如何获取列表的最后一个元素？

Question

How do I get the last element of a list?

Which way is preferred?

alist[-1]
alist[len(alist) - 1]

Answer 1

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.

Answer 2

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.

Answer 3

You can also do:

last_elem = alist.pop()

It depends on what you want to do with your list because the pop() method will delete the last element.

Answer 4

The simplest way to display last element in python is

>>> list[-1:] # returns indexed value
    [3]
>>> list[-1]  # returns value
    3

there are many other method to achieve such a goal but these are short and sweet to use.

Answer 5

In Python, how do you get the last element of a list?

To just get the last element,

• without modifying the list, and
• assuming you know the list has a last element (ie it is nonempty)

pass -1 to the subscript notation:

>>> a_list = ['zero', 'one', 'two', 'three']
>>> a_list[-1]
'three'

Explanation

Indexes and slices can take negative integers as arguments.

I have modified an example from the documentation to indicate which item in a sequence each index references, in this case, in the string "Python" , -1 references the last element, the character, 'n' :

 +---+---+---+---+---+---+
 | P | y | t | h | o | n |
 +---+---+---+---+---+---+
   0   1   2   3   4   5 
  -6  -5  -4  -3  -2  -1

>>> p = 'Python'
>>> p[-1]
'n'

Assignment via iterable unpacking

This method may unnecessarily materialize a second list for the purposes of just getting the last element, but for the sake of completeness (and since it supports any iterable - not just lists):

>>> *head, last = a_list
>>> last
'three'

The variable name, head is bound to the unnecessary newly created list:

>>> head
['zero', 'one', 'two']

If you intend to do nothing with that list, this would be more apropos:

*_, last = a_list

Or, really, if you know it's a list (or at least accepts subscript notation):

last = a_list[-1]

In a function

A commenter said:

I wish Python had a function for first() and last() like Lisp does... it would get rid of a lot of unnecessary lambda functions.

These would be quite simple to define:

def last(a_list):
    return a_list[-1]

def first(a_list):
    return a_list[0]

Or use operator.itemgetter :

>>> import operator
>>> last = operator.itemgetter(-1)
>>> first = operator.itemgetter(0)

In either case:

>>> last(a_list)
'three'
>>> first(a_list)
'zero'

Special cases

If you're doing something more complicated, you may find it more performant to get the last element in slightly different ways.

If you're new to programming, you should avoid this section, because it couples otherwise semantically different parts of algorithms together. If you change your algorithm in one place, it may have an unintended impact on another line of code.

I try to provide caveats and conditions as completely as I can, but I may have missed something. Please comment if you think I'm leaving a caveat out.

Slicing

A slice of a list returns a new list - so we can slice from -1 to the end if we are going to want the element in a new list:

>>> a_slice = a_list[-1:]
>>> a_slice
['three']

This has the upside of not failing if the list is empty:

>>> empty_list = []
>>> tail = empty_list[-1:]
>>> if tail:
...     do_something(tail)

Whereas attempting to access by index raises an IndexError which would need to be handled:

>>> empty_list[-1]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
IndexError: list index out of range

But again, slicing for this purpose should only be done if you need:

• a new list created
• and the new list to be empty if the prior list was empty.

for loops

As a feature of Python, there is no inner scoping in a for loop.

If you're performing a complete iteration over the list already, the last element will still be referenced by the variable name assigned in the loop:

>>> def do_something(arg): pass
>>> for item in a_list:
...     do_something(item)
...     
>>> item
'three'

This is not semantically the last thing in the list. This is semantically the last thing that the name, item , was bound to.

>>> def do_something(arg): raise Exception
>>> for item in a_list:
...     do_something(item)
...
Traceback (most recent call last):
  File "<stdin>", line 2, in <module>
  File "<stdin>", line 1, in do_something
Exception
>>> item
'zero'

Thus this should only be used to get the last element if you

• are already looping, and
• you know the loop will finish (not break or exit due to errors), otherwise it will point to the last element referenced by the loop.

Getting and removing it

We can also mutate our original list by removing and returning the last element:

>>> a_list.pop(-1)
'three'
>>> a_list
['zero', 'one', 'two']

But now the original list is modified.

( -1 is actually the default argument, so list.pop can be used without an index argument):

>>> a_list.pop()
'two'

Only do this if

• you know the list has elements in it, or are prepared to handle the exception if it is empty, and
• you do intend to remove the last element from the list, treating it like a stack.

These are valid use-cases, but not very common.

Saving the rest of the reverse for later:

I don't know why you'd do it, but for completeness, since reversed returns an iterator (which supports the iterator protocol) you can pass its result to next :

>>> next(reversed([1,2,3]))
3

So it's like doing the reverse of this:

>>> next(iter([1,2,3]))
1

But I can't think of a good reason to do this, unless you'll need the rest of the reverse iterator later, which would probably look more like this:

reverse_iterator = reversed([1,2,3])
last_element = next(reverse_iterator)

use_later = list(reverse_iterator)

and now:

>>> use_later
[2, 1]
>>> last_element
3

Answer 6

To prevent IndexError: list index out of range , use this syntax:

mylist = [1, 2, 3, 4]

# With None as default value:
value = mylist and mylist[-1]

# With specified default value (option 1):
value = mylist and mylist[-1] or 'default'

# With specified default value (option 2):
value = mylist[-1] if mylist else 'default'

Answer 7

Another method:

some_list.reverse() 
some_list[0]

Answer 8

lst[-1] is the best approach, but with general iterables, consider more_itertools.last :

Code

import more_itertools as mit


mit.last([0, 1, 2, 3])
# 3

mit.last(iter([1, 2, 3]))
# 3

mit.last([], "some default")
# 'some default'

Answer 9

list[-1] will retrieve the last element of the list without changing the list. list.pop() will retrieve the last element of the list, but it will mutate/change the original list. Usually, mutating the original list is not recommended.

