Finding the index of an item in a list
>>> ["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 atry/except
block which catchesValueError
(probably faster, at least when the list to search is long, and the item is usually present.)
Python: find position of element in array
Have you thought about using Python list's .index(value)
method? It return the index in the list of where the first instance of the value
passed in is found.
How can I get the index of an item in a list in a single step?
How about the List.FindIndex Method:
int index = myList.FindIndex(a => a.Prop == oProp);
This method performs a linear search; therefore, this method is an
O(n) operation, where n is Count.
If the item is not found, it will return -1
Get index in the list of objects by attribute in Python
Use enumerate
when you want both the values and indices in a for
loop:
for index, item in enumerate(my_list):
if item.id == 'specific_id':
break
else:
index = -1
Or, as a generator expression:
index = next((i for i, item in enumerate(my_list) if item.id == 'specific_id'), -1)
Finding the index of an item in a list of lists
Just use enumerate
:
l = [[1,2,3,4],[5,6,7,8,9,10],[11,12,13]]
# e.g.: find the index of the list containing 12
# This returns the first match (i.e. using index 0), if you want all matches
# simply remove the `[0]`
print [i for i, lst in enumerate(l) if 12 in lst][0]
This outputs:
[2]
Edit:
@hlt's comment suggests using the following for more efficient behavior:
next(i for i,v in enumerate(l) if 12 in v)
Find the index of an item in a list of lists
I'd do something like this:
[(i, colour.index(c))
for i, colour in enumerate(colours)
if c in colour]
This will return a list of tuples where the first index is the position in the first list and second index the position in the second list (note: c
is the colour you're looking for, that is, "#660000"
).
For the example in the question, the returned value is:
[(0, 0)]
If you just need to find the first position in which the colour is found in a lazy way you can use this:
next(((i, colour.index(c))
for i, colour in enumerate(colours)
if c in colour),
None)
This will return the tuple for the first element found or None
if no element is found (you can also remove the None
argument above in it will raise a StopIteration
exception if no element is found).
Edit: As @RikPoggi correctly points out, if the number of matches is high, this will introduce some overhead because colour
is iterated twice to find c
. I assumed this to be reasonable for a low number of matches and to have an answer into a single expression. However, to avoid this, you can also define a method using the same idea as follows:
def find(c):
for i, colour in enumerate(colours):
try:
j = colour.index(c)
except ValueError:
continue
yield i, j
matches = [match for match in find('#660000')]
Note that since find
is a generator you can actually use it as in the example above with next
to stop at the first match and skip looking further.
How to find all occurrences of an element in a list
You can use a list comprehension with enumerate
:
indices = [i for i, x in enumerate(my_list) if x == "whatever"]
The iterator enumerate(my_list)
yields pairs (index, item)
for each item in the list. Using i, x
as loop variable target unpacks these pairs into the index i
and the list item x
. We filter down to all x
that match our criterion, and select the indices i
of these elements.
How to find index position of an element in a list when contains returns true
benefit.indexOf(map4)
It either returns an index or -1 if the items is not found.
I strongly recommend wrapping the map in some object and use generics if possible.
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