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How to find all occurrences of an element in a list

index() will give the first occurrence of an item in a list. Is there a neat trick which returns all indices in a list for an element?

18 Answers 18

You can use a list comprehension with enumerate :

indices = [i for i, x in enumerate(my_list) if x == "whatever"] 

The iterator enumerate(my_list) yields pairs (index, item) for each item in the list. Using i, x as loop variable target unpacks these pairs into the index i and the list item x . We filter down to all x that match our criterion, and select the indices i of these elements.

While not a solution for lists directly, numpy really shines for this sort of thing:

import numpy as np values = np.array([1,2,3,1,2,4,5,6,3,2,1]) searchval = 3 ii = np.where(values == searchval)[0] 

This can be significantly faster for lists (arrays) with a large number of elements vs some of the other solutions.

@Hari I get different results from np.where([7, 8, 9, 8] == 8)[0] and np.where(np.array([7, 8, 9, 8]) == 8)[0] ; only the latter works as intended.

Indeed, @AttilatheFun. I am not able to refer to the piece of code that led me to think that numpy where works with list also. Casting as a numpy array is the correct and safe thing to do before using numpy where.

A solution using list.index :

def indices(lst, element): result = [] offset = -1 while True: try: offset = lst.index(element, offset+1) except ValueError: return result result.append(offset) 

It’s much faster than the list comprehension with enumerate , for large lists. It is also much slower than the numpy solution if you already have the array, otherwise the cost of converting outweighs the speed gain (tested on integer lists with 100, 1000 and 10000 elements).

NOTE: A note of caution based on Chris_Rands’ comment: this solution is faster than the list comprehension if the results are sufficiently sparse, but if the list has many instances of the element that is being searched (more than ~15% of the list, on a test with a list of 1000 integers), the list comprehension is faster.

This was a long time ago, I probably used timeit.timeit with randomly generated lists. That’s an important point though, and I suppose that may be why you ask. At the time it didn’t occur to me, but the speed gains are only true if the results are sufficiently sparse. I just tested with a list full of the element to search for, and it’s much slower than the list comprehension.

In [1]: l=[1,2,3,4,3,2,5,6,7] In [2]: [i for i,val in enumerate(l) if val==3] Out[2]: [2, 4] 

more_itertools.locate finds indices for all items that satisfy a condition.

from more_itertools import locate list(locate([0, 1, 1, 0, 1, 0, 0])) # [1, 2, 4] list(locate(['a', 'b', 'c', 'b'], lambda x: x == 'b')) # [1, 3] 

more_itertools is a third-party library > pip install more_itertools .

occurrences = lambda s, lst: (i for i,e in enumerate(lst) if e == s) list(occurrences(1, [1,2,3,1])) # = [0, 3] 

• There’s an answer using np.where to find the indices of a single value, which is not faster than a list-comprehension, if the time to convert a list to an array is included
• The overhead of importing numpy and converting a list to a numpy.array probably makes using numpy a less efficient option for most circumstances. A careful timing analysis would be necessary.
  • In cases where multiple functions/operations will need to be performed on the list , converting the list to an array , and then using numpy functions will likely be a faster option.
  • Using np.where on an array (including the time to convert the list to an array) is slightly slower than a list-comprehension on a list, for finding all indices of all unique elements.
  • This has been tested on an 2M element list with 4 unique values, and the size of the list/array and number of unique elements will have an impact.

  import numpy as np import random # to create test list # create sample list random.seed(365) l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(20)] # convert the list to an array for use with these numpy methods a = np.array(l) # create a dict of each unique entry and the associated indices idx = # print(idx)

  %timeit on a 2M element list with 4 unique str elements

  # create 2M element list random.seed(365) l = [random.choice(['s1', 's2', 's3', 's4']) for _ in range(2000000)] 

  Functions

  def test1(): # np.where: convert list to array and find indices of a single element a = np.array(l) return np.where(a == 's1') def test2(): # list-comprehension: on list l and find indices of a single element return [i for i, x in enumerate(l) if x == "s1"] def test3(): # filter: on list l and find indices of a single element return list(filter(lambda i: l[i]=="s1", range(len(l)))) def test4(): # use np.where and np.unique to find indices of all unique elements: convert list to array a = np.array(l) return def test5(): # list comprehension inside dict comprehension: on list l and find indices of all unique elements return

  Function Call

  %timeit test1() %timeit test2() %timeit test3() %timeit test4() %timeit test5() 

  Result python 3.10.4

  214 ms ± 19.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 85.1 ms ± 1.48 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 146 ms ± 1.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 365 ms ± 11.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 360 ms ± 5.82 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 

  Result python 3.11.0

  209 ms ± 15.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 70.4 ms ± 1.86 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 132 ms ± 4.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each) 371 ms ± 20.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 314 ms ± 15.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each) 

  l=[i for i in range(len(lst)) if lst[i]=='something. '] 

  l=[i for i in xrange(len(lst)) if lst[i]=='something. '] 

  Getting all the occurrences and the position of one or more (identical) items in a list

  With enumerate(alist) you can store the first element (n) that is the index of the list when the element x is equal to what you look for.

