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Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index 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.
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
To get all indexes:
indexes = [i for i, x in enumerate(xs) if x == 'foo']
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"])
A problem will arise if the element is not in the list. This function handles the issue:
# if element is found it returns index of element else returns None
def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']
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
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.
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?
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.
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
me = ["foo", "bar", "baz"]
me.index("bar")
You can apply this for any member of the list to get their index
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')
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]
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]
>>>
And now, for something completely different...
... like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list. It works with strings as well.
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
When pasted into an interactive python window:
Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(the_list, val):
... """Always returns a list containing the indices of val in the_list"""
... retval = []
... last = 0
... while val in the_list[last:]:
... i = the_list[last:].index(val)
... retval.append(last + i)
... last += i + 1
... return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
Update
After another year of heads-down python development, I'm a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.
Something like this:
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
Which, when pasted into an interactive python window yields:
Python 2.7.14 |Anaconda, Inc.| (default, Dec 7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(l, val):
... """Always returns a list containing the indices of val in the_list"""
... return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn't even see that answer, because I didn't understand it. I hope that my somewhat more verbose example will aid understanding.
If the single line of code above still doesn't make sense to you, I highly recommend you Google 'python list comprehension' and take a few minutes to familiarize yourself. It's just one of the many powerful features that make it a joy to use Python to develop code.
Here's a two-liner using Python's index() function:
LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
Output: 1
A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:
>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.
Finding index of item x in list L:
idx = L.index(x) if (x in L) else -1
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
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")
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"])
It just uses the python function array.index() and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function indexOf()).
fruits = ['apple', 'banana', 'cherry']
try:
pos = fruits.index("mango")
except:
pos = -1
In this case "mango" is not present in the list fruits so the pos variable is -1, if I had searched for "cherry" the pos variable would be 2.
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))))))
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()]
Python index() method throws an error if the item was not found. So instead you can make it similar to the indexOf() function of JavaScript which returns -1 if the item was not found:
try:
index = array.index('search_keyword')
except ValueError:
index = -1
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
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
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.
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.)
The list.index(x) function returns the index in the list of the first item whose value is x.
Is there a function, list_func_index(), similar to the index() function that has a function, f(), as a parameter. The function, f() is run on every element, e, of the list until f(e) returns True. Then list_func_index() returns the index of e.
Codewise:
>>> def list_func_index(lst, func):
for i in range(len(lst)):
if func(lst[i]):
return i
raise ValueError('no element making func True')
>>> l = [8,10,4,5,7]
>>> def is_odd(x): return x % 2 != 0
>>> list_func_index(l,is_odd)
3
Is there a more elegant solution? (and a better name for the function)
You could do that in a one-liner using generators:
next(i for i,v in enumerate(l) if is_odd(v))
The nice thing about generators is that they only compute up to the requested amount. So requesting the first two indices is (almost) just as easy:
y = (i for i,v in enumerate(l) if is_odd(v))
x1 = next(y)
x2 = next(y)
Though, expect a StopIteration exception after the last index (that is how generators work). This is also convenient in your "take-first" approach, to know that no such value was found --- the list.index() function would throw ValueError here.
One possibility is the built-in enumerate function:
def index_of_first(lst, pred):
for i,v in enumerate(lst):
if pred(v):
return i
return None
It's typical to refer a function like the one you describe as a "predicate"; it returns true or false for some question. That's why I call it pred in my example.
I also think it would be better form to return None, since that's the real answer to the question. The caller can choose to explode on None, if required.
#Paul's accepted answer is best, but here's a little lateral-thinking variant, mostly for amusement and instruction purposes...:
>>> class X(object):
... def __init__(self, pred): self.pred = pred
... def __eq__(self, other): return self.pred(other)
...
>>> l = [8,10,4,5,7]
>>> def is_odd(x): return x % 2 != 0
...
>>> l.index(X(is_odd))
3
essentially, X's purpose is to change the meaning of "equality" from the normal one to "satisfies this predicate", thereby allowing the use of predicates in all kinds of situations that are defined as checking for equality -- for example, it would also let you code, instead of if any(is_odd(x) for x in l):, the shorter if X(is_odd) in l:, and so forth.
