Suppose I\'ve the following list:
list1 = [1, 2, 33, 51]
^
|
indices 0 1 2 3
How do I obtain the
the best and fast way to obtain 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])
out put is : 'hi'
.
index -1
in show you last index or first index of the end.
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
You can use the list length. The last index will be the length of the list minus one.
len(list1)-1 == 3
a = ['1', '2', '3', '4']
print len(a) - 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
I guess you want
last_index = len(list1) - 1
which would store 3 in last_index
.