:
        if index > len(itrbl) - 1:
            return None
        elif itrbl[index] == x:
            count += 1
            index += 1
            index += 1
    return index - 1
Answered By: apolune

here is a way :
for the example above :


we can define a function find_index

def find_index(lst, value, n):
    for element in lst :
          if element == value :
              c .append (i)
    return c[n]

and if we apply the function :

nth_index = find_index(x, True, 4)
print nth_index

the result is:

Answered By: mzn.rft

I think this should work.

def get_nth_occurrence_of_specific_term(my_list, term, n):
    assert type(n) is int and n > 0
    start = -1
    for i in range(n):
        if term not in my_list[start + 1:]:
            return -1
        start = my_list.index(term, start + 1)
    return start
Answered By: Johnny Woo

You can use next with enumerate and a generator expression. itertools.islice allows you to slice an iterable as required.

from itertools import islice

x = [False,True,True,False,True,False,True,False,False,False,True,False,True]

def get_nth_index(L, val, n):
    """return index of nth instance where value in list equals val"""
    return next(islice((i for i, j in enumerate(L) if j == val), n-1, n), -1)

res = get_nth_index(x, True, 3)  # 4

If the iterator is exhausted, i.e. the nth occurrence of the specified value doesn’t exist, next can return a default value, in this instance -1:

Answered By: jpp

You could use count:

from itertools import count

x = [False, True, True, False, True, False, True, False, False, False, True, False, True]

def nth_index(n, item, iterable):
    counter = count(1)
    return next((i for i, e in enumerate(iterable) if e == item and next(counter) == n), -1)

print(nth_index(3, True, x))



The idea is that due to the short-circuit nature of e == item and next(counter) == n), the expression next(counter) == n only gets evaluated when e == item so you are only counting the elements equals to item.

Answered By: Dani Mesejo