Finding the index of an item in a list

Asked : Nov 17

Viewed : 49 times

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

python list indexing 

Nov 17

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

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)
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

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)
0
>>> [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)
0
>>> next(g)
2

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.)

answered Jan 20


Index of the Element not Present in the List

# vowels list
vowels = ['a', 'e', 'i', 'o', 'u']

# index of 'p' is vowels
index = vowels.index('p')

print('The index of p:', index)

Output

ValueError: 'p' is not in list

answered Jan 20


Find Index of item in List

In the following example, we have taken a List with numbers. Using index() method we will find the index of item 8 in the list.

Python Program

mylist = [21, 5, 8, 52, 21, 87]
item = 8

#search for the item
index = mylist.index(item)

print('The index of', item, 'in the list is:', index)

Output

The index of 8 in the list is: 2

answered Jan 20


As you can see, we have used the above code to find the index of an element in a list. Next, let us look at an example using the other two parameters.
 

vowels = ['a', 'e', 'i', 'o', 'i', 'u']


# Searching for “o” within index 1-5
print(vowels.index("o",1,5))
#Output - 3


# Searching for “o” within index 1-3
print(vowels.index("o",0,2))
#Output - ValueError: 'o' is not in list

answered Jan 20


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