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Finding the index of an item in a list

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index 1?

Are you returning: [1] The lowest index in case there are multiple instances of "bar", [2] All the indices of "bar"?
a) Is it guaranteed that item is in the list, or else how we should handle the error case? (return None/ raise ValueError) b) Are list entries guaranteed to be unique, and should we return the first index of a match, or all indexes?
View the answers with numpy integration, numpy arrays are far more efficient than Python lists. If the list is short it's no problem making a copy of it from a Python list, if it isn't then perhaps you should consider storing the elements in numpy array in the first place.

C
Community
>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can't remember the last time I used it in anger. It's been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]]) Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item. The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don't know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514
 

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they're more generalizable. So if you're considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item's not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

Check for it first with item in my_list (clean, readable approach), or Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)


index returns the first item whose value is "bar". If "bar" exists twice at list, you'll never find the key for the second "bar". See documentation: docs.python.org/3/tutorial/datastructures.html
If you're only searching for one element (the first), I found that index() is just under 90% faster than list comprehension against lists of integers.
What data structure should be used if the list is very long?
@izhang: Some auxillary index, like an {element -> list_index} dict, if the elements are hashable, and the position in the list matters.
sequence1 = sorted(sequence2, key=.sequence3.index) is a very handy idiom. You may use index more often if that's in your repertoire.
d
davidavr

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
 ...

 |
 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value
 |

which will often lead you to the method you are looking for.


bpython is a nice user-friendly way to read the docs in an interactive fashion.
@davidavr yes, but then the rest of us who just want to google it instead of scrolling through the help docs wouldn't have this nice, central, ranked set of options. :)
B
Boris Verkhovskiy

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(['foo', 'bar', 'baz']):
    if j == 'bar':
        print(i)

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar']

Here's also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ['foo', 'bar', 'baz']) if j == 'bar']

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(['foo', 'bar', 'baz']*500) if j == 'bar']"
10000 loops, best of 3: 196 usec per loop

Enumeration works better than the index-based methods for me, since I'm looking to gather the indices of strings using 'startswith" , and I need to gather multiple occurrences. Or is there a way to use index with "startswith" that I couldn't figure out
In my hands, the enumerate version is consistently slightly faster. Some implementation details may have changed since the measurement above was posted.
This was already answered since '11: stackoverflow.com/questions/6294179/…
A
Arkistarvh Kltzuonstev

To get all indexes:

indexes = [i for i,x in enumerate(xs) if x == 'foo']

There's already another question for this, added in '11: stackoverflow.com/questions/6294179/…
H
HongboZhu

index() returns the first index of value!

| index(...) | L.index(value, [start, [stop]]) -> integer -- return first index of value

def all_indices(value, qlist):
    indices = []
    idx = -1
    while True:
        try:
            idx = qlist.index(value, idx+1)
            indices.append(idx)
        except ValueError:
            break
    return indices

all_indices("foo", ["foo","bar","baz","foo"])

And if doesn't exist in the list?
Not-exist item will raise ValueError
This answer would fit better here: stackoverflow.com/questions/6294179/…
t
tanzil

A problem will arise if the element is not in the list. This function handles the issue:

# if element is found it returns index of element else returns None

def find_element_in_list(element, list_element):
    try:
        index_element = list_element.index(element)
        return index_element
    except ValueError:
        return None

s
savinson
a = ["foo","bar","baz",'bar','any','much']

indexes = [index for index in range(len(a)) if a[index] == 'bar']

u
user3670684

You have to set a condition to check if the element you're searching is in the list

if 'your_element' in mylist:
    print mylist.index('your_element')
else:
    print None

This helps us to avoid try catch!
However, it might double the complexity. Did anybody check?
@stefanct Time complexity is still linear but it will iterate through the list twice.
@ApproachingDarknessFish That is obviously what I meant. Even if pedantically it is the same order of complexity, iterating twice might be a severe disadvantage in many use cases thus I brought it up. And we still don't know the answer...
@stefanct this likely does double the complexity, I believe the in operator on a list has linear runtime. @ApproachingDarknessFish stated it would iterate twice which answers your question, and is right in saying that doubling the linear complexity is not a huge deal. I wouldn't call iterating over a list twice a severe disadvantage in many use cases, as complexity theory tells us that O(n) + O(n) -> O(2*n) -> O(n), ie- the change is typically neglibile.
P
Peter Mortensen

If you want all indexes, then you can use NumPy:

import numpy as np

array = [1, 2, 1, 3, 4, 5, 1]
item = 1
np_array = np.array(array)
item_index = np.where(np_array==item)
print item_index
# Out: (array([0, 2, 6], dtype=int64),)

It is clear, readable solution.


