>>> ["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.)
When you write [x]*3
you get, essentially, the list [x, x, x]
. That is, a list with 3 references to the same x
. When you then modify this single x
it is visible via all three references to it:
x = [1] * 4
l = [x] * 3
print(f"id(x): {id(x)}")
# id(x): 140560897920048
print(
f"id(l[0]): {id(l[0])}\n"
f"id(l[1]): {id(l[1])}\n"
f"id(l[2]): {id(l[2])}"
)
# id(l[0]): 140560897920048
# id(l[1]): 140560897920048
# id(l[2]): 140560897920048
x[0] = 42
print(f"x: {x}")
# x: [42, 1, 1, 1]
print(f"l: {l}")
# l: [[42, 1, 1, 1], [42, 1, 1, 1], [42, 1, 1, 1]]
To fix it, you need to make sure that you create a new list at each position. One way to do it is
[[1]*4 for _ in range(3)]
which will reevaluate [1]*4
each time instead of evaluating it once and making 3 references to 1 list.
You might wonder why *
can't make independent objects the way the list comprehension does. That's because the multiplication operator *
operates on objects, without seeing expressions. When you use *
to multiply [[1] * 4]
by 3, *
only sees the 1-element list [[1] * 4]
evaluates to, not the [[1] * 4
expression text. *
has no idea how to make copies of that element, no idea how to reevaluate [[1] * 4]
, and no idea you even want copies, and in general, there might not even be a way to copy the element.
The only option *
has is to make new references to the existing sublist instead of trying to make new sublists. Anything else would be inconsistent or require major redesigning of fundamental language design decisions.
In contrast, a list comprehension reevaluates the element expression on every iteration. [[1] * 4 for n in range(3)]
reevaluates [1] * 4
every time for the same reason [x**2 for x in range(3)]
reevaluates x**2
every time. Every evaluation of [1] * 4
generates a new list, so the list comprehension does what you wanted.
Incidentally, [1] * 4
also doesn't copy the elements of [1]
, but that doesn't matter, since integers are immutable. You can't do something like 1.value = 2
and turn a 1 into a 2.
Best Answer
Let me see if i understand you correctly.
You have seen the browse for document link in publishing sites, and you want this functionality in your standard list.
Unfortuately you cannot do this, the standard data types are:
http://office.microsoft.com/en-us/windows-sharepoint-services-help/create-a-site-column-HA010157769.aspx
So you are left with just plain old Hyperlink, which doesn't have the browse for document box.
If you try to create a site column with the datatype Publishing Hyperlink, and add it to your list, you should get an error message when you try to browse
"Cannot complete action"