>>> ["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
There are several things wrong here.
First off, you aren't initializing the iterator, like other's have said:
Do this and your code will be fine. But your code is done in a bad manner.
Why are you allocating the list from the heap? Look at your code: you have a memory leak. You aren't calling
delete r_list
anywhere. This is why you should use smart pointers (std::unique_ptr
,std::shared_ptr
if you have C++11, boost equivalents otherwise :boost::scoped_ptr
andboost::shared_ptr
)But better yet, just do it on the stack:
In addition, using the iterator to insert is going about things the long way. Just use push front() or push back():
Another thing: if your list outlives the vertex you've constructed, it will be pointing to something invalid.
For example:
One solution is to store pointers to heap-allocated vertices:
Now even after the function the list is pointing to a valid heap-allocated vertex. This now has the problem that when you're done using the list, you need to go through the lsit and call
delete
on each element. This problem is assisted by using the Boost Pointer Container Library.The best way, though, is to just store vertices themselves (rather than pointers to them):
If you give vertex a constructor, you can even just construct them in-place:
(these are now outside your problem)
Firstly, NULL is generally only used when dealing with pointers. Since
distance
andparent
are not pointers, use0
to initialize them, rather thanNULL
:Secondly, use
constants
rather than#define
:Lastly, give your enum a name, or place it in a namespace:
Hope these help!