What about lists of strings, lists of non-numeric objects, etc... ?
This answer should be better posted here: stackoverflow.com/questions/6294179/…
This is the best one I have read. numpy arrays are far more efficient than Python lists. If the list is short it's no problem making a copy of it from a Python list, if it isn't then perhaps the developer should consider storing the elements in numpy array in the first place.
P
Peter Badida

All of the proposed functions here reproduce inherent language behavior but obscure what's going on.

[i for i in range(len(mylist)) if mylist[i]==myterm]  # get the indices

[each for each in mylist if each==myterm]             # get the items

mylist.index(myterm) if myterm in mylist else None    # get the first index and fail quietly

Why write a function with exception handling if the language provides the methods to do what you want itself?


The 3rd method iterates twice over the list, right?
Re: "All of the proposed functions here": At the time of writing perhaps, but you ought to check newer answers to see if it is still true.
C
Community

Finding the index of an item given a list containing it in Python For a list ["foo", "bar", "baz"] and an item in the list "bar", what's the cleanest way to get its index (1) in Python?

Well, sure, there's the index method, which returns the index of the first occurrence:

>>> l = ["foo", "bar", "baz"]
>>> l.index('bar')
1

There are a couple of issues with this method:

if the value isn't in the list, you'll get a ValueError

if more than one of the value is in the list, you only get the index for the first one

No values

If the value could be missing, you need to catch the ValueError.

You can do so with a reusable definition like this:

def index(a_list, value):
    try:
        return a_list.index(value)
    except ValueError:
        return None

And use it like this:

>>> print(index(l, 'quux'))
None
>>> print(index(l, 'bar'))
1

And the downside of this is that you will probably have a check for if the returned value is or is not None:

result = index(a_list, value)
if result is not None:
    do_something(result)

More than one value in the list

If you could have more occurrences, you'll not get complete information with list.index:

>>> l.append('bar')
>>> l
['foo', 'bar', 'baz', 'bar']
>>> l.index('bar')              # nothing at index 3?
1

You might enumerate into a list comprehension the indexes:

>>> [index for index, v in enumerate(l) if v == 'bar']
[1, 3]
>>> [index for index, v in enumerate(l) if v == 'boink']
[]

If you have no occurrences, you can check for that with boolean check of the result, or just do nothing if you loop over the results:

indexes = [index for index, v in enumerate(l) if v == 'boink']
for index in indexes:
    do_something(index)

Better data munging with pandas

If you have pandas, you can easily get this information with a Series object:

>>> import pandas as pd
>>> series = pd.Series(l)
>>> series
0    foo
1    bar
2    baz
3    bar
dtype: object

A comparison check will return a series of booleans:

>>> series == 'bar'
0    False
1     True
2    False
3     True
dtype: bool

Pass that series of booleans to the series via subscript notation, and you get just the matching members:

>>> series[series == 'bar']
1    bar
3    bar
dtype: object

If you want just the indexes, the index attribute returns a series of integers:

>>> series[series == 'bar'].index
Int64Index([1, 3], dtype='int64')

And if you want them in a list or tuple, just pass them to the constructor:

>>> list(series[series == 'bar'].index)
[1, 3]

Yes, you could use a list comprehension with enumerate too, but that's just not as elegant, in my opinion - you're doing tests for equality in Python, instead of letting builtin code written in C handle it:

>>> [i for i, value in enumerate(l) if value == 'bar']
[1, 3]

Is this an XY problem?

The XY problem is asking about your attempted solution rather than your actual problem.

Why do you think you need the index given an element in a list?

If you already know the value, why do you care where it is in a list?

If the value isn't there, catching the ValueError is rather verbose - and I prefer to avoid that.

I'm usually iterating over the list anyways, so I'll usually keep a pointer to any interesting information, getting the index with enumerate.

If you're munging data, you should probably be using pandas - which has far more elegant tools than the pure Python workarounds I've shown.

