How can I find the index for a given item in a list?

Question

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

• Ṃųỻịgǻňạcểơửṩ: Are you returning: [1] The lowest index in case there are multiple instances of "bar", [2] All the indices of "bar"?
• smci: 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?
• Athanassios: 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.
• Karl Knechtel: I’m voting to close this question (in protest) because there are already 42 undeleted answers (and 16 more deleted) for a simple, one-liner reference question that almost all have the same built-in function at their core (as they should, because it's the only reasonable and sane approach to the problem and everything surrounding it is just error-checking or creatively re-interpreting the specification, which still only leaves one other reasonable, sane approach to the expanded problem).
• Karl Knechtel: There is no realistic chance of a better approach becoming possible in future versions of Python, because the existing approach is already just calling a single, built-in method on the list - as simple as it gets.

Answer 1

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

See the documentation for the built-in .index() method of the list:

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.

Caveats

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.

This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the start and end parameters can be used to narrow the search.

For example:

>>> 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

The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.

Only the index of the first match is returned

A call to index searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, index cannot solve the problem:

>>> [1, 1].index(1) # the `1` index is not found.
0

Instead, use a list comprehension or generator expression to do the search, with enumerate to get indices:

>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2

The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.

Raises an exception if there is no match

As noted in the documentation above, using .index will raise an exception if the searched-for value is not in the list:

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

If this is a concern, either explicitly check first using item in my_list, or handle the exception with try/except as appropriate.

The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.

• mirapole: 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
• slybloty: 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.
• izhang05: What data structure should be used if the list is very long?
• Alex Coventry: @izhang: Some auxillary index, like an {element -> list_index} dict, if the elements are hashable, and the position in the list matters.
• Alex Coventry: @jvel07, see the list/generator comprehension examples in my answer.

Answer 2

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

• Tupelo Thistlehead: 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
• Alex Coventry: In my hands, the enumerate version is consistently slightly faster. Some implementation details may have changed since the measurement above was posted.
• Cristik: This was already answered since '11: stackoverflow.com/questions/6294179/…
• UnusualWays: In Python 3 izip should be replaced by the built in zip. See here
• Brian Keith: This is a good solution and it's much more flexible than the accepted solution. For example, if you only are expecting to have 1 value in the list, you can add a if statement to raise an exception if len([i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar']) > 1 otherwise you could just return [i for i, j in enumerate(['foo', 'bar', 'baz']) if j == 'bar'][0]

Answer 3

To get all indexes:

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

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

Answer 4

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"])

• Peter Mortensen: And if doesn't exist in the list?
• Nam G VU: Not-exist item will raise ValueError
• Cristik: This answer would fit better here: stackoverflow.com/questions/6294179/…

Answer 5

a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']

Answer 6

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

# If 'element' is found, it returns the index of 'element', else it returns 'None'
def find_element_in_list(element, list_element):
 try:
 index_element = list_element.index(element)
 return index_element
 except ValueError:
 return None

Answer 7

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

• devssh: This helps us to avoid try catch!
• stefanct: However, it might double the complexity. Did anybody check?
• ApproachingDarknessFish: @stefanct Time complexity is still linear but it will iterate through the list twice.
• stefanct: @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...
• Matthew Strasiotto: @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.

Answer 8

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.

• András Aszódi: What about lists of strings, lists of non-numeric objects, etc... ?
• Cristik: This answer should be better posted here: stackoverflow.com/questions/6294179/…
• Athanassios: 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.

Answer 9

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.

Answer 10

All of the proposed functions here reproduce inherent language behavior, but it 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?

• Eric Duminil: The 3rd method iterates twice over the list, right?
• Peter Mortensen: 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.

Answer 11

me = ["foo", "bar", "baz"]
me.index("bar") 

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

Answer 12

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

• Cristik: This answer should be better posted here: stackoverflow.com/questions/6294179/…

Answer 13

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')

• Cristik: This answer should be better posted here: stackoverflow.com/questions/6294179/…
• ggorlen: enumerate(xs) is clearer than zip(xs, range(len(xs)). Also, this doesn't answer the question.

