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

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)
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.)

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.

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

It"s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the "joiner" must be strings.

For example:

"_".join(["welcome", "to", "stack", "overflow"])
"_".join(("welcome", "to", "stack", "overflow"))

"welcome_to_stack_overflow"

Using something other than strings will raise the following error:

TypeError: sequence item 0: expected str instance, int found

This was discussed in the String methods... finally thread in the Python-Dev achive, and was accepted by Guido. This thread began in Jun 1999, and str.join was included in Python 1.6 which was released in Sep 2000 (and supported Unicode). Python 2.0 (supported str methods including join) was released in Oct 2000.

• There were four options proposed in this thread:
  • str.join(seq)
  • seq.join(str)
  • seq.reduce(str)
  • join as a built-in function
• Guido wanted to support not only lists and tuples, but all sequences/iterables.
• seq.reduce(str) is difficult for newcomers.
• seq.join(str) introduces unexpected dependency from sequences to str/unicode.
• join() as a built-in function would support only specific data types. So using a built-in namespace is not good. If join() supports many datatypes, creating an optimized implementation would be difficult, if implemented using the __add__ method then it would ve O(n²).
• The separator string (sep) should not be omitted. Explicit is better than implicit.

Here are some additional thoughts (my own, and my friend"s):

• Unicode support was coming, but it was not final. At that time UTF-8 was the most likely about to replace UCS2/4. To calculate total buffer length of UTF-8 strings it needs to know character coding rule.
• At that time, Python had already decided on a common sequence interface rule where a user could create a sequence-like (iterable) class. But Python didn"t support extending built-in types until 2.2. At that time it was difficult to provide basic iterable class (which is mentioned in another comment).

Guido"s decision is recorded in a historical mail, deciding on str.join(seq):

Funny, but it does seem right! Barry, go for it...
Guido van Rossum

Because the join() method is in the string class, instead of the list class?

I agree it looks funny.

See http://www.faqs.org/docs/diveintopython/odbchelper_join.html:

Historical note. When I first learned Python, I expected join to be a method of a list, which would take the delimiter as an argument. Lots of people feel the same way, and there’s a story behind the join method. Prior to Python 1.6, strings didn’t have all these useful methods. There was a separate string module which contained all the string functions; each function took a string as its first argument. The functions were deemed important enough to put onto the strings themselves, which made sense for functions like lower, upper, and split. But many hard-core Python programmers objected to the new join method, arguing that it should be a method of the list instead, or that it shouldn’t move at all but simply stay a part of the old string module (which still has lots of useful stuff in it). I use the new join method exclusively, but you will see code written either way, and if it really bothers you, you can use the old string.join function instead.

--- Mark Pilgrim, Dive into Python