By Eugene M

2008-10-07 01:39:38 8 Comments

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


@Badri Paudel 2020-08-28 09:35:53

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))
    print("That word does not exist") 

@Caveman 2020-06-22 16:02:01

If you need only one this one reads well...

some_list = ['list', 'of', 'something you like']

# StopIteration is raised if it doesn't find
index_of_value_you_like = next(
    for i, value in enumerate(some_list)
    if value == 'something you like')

... but for all indices then...

all_values_that_match_your_criteria = [
    for i, instance in enumerate(some_list_of_classes)
    if instance.matches_your_criteria()]

@Vlad Bezden 2020-04-03 16:50:22

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 (
      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}")


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

@Coder123 2015-07-05 13:12:19

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

@Alex Coventry 2008-10-07 01:40:49

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

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

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)
>>> [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)
>>> next(g)

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

  1. Check for it first with item in my_list (clean, readable approach), or
  2. 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.)

@mpoletto 2018-01-30 04:51:44

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:

@slybloty 2019-09-19 20:13:53

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 2020-02-22 20:36:31

What data structure should be used if the list is very long?

@Alex Coventry 2020-02-24 04:30:03

@izhang: Some auxillary index, like an {element -> list_index} dict, if the elements are hashable, and the position in the list matters.

@TerryA 2013-06-19 22:31:52

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

@Tupelo Thistlehead 2017-10-26 19:15:53

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 2017-11-17 18:43:32

In my hands, the enumerate version is consistently slightly faster. Some implementation details may have changed since the measurement above was posted.

@Cristik 2019-02-10 10:55:37

This was already answered since '11:…

@Viv 2020-07-31 16:48:15

such a saver of an answer! thank you!!.. I tried for 2 days to get the index of a nested dictionary before understanding we could use enumerate

@Hamed Baatour 2018-03-04 08:39:34

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:

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

@Sapphire_Brick 2019-10-29 22:44:21

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.

@FMc 2013-06-25 15:07:55

To get all indexes:

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

@Cristik 2019-02-10 10:56:16

There's already another question for this, added in '11:…

@sahasrara62 2019-03-18 09:32:09

using dictionary , where process the list first and then add the index to it

from collections import defaultdict

index_dict = defaultdict(list)    
word_list =  ['foo','bar','baz','bar','any', 'foo', 'much']

for word_index in range(len(word_list)) :

word_index_to_find = 'foo'       

# output :  [0, 5]

@Ketan 2018-05-25 21:56:07

Finding index of item x in list L:

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

@Cristik 2019-02-10 11:00:50

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

@Ketan 2019-02-25 08:13:33

Agree. If lists are considerably long, I'd go for something else. Shouldn't be a big deal for small to medium sized lists though.

@Siddharth Satpathy 2018-11-14 18:49:26

Let’s give the name lst to the list that you have. One can convert the list lst to a numpy array. And, then use numpy.where to get the index of the chosen item in the list. Following is the way in which you will implement it.

import numpy as np

lst = ["foo", "bar", "baz"]  #lst: : 'list' data type
print np.where( np.array(lst) == 'bar')[0][0]

>>> 1

@Caveman 2020-06-22 16:11:18

Does not work if the item is an instance of a class

@Giovanni G. PY 2017-08-08 05:01:01

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"))


[1, 3, 5, 7]


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



@Cristik 2019-02-10 10:59:07

This answer should be better posted here:…

@pylang 2018-09-25 15:47:56

As indicated by @TerryA, many answers discuss how to find one index.

more_itertools is a third-party library with tools to locate multiple indices within an iterable.


import more_itertools as mit

iterable = ["foo", "bar", "baz", "ham", "foo", "bar", "baz"]


Find indices of multiple observations:

list(mit.locate(iterable, lambda x: x == "bar"))
# [1, 5]

Test multiple items:

list(mit.locate(iterable, lambda x: x in {"bar", "ham"}))
# [1, 3, 5]

See also more options with more_itertools.locate. Install via > pip install more_itertools.

@FatihAkici 2018-09-10 18:51:04

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):

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

@MrWonderful 2014-12-30 21:03:10

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]


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.

