Finding the index of an item given a list containing it in Python

Eugene M Source

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?



answered 9 years ago Alex Coventry #1

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

Reference: Data Structures > More on Lists

answered 9 years ago davidavr #2

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.

answered 7 years ago HongboZhu #3

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:
            idx = qlist.index(value, idx+1)
        except ValueError:
    return indices

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

answered 6 years ago savinson #4

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

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

answered 5 years ago tanzil #5

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):
        index_element = list_element.index(element)
        return index_element
    except ValueError:
        return None

answered 5 years ago Graham Giller #6

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?

answered 5 years ago octoback #7

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]

answered 5 years ago Mathitis2Software #8

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]

answered 5 years ago TerryA #9

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':

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

answered 5 years ago FMc #10

To get all indexes:

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

answered 4 years ago bvanlew #11

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']
>>> 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.

answered 4 years ago user3670684 #12

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')
    print None

answered 3 years ago MrWonderful #13

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]

answered 3 years ago dylankb #14

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
            i +=1
    return -1

answered 3 years ago Coder123 #15

name ="bar"
list = [["foo", 1], ["bar", 2], ["baz", 3]]
for item in list:
    location= new_list.index(name)
print (location)

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

answered 2 years ago Arnaldo P. Figueira Figueira #16

all indexes with 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')

answered 2 years ago rbrisuda #17

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.

answered 7 months ago Giovanni Gianni #18

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 arg and return the position of the item in the liste, like we saw before.

def findindex(item2find,listOrString):
  "Search indexes of an item (arg.1) contained in a list or a string (arg.2)"
  return [n for n,item in enumerate(listOrString) if item==item2find]

indexes1 = findindex("1","010101010")


[1, 3, 5, 7]

answered 7 months ago jihed gasmi #19

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)

answered 7 months ago Aaron Hall #20

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')

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):
        return a_list.index(value)
    except ValueError:
        return None

And use it like this:

>>> print(index(l, 'quux'))
>>> print(index(l, 'bar'))

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:

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?

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:

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/ it seems to be using it like an ordered mapping:

key_list[key_list.index(old)] = new


del key_list[key_list.index(key)]

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

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

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

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

In Lib/

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.

answered 2 months ago mpoletto #21

For those caming from another language like me, maybe with a simple loop it's easier to understand and use it:

mylist = ["foo", "bar", "baz", "bar"]
newlist = enumerate(mylist)
for index, item in newlist:
  if item == "bar":
    print(index, item)

Thankfull for, that helped me to understand.

answered 2 weeks ago Hamed Baatour #22

Python index() method throws an error if the item was not found, which sucks!

So instead you can make it similar to the indexOf() function of Javascript which returns -1 if the item was not found:

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

answered 2 weeks ago Ankit Gupta #23

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))))))

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