Worth using? Not when a more explicit approach like that taken by #Paul is just as handy (especially when changed to use the new, shiny built-in next function rather than the older, less appropriate .next method, as I suggest in a comment to that answer), but there are other situations where it (or other variants of the idea "tweak the meaning of equality", and maybe other comparators and/or hashing) may be appropriate. Mostly, worth knowing about the idea, to avoid having to invent it from scratch one day;-).
Not one single function, but you can do it pretty easily:
>>> test = lambda c: c == 'x'
>>> data = ['a', 'b', 'c', 'x', 'y', 'z', 'x']
>>> map(test, data).index(True)
3
>>>
If you don't want to evaluate the entire list at once you can use itertools, but it's not as pretty:
>>> from itertools import imap, ifilter
>>> from operator import itemgetter
>>> test = lambda c: c == 'x'
>>> data = ['a', 'b', 'c', 'x', 'y', 'z']
>>> ifilter(itemgetter(1), enumerate(imap(test, data))).next()[0]
3
>>>
Just using a generator expression is probably more readable than itertools though.
Note in Python3, map and filter return lazy iterators and you can just use:
from operator import itemgetter
test = lambda c: c == 'x'
data = ['a', 'b', 'c', 'x', 'y', 'z']
next(filter(itemgetter(1), enumerate(map(test, data))))[0] # 3
A variation on Alex's answer. This avoids having to type X every time you want to use is_odd or whichever predicate
>>> class X(object):
... def __init__(self, pred): self.pred = pred
... def __eq__(self, other): return self.pred(other)
...
>>> L = [8,10,4,5,7]
>>> is_odd = X(lambda x: x%2 != 0)
>>> L.index(is_odd)
3
>>> less_than_six = X(lambda x: x<6)
>>> L.index(less_than_six)
2
you could do this with a list-comprehension:
l = [8,10,4,5,7]
filterl = [a for a in l if a % 2 != 0]
Then filterl will return all members of the list fulfilling the expression a % 2 != 0. I would say a more elegant method...
Intuitive one-liner solution:
i = list(map(lambda value: value > 0, data)).index(True)
Explanation:
we use map function to create a list containing True or False based on if each element in our list pass the condition in the lambda or not.
then we convert the map output to list
then using the index function, we get the index of the first true which is the same index of the first value passing the condition.
Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index 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.
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
To get all indexes:
indexes = [i for i, x in enumerate(xs) if x == 'foo']
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"])
A problem will arise if the element is not in the list. This function handles the issue:
# if element is found it returns index of element else returns None
def find_element_in_list(element, list_element):
try:
index_element = list_element.index(element)
return index_element
except ValueError:
return None
a = ["foo","bar","baz",'bar','any','much']
indexes = [index for index in range(len(a)) if a[index] == 'bar']
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
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.
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?
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.
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
me = ["foo", "bar", "baz"]
me.index("bar")
You can apply this for any member of the list to get their index
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')
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]
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]
>>>
And now, for something completely different...
... like confirming the existence of the item before getting the index. The nice thing about this approach is the function always returns a list of indices -- even if it is an empty list. It works with strings as well.
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
retval = []
last = 0
while val in l[last:]:
i = l[last:].index(val)
retval.append(last + i)
last += i + 1
return retval
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
When pasted into an interactive python window:
Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54)
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(the_list, val):
... """Always returns a list containing the indices of val in the_list"""
... retval = []
... last = 0
... while val in the_list[last:]:
... i = the_list[last:].index(val)
... retval.append(last + i)
... last += i + 1
... return retval
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
Update
After another year of heads-down python development, I'm a bit embarrassed by my original answer, so to set the record straight, one can certainly use the above code; however, the much more idiomatic way to get the same behavior would be to use list comprehension, along with the enumerate() function.