Answer 14

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]

Answer 15

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]
>>> 

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]
>>> 

(This is exactly what FMc suggested in his earlier answer. 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 research list comprehensions 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.

Answer 16

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]
>>> 

• Cristik: This answer should be better posted here: stackoverflow.com/questions/6294179/…
• ggorlen: What does this have to do with the question?

Answer 17

Here's a two-liner using Python's index() function:

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

Output: 1

Answer 18

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:

def indexof(array, elem):
try:
 return array.index(elem)
except ValueError:
 return -1

• Sapphire_Brick: 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.
• Sergio Abreu: -1 is not a weird result in java/javascript. It is a language convenction of "not found in list". It is possible to use this java intelligence in Python doing a simple verification: if theindex > -1: or if theindex >= 0: which does the same.

Answer 19

A variant on the answers from FMc 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.

• Cristik: This answer should be better posted here: stackoverflow.com/questions/6294179/…

Answer 20

Finding index of item x in list L:

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

• Cristik: This iterates the array twice, thus it could result in performance issues for large arrays.
• Ketan: @Cristik - Correct. Not suitable if there is no reasonably low upper bound available for the list length.
• Ketan: Should be used only for non-repetitive tasks/deployments, or if the list length is relatively small enough to not affect overall performance noticeably.

Answer 21

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

Answer 22

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") 

Expected output of above code: Index of the item is 0

A more efficient alternative way:

Since calling item_to_find in list and list.index(item_to_find) is doing two searches, it can be shortened with only one operation using try, except. Since it throws ValueError when the item is not in the list, we can use that message to print for user information to indicate what went wrong.

Following code can be used as an enhancement.

list = ["foo", "bar", "baz"]
item_to_find = "fooz"
try:
 index = list.index(item_to_find)
 print("Index of the item is " + str(index))
except ValueError as error:
 print(f'Item couldn\'t be found because {error}.')

Expected output of the above code is: Item couldn't be found because 'fooz' is not in list

• ggorlen: Calling a variable list overwrites a builtin function. Calling in, then index means you're doing two searches. Better to try/except .index() as suggested in other threads.

Answer 23

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"])

• Rolf of Saxony: Works nicely for integers and floats too, as well as finding all occurrences
• Peter Mortensen: Why is list comprehension the best option?

Answer 24

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 returns -1 if it is not found in the list (like in 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.

Answer 25

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))))))

• y2k-shubham: This answer feels at home for Scala / functional-programming enthusiasts
• Caveman: When only a single value is needed in a list that has many matches this one takes long.

Answer 26

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

Answer 27

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()]

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

Answer 28

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.)

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

Answer 29

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]

Answer 30

If you are going to find an index once then using "index" method is fine. However, if you are going to search your data more than once then I recommend using bisect module. Keep in mind that using bisect module data must be sorted. So you sort data once and then you can use bisect. Using bisect module on my machine is about 20 times faster than using index method.

Here is an example of code using Python 3.8 and above syntax:

import bisect
from timeit import timeit
def bisect_search(container, value):
 return (
 index 
 if (index := bisect.bisect_left(container, value)) < len(container) 
 and container[index] == value else -1
 )
data = list(range(1000))
# value to search
value = 666
# times to test
ttt = 1000
t1 = timeit(lambda: data.index(value), number=ttt)
t2 = timeit(lambda: bisect_search(data, value), number=ttt)
print(f"{t1=:.4f}, {t2=:.4f}, diffs {t1/t2=:.2f}")

Output:

t1=0.0400, t2=0.0020, diffs t1/t2=19.60

• ggorlen: "Using bisect module on my machine is about 20 times faster than using index method." is a somewhat inaccurate way to describe the relationship between the two algorithms. It's not a linear relationship, so on small lists of, say, 10 elements, both algorithms should perform about the same. On slightly larger lists, you may begin to notice a difference. On massive lists, binary search may be thousands of times faster.