2008-10-07 01:39
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在 Python 中查找给定包含该项的列表的项索引

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


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  • csdnceshi71
    Memor.の 2008-10-07 01:40
    >>> ["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 docstring for it:

    >>> print(list.index.__doc__)
    L.index(value, [start, [stop]]) -> integer -- return first index of value.
    Raises ValueError if the value is not present.

    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.)
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  • csdnceshi60
    ℡Wang Yan 2008-10-07 13:19

    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.

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  • csdnceshi52
    妄徒之命 2011-08-30 09:40

    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"])
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  • csdnceshi63
    elliott.david 2012-08-21 12:01
    a = ["foo","bar","baz",'bar','any','much']
    indexes = [index for index in range(len(a)) if a[index] == 'bar']
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  • csdnceshi55
    ~Onlooker 2013-04-16 10:19

    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
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  • csdnceshi63
    elliott.david 2013-05-16 16:45

    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?

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  • csdnceshi53
    Lotus@ 2013-05-29 07:17

    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]
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  • csdnceshi52
    妄徒之命 2013-05-29 19:17

    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]
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  • weixin_41568208
    北城已荒凉 2013-06-19 22:31

    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
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  • csdnceshi69
    YaoRaoLov 2013-06-25 15:07

    To get all indexes:

     indexes = [i for i,x in enumerate(xs) if x == 'foo']
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  • weixin_41568183
    零零乙 2014-03-28 09:11

    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.

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  • csdnceshi55
    ~Onlooker 2014-05-26 04:26

    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
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  • csdnceshi76
    斗士狗 2014-12-30 21:03

    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.

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  • csdnceshi53
    Lotus@ 2015-05-17 03:21

    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
                i +=1
        return -1
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  • weixin_41568184
    叼花硬汉 2015-07-05 13:12
    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

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  • csdnceshi71
    Memor.の 2015-11-11 05:16

    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')
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  • weixin_41568110
    七度&光 2015-11-17 19:05

    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.

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  • weixin_41568127
    ?yb? 2017-08-08 05:01

    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:


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  • csdnceshi60
    ℡Wang Yan 2017-08-12 20:01

    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.)

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  • csdnceshi68
    local-host 2017-08-22 03:08

    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/mailbox.py 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/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.

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  • weixin_41568196
    撒拉嘿哟木头 2018-01-30 21:10

    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.

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  • weixin_41568127
    ?yb? 2018-03-04 08:39

    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
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  • csdnceshi51
    旧行李 2018-03-07 19:09

    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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  • csdnceshi77
    狐狸.fox 2018-05-25 21:56

    Finding index of item x in list L:

    idx = L.index(x) if (x in L) else -1
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  • csdnceshi53
    Lotus@ 2018-09-10 18:51

    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]
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  • csdnceshi72
    谁还没个明天 2018-09-25 15:47

    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.

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  • csdnceshi70
    笑故挽风 2018-10-04 05:19

    There is two possibility if the list has no repeated items that you need to check the index for

     eg: li=[10,20,30] # here need to get index of 20 means
         li.index(20) # will work properly because 20 is not repeated

    if its repeated means it will give you only the first index

    If you need to get all the index where the item is present means

    eg: li=[10,20,30,20,40, 50, 10] # here need to get index of 20 means its have 2 index (1,3) 

    to get that you need to do it like

     li=[10,20,30,20,40, 50, 10]
     [i for i, e in enumerate(li) if e == 20]

    then you will get a list of index as o/p like [1,3]

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