>>> ["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
- Check for it first with
item in my_list
(clean, readable approach), or
- 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.)
Best Answer
The internal name of a field may not match the display name; especially if you've changed the name after creating the list. Try debugging and taking a look at the names of the fields in the
SPListItemCollection
and see if you can figure out which one it is there.