How to obtain the last index of a list?
Question:
Suppose I’ve the following list:
list1 = [1, 2, 33, 51]
^
|
indices 0 1 2 3
How do I obtain the last index, which in this case would be 3, of that list?
Answers:
Did you mean len(list1)-1?
If you’re searching for other method, you can try list1.index(list1[-1]), but I don’t recommend this one. You will have to be sure, that the list contains NO duplicates.
I guess you want
last_index = len(list1) - 1
which would store 3 in last_index.
You can use the list length. The last index will be the length of the list minus one.
len(list1)-1 == 3
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
a = ['1', '2', '3', '4']
print len(a) - 1
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.
all above answers is correct but however
a = [];
len(list1) - 1 # where 0 - 1 = -1
to be more precisely
a = [];
index = len(a) - 1 if a else None;
if index == None : raise Exception("Empty Array")
since arrays is starting with 0
list1[-1]
will return the last index of your list.
If you use minus before the array index it will start counting downwards from the end. list1[-2] would return the second to last index etc.
Important to mention that -0 just returns the "first" (0th) index of the list because -0 and 0 are the same number,
Suppose I’ve the following list:
list1 = [1, 2, 33, 51]
^
|
indices 0 1 2 3
How do I obtain the last index, which in this case would be 3, of that list?
Did you mean len(list1)-1?
If you’re searching for other method, you can try list1.index(list1[-1]), but I don’t recommend this one. You will have to be sure, that the list contains NO duplicates.
I guess you want
last_index = len(list1) - 1
which would store 3 in last_index.
You can use the list length. The last index will be the length of the list minus one.
len(list1)-1 == 3
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
a = ['1', '2', '3', '4']
print len(a) - 1
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.
all above answers is correct but however
a = [];
len(list1) - 1 # where 0 - 1 = -1
to be more precisely
a = [];
index = len(a) - 1 if a else None;
if index == None : raise Exception("Empty Array")
since arrays is starting with 0
list1[-1]
will return the last index of your list.
If you use minus before the array index it will start counting downwards from the end. list1[-2] would return the second to last index etc.
Important to mention that -0 just returns the "first" (0th) index of the list because -0 and 0 are the same number,