Alternatively, if, for some reason, you're looking for something less pythonic, you could use list[len(list)-1] , assuming the list is not empty.

Answer 10

如果您不想在列表为空时获取 IndexError，也可以使用下面的代码。

next(reversed(some_list), None)

Answer 11

Ok, but what about common in almost every language way items[len(items) - 1] ? This is IMO the easiest way to get last element, because it does not require anything pythonic knowledge.

Answer 12

Here is the solution for your query.

a=["first","second from last","last"] # A sample list
print(a[0]) #prints the first item in the list because the index of the list always starts from 0.
print(a[-1]) #prints the last item in the list.
print(a[-2]) #prints the last second item in the list.

Output:

>>> first
>>> last
>>> second from last

Answer 13

Strange that nobody posted this yet:

>>> l = [1, 2, 3]
>>> *x, last_elem = l
>>> last_elem
3
>>> 

Just unpack.

Answer 14

Pythonic Way

So lets consider that we have a list a = [1,2,3,4] , in Python List can be manipulated to give us part of it or a element of it, using the following command one can easily get the last element.

print(a[-1])

Answer 15

Accessing the last element from the list in Python:

1: Access the last element with negative indexing -1

>> data = ['s','t','a','c','k','o','v','e','r','f','l','o','w']
>> data[-1]
'w'

2. Access the last element with pop() method

>> data = ['s','t','a','c','k','o','v','e','r','f','l','o','w']
>> data.pop()
'w'

However, pop method will remove the last element from the list.

Answer 16

To avoid "IndexError: list index out of range", you can use this piece of code.

list_values = [12, 112, 443]

def getLastElement(lst):
    if len(lst) == 0:
        return 0
    else:
        return lst[-1]

print(getLastElement(list_values))

Answer 17

You can also use the length to get the last element:

last_elem = arr[len(arr) - 1]

If the list is empty, you'll get an IndexError exception, but you also get that with arr[-1] .

Answer 18

the below is the code if IndexError is not needed when list is empty. next(reversed(some_list), None)

Answer 19

If you do my_list[-1] this returns the last element of the list. Negative sequence indexes represent positions from the end of the array. Negative indexing means beginning from the end, -1 refers to the last item, -2 refers to the second-last item, etc.

Answer 20

You will just need to take the and put [-1] index. For example:

list=[0,1,2]
last_index=list[-1]
print(last_index)

You will get 2 as the output.

Answer 21

array=[1,2,3,4,5,6,7]
last_element= array[len(array)-1]
last_element

Another simple solution

Answer 22

Couldn't find any answer mentioning this. So adding.

You could try some_list[~0] also.

That's the tilde symbol

Answer 23

You could use it with next and iter with [::-1] :

>>> a = [1, 2, 3]
>>> next(iter(a[::-1]))
3
>>> 

Answer 24

If you use negative numbers, it will start giving you elements from last if the list Example

lst=[1,3,5,7,9]
print(lst[-1])

Result

9

Answer 25

You can use ~ operator to get the ith element from end (indexed from 0).

lst=[1,3,5,7,9]
print(lst[~0])

Answer 26

What is a List in Python?

The list is one of the most commonly used data types in Python. A list is a collection of elements that can be of any data type. A single list can contain a combination of numbers, strings, nested lists, etc.

How to Get the Last Element of a List in Python

---

Method 1 – By Iterating all the elements in a list

numbers_list = [1, 2, 3, 4, 5, 6, 7]

# using loop method to print last element 
for i in range(0, len(number_list)):
  
    if i == (len(number_list)-1):
        print ("The last element of list is : ", number_list[i])

# Output
The last element of list is : 7

Method 2 – Using the reverse() method

numbers_list = [1, 2, 3, 4, 5, 6, 7]

# using reverse method to print last element
number_list.reverse()
print("The last element of list using reverse method are :", number_list[0])

# Output
The last element of list using reverse method are : 7

Method 3 – Using pop() method

numbers_list = [1, 2, 3, 4, 5, 6, 7]

# using pop()  method to print last element
print("The last element of list using reverse method are :", number_list.pop())

# Output
The last element of list using reverse method are : 7

Method 4 – Using Negative Indexing [] Operator

numbers_list = [1, 2, 3, 4, 5, 6, 7]

# using length-1 to print last element
print("The last element of list using length-1 method are :", number_list[len(number_list) -1])

# using [] operator to print last element
print("The last element of list using reverse method are :", number_list[-1])

# Output
The last element of list using length-1 method are : 7
The last element of list using reverse method are : 7

Method 5 – Using itemgetter()

import operator

numbers_list = [1, 2, 3, 4, 5, 6, 7]

getLastElement = operator.itemgetter(-1)


# using [] operator to print last element
print("The last element of list using reverse method are :", getLastElement(number_list))

# Output
The last element of list using reverse method are : 7

Answer 27

METHOD 1:

L = [8, 23, 45, 12, 78]
print(L[len(L)-1])

METHOD 2:

L = [8, 23, 45, 12, 78]
print(L[-1])

METHOD 3:

L = [8, 23, 45, 12, 78]
L.reverse() 
print(L[0])

METHOD 4:

L = [8, 23, 45, 12, 78]
print(L[~0])

METHOD 5:

L = [8, 23, 45, 12, 78]
print(L.pop())

All are outputting 78