Something like this:
def indices(l, val):
"""Always returns a list containing the indices of val in the_list"""
return [index for index, value in enumerate(l) if value == val]
l = ['bar','foo','bar','baz','bar','bar']
q = 'bar'
print indices(l,q)
print indices(l,'bat')
print indices('abcdaababb','a')
Which, when pasted into an interactive python window yields:
Python 2.7.14 |Anaconda, Inc.| (default, Dec 7 2017, 11:07:58)
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> def indices(l, val):
... """Always returns a list containing the indices of val in the_list"""
... return [index for index, value in enumerate(l) if value == val]
...
>>> l = ['bar','foo','bar','baz','bar','bar']
>>> q = 'bar'
>>> print indices(l,q)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>>
And now, after reviewing this question and all the answers, I realize that this is exactly what FMc suggested in his earlier answer. At the time I originally answered this question, I didn't even see that answer, because I didn't understand it. I hope that my somewhat more verbose example will aid understanding.
If the single line of code above still doesn't make sense to you, I highly recommend you Google 'python list comprehension' and take a few minutes to familiarize yourself. It's just one of the many powerful features that make it a joy to use Python to develop code.
Here's a two-liner using Python's index() function:
LIST = ['foo' ,'boo', 'shoo']
print(LIST.index('boo'))
Output: 1
A variant on the answer from FMc and user7177 will give a dict that can return all indices for any entry:
>>> a = ['foo','bar','baz','bar','any', 'foo', 'much']
>>> l = dict(zip(set(a), map(lambda y: [i for i,z in enumerate(a) if z is y ], set(a))))
>>> l['foo']
[0, 5]
>>> l ['much']
[6]
>>> l
{'baz': [2], 'foo': [0, 5], 'bar': [1, 3], 'any': [4], 'much': [6]}
>>>
You could also use this as a one liner to get all indices for a single entry. There are no guarantees for efficiency, though I did use set(a) to reduce the number of times the lambda is called.
Finding index of item x in list L:
idx = L.index(x) if (x in L) else -1
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
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")
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"])
It just uses the python function array.index() and with a simple Try / Except it returns the position of the record if it is found in the list and return -1 if it is not found in the list (like on JavaScript with the function indexOf()).
fruits = ['apple', 'banana', 'cherry']
try:
pos = fruits.index("mango")
except:
pos = -1
In this case "mango" is not present in the list fruits so the pos variable is -1, if I had searched for "cherry" the pos variable would be 2.
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))))))
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()]
Python index() method throws an error if the item was not found. So instead you can make it similar to the indexOf() function of JavaScript which returns -1 if the item was not found:
try:
index = array.index('search_keyword')
except ValueError:
index = -1
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
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
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.
Since Python lists are zero-based, we can use the zip built-in function as follows:
>>> [i for i,j in zip(range(len(haystack)), haystack) if j == 'needle' ]
where "haystack" is the list in question and "needle" is the item to look for.
(Note: Here we are iterating using i to get the indexes, but if we need rather to focus on the items we can switch to j.)
I read the documentation on next() and I understand it abstractly. From what I understand, next() is used as a reference to an iterable object and makes python cycle to the next iterable object sequentially. Makes sense! My question is, how is this useful outside the context of the builtin for loop? When would someone ever need to use next() directly? Can someone provide a simplistic example? Thanks mates!
As luck would have it, I wrote one yesterday:
def skip_letters(f, skip=" "):
"""Wrapper function to skip specified characters when encrypting."""
def func(plain, *args, **kwargs):
gen = f(p for p in plain if p not in skip, *args, **kwargs)
for p in plain:
if p in skip:
yield p
else:
yield next(gen)
return func
This uses next to get the return values from the generator function f, but interspersed with other values. This allows some values to be passed through the generator, but others to be yielded straight out.
There are many places where we can use next, for eg.