I do not recall needing list.index, myself. However, I have looked through the Python standard library, and I see some excellent uses for it.

There are many, many uses for it in idlelib, for GUI and text parsing.

The keyword module uses it to find comment markers in the module to automatically regenerate the list of keywords in it via metaprogramming.

In Lib/mailbox.py it seems to be using it like an ordered mapping:

key_list[key_list.index(old)] = new

and

del key_list[key_list.index(key)]

In Lib/http/cookiejar.py, seems to be used to get the next month:

mon = MONTHS_LOWER.index(mon.lower())+1

In Lib/tarfile.py similar to distutils to get a slice up to an item:

members = members[:members.index(tarinfo)]

In Lib/pickletools.py:

numtopop = before.index(markobject)

What these usages seem to have in common is that they seem to operate on lists of constrained sizes (important because of O(n) lookup time for list.index), and they're mostly used in parsing (and UI in the case of Idle).

While there are use-cases for it, they are fairly uncommon. If you find yourself looking for this answer, ask yourself if what you're doing is the most direct usage of the tools provided by the language for your use-case.


S
Sociopath

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"))

Output

[1, 3, 5, 7]

Simple

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

Output:

0
4

This answer should be better posted here: stackoverflow.com/questions/6294179/…
P
Peter Mortensen

All indexes with the zip function:

get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]

print get_indexes(2, [1, 2, 3, 4, 5, 6, 3, 2, 3, 2])
print get_indexes('f', 'xsfhhttytffsafweef')

This answer should be better posted here: stackoverflow.com/questions/6294179/…
k
kiriloff

Simply you can go with

a = [['hand', 'head'], ['phone', 'wallet'], ['lost', 'stock']]
b = ['phone', 'lost']

res = [[x[0] for x in a].index(y) for y in b]

M
Mathitis2Software

Another option

>>> a = ['red', 'blue', 'green', 'red']
>>> b = 'red'
>>> offset = 0;
>>> indices = list()
>>> for i in range(a.count(b)):
...     indices.append(a.index(b,offset))
...     offset = indices[-1]+1
... 
>>> indices
[0, 3]
>>> 

This answer should be better posted here: stackoverflow.com/questions/6294179/…
A
Ayeni Lawrence
me = ["foo", "bar", "baz"]
me.index("bar") 

You can apply this for any member of the list to get their index


M
MrWonderful

And now, for something completely different...

... like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list. It works with strings as well.

def indices(l, val):
    """Always returns a list containing the indices of val in the_list"""
    retval = []
    last = 0
    while val in l[last:]:
            i = l[last:].index(val)
            retval.append(last + i)
            last += i + 1   
    return retval

l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')

When pasted into an interactive python window:

Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(the_list, val):
...     """Always returns a list containing the indices of val in the_list"""
...     retval = []
...     last = 0
...     while val in the_list[last:]:
...             i = the_list[last:].index(val)
...             retval.append(last + i)
...             last += i + 1   
...     return retval
... 
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>> 

Update

After another year of heads-down python development, I'm a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.

Something like this:

def indices(l, val):
    """Always returns a list containing the indices of val in the_list"""
    return [index for index, value in enumerate(l) if value == val]

l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')

Which, when pasted into an interactive python window yields:

Python 2.7.14 |Anaconda, Inc.| (default, Dec  7 2017, 11:07:58) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(l, val):
...     """Always returns a list containing the indices of val in the_list"""
...     return [index for index, value in enumerate(l) if value == val]
... 
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>> 

And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn't even see that answer, because I didn't understand it. I hope that my somewhat more verbose example will aid understanding.

If the single line of code above still doesn't make sense to you, I highly recommend you Google 'python list comprehension' and take a few minutes to familiarize yourself. It's just one of the many powerful features that make it a joy to use Python to develop code.


b
bvanlew

A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:

>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>> 

You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.


This answer should be better posted here: stackoverflow.com/questions/6294179/…
L
LunaticXXD10

My friend, I have made the easiest code to solve your question. While you were receiving gigantic lines of codes, I am here to cater you a two line code which is all due to the help of index() function in python.

LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))

Output:

1

I Hope I have given you the best and the simplest answer which might help you greatly.


