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

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.

I'm honestly surprised no one has mentioned enumerate() yet:

for i, j in enumerate(['foo', 'bar', 'baz']):
    if j == 'bar':
        print i

This can be more useful than index if there are duplicates in the list, because index() 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 == 'foo']

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 == 'foo']

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 == 'foo']"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(['foo', 'bar', 'baz']*500) if j == 'foo']"
10000 loops, best of 3: 196 usec per loop

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"])

To get all indexes:

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

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

b = [item for item in range(len(a)) if a[item] == 'bar']

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 -1

def find_element_in_list(element, list_element):
    try:
        index_element = list_element.index(element)
        return index_element
    except ValueError:
        return None

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

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?

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.

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]

all indexes with zip function

get_indexes = lambda x, xs: [i for (y, i) in zip(xs, range(len(xs))) if x == y]

print get_indexes(2,[1,2,3,4,5,6,3,2,3,2])
print get_indexes('f','xsfhhttytffsafweef')

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]
>>> 

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.

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

And now, for something completely different, checking for 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 l
    """
    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(l, val):
...     """always returns a list containing the indices of val in l
...     """
...     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)
[0, 2, 4, 5]
>>> print indices(l,'bat')
[]
>>> print indices('abcdaababb','a')
[0, 4, 5, 7]
>>> 

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