@jihed gasmi 2017-08-12 20:01:08

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.)

@Giovanni G. PY 2017-12-27 07:23:34

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

@mpoletto 2018-01-30 21:10:12

For those coming 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)

I am thankful for So what exactly does enumerate do?. That helped me to understand.

@Ankit Gupta 2018-03-07 19:09:44

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 2018-08-21 05:13:12

This answer feels at home for Scala / functional-programming enthusiasts

@Caveman 2020-06-22 16:12:23

When only a single value is needed in a list that has many matches this one takes long.

@Arnaldo P. Figueira Figueira 2015-11-11 05:16:38

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 2019-02-10 10:58:40

This answer should be better posted here:…

@rbrisuda 2015-11-17 19:05:23

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.

@Laryx Decidua 2016-10-12 14:55:58

What about lists of strings, lists of non-numeric objects, etc... ?

@Cristik 2019-02-10 10:58:13

This answer should be better posted here:…

@Athanassios 2020-01-28 12:23:12

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.

@dylankb 2015-05-17 03:21:00

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

@savinson 2012-08-21 12:01:54

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

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

@Cristik 2019-02-10 10:57:26

This answer should be better posted here:…

@Aaron Hall 2017-08-22 03:08:59

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.

@tanzil 2013-04-16 10:19:36

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

@Graham Giller 2013-05-16 16:45:29

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?

@Eric Duminil 2017-02-05 13:59:54

The 3rd method iterates twice over the list, right?

@Peter Mortensen 2018-06-04 20:19:50

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.

@user3670684 2014-05-26 04:26:52

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

@devssh 2018-09-10 07:45:39

This helps us to avoid try catch!

@stefanct 2019-09-06 15:58:35

However, it might double the complexity. Did anybody check?

@ApproachingDarknessFish 2020-01-28 20:55:03

@stefanct Time complexity is still linear but it will iterate through the list twice.

@stefanct 2020-01-29 01:50:05

@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...

@bvanlew 2014-03-28 09:11:57

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.

@Cristik 2019-02-10 10:59:16

This answer should be better posted here:…

@Mathitis2Software 2013-05-29 19:17:21

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 2019-02-10 10:58:53

This answer should be better posted here:…

@kiriloff 2013-05-29 07:17:15

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]

@HongboZhu 2011-08-30 09:40:54

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"])

@Peter Mortensen 2018-06-04 20:16:56

And if doesn't exist in the list?

@Nam G VU 2018-08-13 05:29:28

Not-exist item will raise ValueError

@Cristik 2019-02-10 10:56:42

This answer would fit better here:…

Related Questions

Sponsored Content

25 Answered Questions

[SOLVED] How can I count the occurrences of a list item?

  • 2010-04-08 13:30:00
  • weakish
  • 1786555 View
  • 1594 Score
  • 25 Answer
  • Tags:   python list count

42 Answered Questions

[SOLVED] How to make a flat list out of list of lists?

44 Answered Questions

20 Answered Questions

[SOLVED] Accessing the index in 'for' loops?

  • 2009-02-06 22:47:54
  • Joan Venge
  • 2227813 View
  • 3766 Score
  • 20 Answer
  • Tags:   python loops list

24 Answered Questions

[SOLVED] How do I concatenate two lists in Python?

12 Answered Questions

[SOLVED] Getting the last element of a list

  • 2009-05-30 19:28:53
  • Janusz
  • 2251000 View
  • 2167 Score
  • 12 Answer
  • Tags:   python list indexing

9 Answered Questions

[SOLVED] How do I get the number of elements in a list?

  • 2009-11-11 00:30:54
  • y2k
  • 3313562 View
  • 1976 Score
  • 9 Answer
  • Tags:   python list

18 Answered Questions

[SOLVED] How to remove an element from a list by index

  • 2009-03-09 18:16:11
  • Joan Venge
  • 2715805 View
  • 1575 Score
  • 18 Answer
  • Tags:   python list

21 Answered Questions

[SOLVED] How do I list all files of a directory?

  • 2010-07-08 19:31:22
  • duhhunjonn
  • 4691066 View
  • 3470 Score
  • 21 Answer
  • Tags:   python directory

Sponsored Content