Drop the header while reading a file.
with open(filename) as f:
next(f) #drop the first line
#now do something with rest of the lines
Iterator based implementation of zip(seq, seq[1:])(from pairwise recipe iterools):
from itertools import tee, izip
it1, it2 = tee(seq)
next(it2)
izip(it1, it2)
Get the first item that satisfies a condition:
next(x for x in seq if x % 100)
Creating a dictionary using adjacent items as key-value:
>>> it = iter(['a', 1, 'b', 2, 'c', '3'])
>>> {k: next(it) for k in it}
{'a': 1, 'c': '3', 'b': 2}
next is useful in many different ways, even outside of a for-loop. For example, if you have an iterable of objects and you want the first that meets a condition, you can give it a generator expression like so:
>>> lst = [1, 2, 'a', 'b']
>>> # Get the first item in lst that is a string
>>> next(x for x in lst if isinstance(x, str))
'a'
>>> # Get the fist item in lst that != 1
>>> lst = [1, 1, 1, 2, 1, 1, 3]
>>> next(x for x in lst if x != 1)
2
>>>
This question already has answers here:
Finding the index of an item in a list
(43 answers)
Closed 9 years ago.
What is a good way to find the index of an element in a list in Python?
Note that the list may not be sorted.
Is there a way to specify what comparison operator to use?
From Dive Into Python:
>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> li.index("example")
5
If you just want to find out if an element is contained in the list or not:
>>> li
['a', 'b', 'new', 'mpilgrim', 'z', 'example', 'new', 'two', 'elements']
>>> 'example' in li
True
>>> 'damn' in li
False
The best way is probably to use the list method .index.
For the objects in the list, you can do something like:
def __eq__(self, other):
return self.Value == other.Value
with any special processing you need.
You can also use a for/in statement with enumerate(arr)
Example of finding the index of an item that has value > 100.
for index, item in enumerate(arr):
if item > 100:
return index, item
Source
Here is another way using list comprehension (some people might find it debatable). It is very approachable for simple tests, e.g. comparisons on object attributes (which I need a lot):
el = [x for x in mylist if x.attr == "foo"][0]
Of course this assumes the existence (and, actually, uniqueness) of a suitable element in the list.
assuming you want to find a value in a numpy array,
I guess something like this might work:
Numpy.where(arr=="value")[0]
There is the index method, i = array.index(value), but I don't think you can specify a custom comparison operator. It wouldn't be hard to write your own function to do so, though:
def custom_index(array, compare_function):
for i, v in enumerate(array):
if compare_function(v):
return i
I use function for returning index for the matching element (Python 2.6):
def index(l, f):
return next((i for i in xrange(len(l)) if f(l[i])), None)
Then use it via lambda function for retrieving needed element by any required equation e.g. by using element name.
element = mylist[index(mylist, lambda item: item["name"] == "my name")]
If i need to use it in several places in my code i just define specific find function e.g. for finding element by name:
def find_name(l, name):
return l[index(l, lambda item: item["name"] == name)]
And then it is quite easy and readable:
element = find_name(mylist,"my name")
The index method of a list will do this for you. If you want to guarantee order, sort the list first using sorted(). Sorted accepts a cmp or key parameter to dictate how the sorting will happen:
a = [5, 4, 3]
print sorted(a).index(5)
Or:
a = ['one', 'aardvark', 'a']
print sorted(a, key=len).index('a')
how's this one?
def global_index(lst, test):
return ( pair[0] for pair in zip(range(len(lst)), lst) if test(pair[1]) )
Usage:
>>> global_index([1, 2, 3, 4, 5, 6], lambda x: x>3)
<generator object <genexpr> at ...>
>>> list(_)
[3, 4, 5]
I found this by adapting some tutos. Thanks to google, and to all of you ;)
def findall(L, test):
i=0
indices = []
while(True):
try:
# next value in list passing the test
nextvalue = filter(test, L[i:])[0]
# add index of this value in the index list,
# by searching the value in L[i:]
indices.append(L.index(nextvalue, i))
# iterate i, that is the next index from where to search
i=indices[-1]+1
#when there is no further "good value", filter returns [],
# hence there is an out of range exeption
except IndexError:
return indices
A very simple use:
a = [0,0,2,1]
ind = findall(a, lambda x:x>0))
[2, 3]
P.S. scuse my english