len(list1)-1 is definitely the way to go, but if you absolutely need a list that has a function that returns the last index, you could create a class that inherits from list.
class MyList(list):
def last_index(self):
return len(self)-1
>>> l=MyList([1, 2, 33, 51])
>>> l.last_index()
3
Let's say v is a list or a tuple.
To find the last occurrence of value in v we would need to compute len(v) - 1 - v[::-1].index(value)
Why is this? Why must we subtract the last term from len(v) - 1? Why does simply writing v[::-1].index(value) give the wrong result?
In fact, what does v[::-1] actually do? Doesn't it reverse the list/tuple? If it does reverse it, then v[::-1].index(value) should give the last occurrence of value in v, but for some reason it does not work like that.
len(list1)-1 is definitely the way to go, but if you absolutely need a list that has a function that returns the last index, you could create a class that inherits from list.
class MyList(list):
def last_index(self):
return len(self)-1
>>> l=MyList([1, 2, 33, 51])
>>> l.last_index()
3
The best and fast way to obtain the content of the last index of a list is using -1 for number of index ,
for example:
my_list = [0, 1, 'test', 2, 'hi']
print(my_list[-1])
Output is: 'hi'.
Index -1 shows you the last index or first index of the end.
But if you want to get only the last index, you can obtain it with this function:
def last_index(input_list:list) -> int:
return len(input_list) - 1
In this case, the input is the list, and the output will be an integer which is the last index number.
Videos
To get the last index of a list
len(mylist)-1
To get the last element of a list you can simply use a negative index.
mylist[-1]
You can just access the index variable after the loop. No need to introduce another variable (highest in your case).
for index, value in enumerate(long_list):
# do stuff with index and value
return index
Sequences have a method index(value) which returns index of first occurrence - in your case this would be verts.index(value).
You can run it on verts[::-1] to find out the last index. Here, this would be len(verts) - 1 - verts[::-1].index(value)
Perhaps the two most efficient ways to find the last index:
def rindex(lst, value):
lst.reverse()
i = lst.index(value)
lst.reverse()
return len(lst) - i - 1
def rindex(lst, value):
return len(lst) - operator.indexOf(reversed(lst), value) - 1
Both take only O(1) extra space and the two in-place reversals of the first solution are much faster than creating a reverse copy. Let's compare it with the other solutions posted previously:
def rindex(lst, value):
return len(lst) - lst[::-1].index(value) - 1
def rindex(lst, value):
return len(lst) - next(i for i, val in enumerate(reversed(lst)) if val == value) - 1
Benchmark results, my solutions are the red and green ones:
This is for searching a number in a list of a million numbers. The x-axis is for the location of the searched element: 0% means it's at the start of the list, 100% means it's at the end of the list. All solutions are fastest at location 100%, with the two reversed solutions taking pretty much no time for that, the double-reverse solution taking a little time, and the reverse-copy taking a lot of time.
A closer look at the right end:
At location 100%, the reverse-copy solution and the double-reverse solution spend all their time on the reversals (index() is instant), so we see that the two in-place reversals are about seven times as fast as creating the reverse copy.
The above was with lst = list(range(1_000_000, 2_000_001)), which pretty much creates the int objects sequentially in memory, which is extremely cache-friendly. Let's do it again after shuffling the list with random.shuffle(lst) (probably less realistic, but interesting):
All got a lot slower, as expected. The reverse-copy solution suffers the most, at 100% it now takes about 32 times (!) as long as the double-reverse solution. And the enumerate-solution is now second-fastest only after location 98%.
Overall I like the operator.indexOf solution best, as it's the fastest one for the last half or quarter of all locations, which are perhaps the more interesting locations if you're actually doing rindex for something. And it's only a bit slower than the double-reverse solution in earlier locations.
All benchmarks done with CPython 3.9.0 64-bit on Windows 10 Pro 1903 64-bit.
some_list[-1] is the shortest and most Pythonic.
In fact, you can do much more with this syntax. The some_list[-n] syntax gets the nth-to-last element. So some_list[-1] gets the last element, some_list[-2] gets the second to last, etc, all the way down to some_list[-len(some_list)], which gives you the first element.
You can also set list elements in this way. For instance:
>>> some_list = [1, 2, 3]
>>> some_list[-1] = 5 # Set the last element
>>> some_list[-2] = 3 # Set the second to last element
>>> some_list
[1, 3, 5]
Note that getting a list item by index will raise an IndexError if the expected item doesn't exist. This means that some_list[-1] will raise an exception if some_list is empty, because an empty list can't have a last element.
If your str() or list() objects might end up being empty as so: astr = '' or alist = [], then you might want to use alist[-1:] instead of alist[-1] for object "sameness".
The significance of this is:
alist = []
alist[-1] # will generate an IndexError exception whereas
alist[-1:] # will return an empty list
astr = ''
astr[-1] # will generate an IndexError exception whereas
astr[-1:] # will return an empty str
Where the distinction being made is that returning an empty list object or empty str object is more "last element"-like then an exception object.