  >>> alist = ['foo', 'spam', 'egg', 'foo'] >>> foo_indexes = [n for n,x in enumerate(alist) if x=='foo'] >>> foo_indexes [0, 3] >>> 

  Let’s make our function findindex

  This function takes the item and the list as arguments and return the position of the item in the list, like we saw before.

  def indexlist(item2find, list_or_string): "Returns all indexes of an item in a list or a string" return [n for n,item in enumerate(list_or_string) if item==item2find] print(indexlist("1", "010101010")) 

  Simple

  for n, i in enumerate([1, 2, 3, 4, 1]): if i == 1: print(n) 

  One more solution(sorry if duplicates) for all occurrences:

  values = [1,2,3,1,2,4,5,6,3,2,1] map(lambda val: (val, [i for i in xrange(len(values)) if values[i] == val]), values) 

  >>> q = ['Yeehaw', 'Yeehaw', 'Googol', 'B9', 'Googol', 'NSM', 'B9', 'NSM', 'Dont Ask', 'Googol'] >>> filter(lambda i: q[i]=="Googol", range(len(q))) [2, 4, 9] 

  If you need to search for all element’s positions between certain indices, you can state them:

  [i for i,x in enumerate([1,2,3,2]) if x==2 & 2 [3] 

  You can create a defaultdict

  from collections import defaultdict d1 = defaultdict(int) # defaults to 0 values for keys unq = set(lst1) # lst1 = [1, 2, 2, 3, 4, 1, 2, 7] for each in unq: d1[each] = lst1.count(each) else: print(d1) 

  A dynamic list comprehension based solution incase we do not know in advance which element:

  You can think of all other ways along the same lines as well but with index() you can find only one index although you can set occurrence number yourself.

  Using a for-loop :

  • Answers with enumerate and a list comprehension are more pythonic, but not necessarily faster. However, this answer is aimed at students who may not be allowed to use some of those built-in functions.
  • create an empty list, indices
  • create the loop with for i in range(len(x)): , which essentially iterates through a list of index locations [0, 1, 2, 3, . len(x)-1]
  • in the loop, add any i , where x[i] is a match to value , to indices
    • x[i] accesses the list by index

    def get_indices(x: list, value: int) -> list: indices = list() for i in range(len(x)): if x[i] == value: indices.append(i) return indices n = [1, 2, 3, -50, -60, 0, 6, 9, -60, -60] print(get_indices(n, -60)) >>> [4, 8, 9] 

    • The functions, get_indices , are implemented with type hints. In this case, the list, n , is a bunch of int s, therefore we search for value , also defined as an int .

    Using a while-loop and .index :

    def get_indices(x: list, value: int) -> list: indices = list() i = 0 while True: try: # find an occurrence of value and update i to that index i = x.index(value, i) # add i to the list indices.append(i) # advance i by 1 i += 1 except ValueError as e: break return indices print(get_indices(n, -60)) >>> [4, 8, 9] 

    Your self-define get_indeices is a bit faster(~15%) than normal list comprehension. I am trying to figure it out.

    If you are using Python 2, you can achieve the same functionality with this:

    f = lambda my_list, value:filter(lambda x: my_list[x] == value, range(len(my_list))) 

    Where my_list is the list you want to get the indexes of, and value is the value searched. Usage:

    Create a generator

    Generators are fast and use a tiny memory footprint. They give you flexibility in how you use the result.

    def indices(iter, val): """Generator: Returns all indices of val in iter Raises a ValueError if no val does not occur in iter Passes on the AttributeError if iter does not have an index method (e.g. is a set) """ i = -1 NotFound = False while not NotFound: try: i = iter.index(val, i+1) except ValueError: NotFound = True else: yield i if i == -1: raise ValueError("No occurrences of in ".format(v = val, i = iter)) 

    The above code can be use to create a list of the indices: list(indices(input,value)) ; use them as dictionary keys: dict(indices(input,value)) ; sum them: sum(indices(input,value)) ; in a for loop for index_ in indices(input,value): ; . etc. without creating an interim list/tuple or similar.

    In a for loop you will get your next index back when you call for it, without waiting for all the others to be calculated first. That means: if you break out of the loop for some reason you save the time needed to find indices you never needed.

    How it works

    • Call .index on the input iter to find the next occurrence of val
    • Use the second parameter to .index to start at the point after the last found occurrence
    • Yield the index
    • Repeat until index raises a ValueError

    Alternative versions

    I tried four different versions for flow control; two EAFP (using try — except ) and two TBYL (with a logical test in the while statement):

    1. «WhileTrueBreak»: while True: . except ValueError: break . Surprisingly, this was usually a touch slower than option 2 and (IMV) less readable
    2. «WhileErrFalse»: Using a bool variable err to identify when a ValueError is raised. This is generally the fastest and more readable than 1
    3. «RemainingSlice»: Check whether val is in the remaining part of the input using slicing: while val in iter[i:] . Unsurprisingly, this does not scale well
    4. «LastOccurrence»: Check first where the last occurrence is, keep going while i < last

    The overall performance differences between 1,2 and 4 are negligible, so it comes down to personal style and preference. Given that .index uses ValueError to let you know it didn’t find anything, rather than e.g. returning None , an EAFP-approach seems fitting to me.

    Here are the 4 code variants and results from timeit (in milliseconds) for different lengths of input and sparsity of matches

    @version("WhileTrueBreak", versions) def indices2(iter, val): i = -1 while True: try: i = iter.index(val, i+1) except ValueError: break else: yield i @version("WhileErrFalse", versions) def indices5(iter, val): i = -1 err = False while not err: try: i = iter.index(val, i+1) except ValueError: err = True else: yield i @version("RemainingSlice", versions) def indices1(iter, val): i = 0 while val in iter[i:]: i = iter.index(val, i) yield i i += 1 @version("LastOccurrence", versions) def indices4(iter,val): i = 0 last = len(iter) - tuple(reversed(iter)).index(val) while i < last: i = iter.index(val, i) yield i i += 1 

    Length: 100, Ocurrences: 4.0% Length: 1000, Ocurrences: 1.2% Length: 10000, Ocurrences: 2.05% Length: 100000, Ocurrences: 1.977% Length: 100000, Ocurrences: 9.873% Length: 100000, Ocurrences: 25.058% Length: 100000, Ocurrences: 49.698%

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