K
Ketan

Finding index of item x in list L:

idx = L.index(x) if (x in L) else -1

This iterates the array twice, thus it could result in performance issues for large arrays.
t
totallyhuman

This solution is not as powerful as others, but if you're a beginner and only know about forloops it's still possible to find the first index of an item while avoiding the ValueError:

def find_element(p,t):
    i = 0
    for e in p:
        if e == t:
            return i
        else:
            i +=1
    return -1

B
Badri Paudel

There is a chance that that value may not be present so to avoid this ValueError, we can check if that actually exists in the list .

list =  ["foo", "bar", "baz"]

item_to_find = "foo"

if item_to_find in list:
      index = list.index(item_to_find)
      print("Index of the item is " + str(index))
else:
    print("That word does not exist") 

G
Giacomo Casadei

It just uses the python function array.index() and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function indexOf()).

fruits = ['apple', 'banana', 'cherry']

try:
  pos = fruits.index("mango")
except:
  pos = -1

In this case "mango" is not present in the list fruits so the pos variable is -1, if I had searched for "cherry" the pos variable would be 2.


P
Peter Mortensen

There is a more functional answer to this.

list(filter(lambda x: x[1]=="bar",enumerate(["foo", "bar", "baz", "bar", "baz", "bar", "a", "b", "c"])))

More generic form:

def get_index_of(lst, element):
    return list(map(lambda x: x[0],\
       (list(filter(lambda x: x[1]==element, enumerate(lst))))))

This answer feels at home for Scala / functional-programming enthusiasts
When only a single value is needed in a list that has many matches this one takes long.
A
Arkistarvh Kltzuonstev

Python index() method throws an error if the item was not found. So instead you can make it similar to the indexOf() function of JavaScript which returns -1 if the item was not found:

try:
    index = array.index('search_keyword')
except ValueError:
    index = -1

however, JavaScript has the philosophy that weird results are better than errors, so it makes sense to return -1, but in Python, it can make a hard to track down bug, since -1 returns an item from the end of the list.
S
Sociopath
name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
new_list=[]
for item in list:
    new_list.append(item[0])
print(new_list)
try:
    location= new_list.index(name)
except:
    location=-1
print (location)

This accounts for if the string is not in the list too, if it isn't in the list then location = -1


C
Caveman

For one comparable

# Throws ValueError if nothing is found
some_list = ['foo', 'bar', 'baz'].index('baz')
# some_list == 2

Custom predicate

some_list = [item1, item2, item3]

# Throws StopIteration if nothing is found
# *unless* you provide a second parameter to `next`
index_of_value_you_like = next(
    i for i, item in enumerate(some_list)
    if item.matches_your_criteria())

Finding index of all items by predicate

index_of_staff_members = [
    i for i, user in enumerate(users)
    if user.is_staff()]

idx = next((i for i, v in enumerate(ls) if v == chk), -1) to get the behavior similar to str.index(chk).
@tejasvi88 Decided to put some extra work into the answer
K
Kofi

List comprehension would be the best option to acquire a compact implementation in finding the index of an item in a list.

a_list = ["a", "b", "a"]
print([index for (index , item) in enumerate(a_list) if item == "a"])

P
Peter Mortensen

Since Python lists are zero-based, we can use the zip built-in function as follows:

>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]

where "haystack" is the list in question and "needle" is the item to look for.

(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)


[i for i,j in enumerate(haystack) if j==‘needle’] is more compact and readable, I think.
F
FatihAkici

If performance is of concern:

It is mentioned in numerous answers that the built-in method of list.index(item) method is an O(n) algorithm. It is fine if you need to perform this once. But if you need to access the indices of elements a number of times, it makes more sense to first create a dictionary (O(n)) of item-index pairs, and then access the index at O(1) every time you need it.

If you are sure that the items in your list are never repeated, you can easily:

myList = ["foo", "bar", "baz"]

# Create the dictionary
myDict = dict((e,i) for i,e in enumerate(myList))

# Lookup
myDict["bar"] # Returns 1
# myDict.get("blah") if you don't want an error to be raised if element not found.

If you may have duplicate elements, and need to return all of their indices:

from collections import defaultdict as dd
myList = ["foo", "bar", "bar", "baz", "foo"]

# Create the dictionary
myDict = dd(list)
for i,e in enumerate(myList):
    myDict[e].append(i)

# Lookup
myDict["foo"] # Returns [0, 4]