I'm currently learning about the pop() function in Python and have a question.
>>> a = [1,2,3,4]
>>> a.pop(3) #or a.pop()
4
>>> print(a)
[1,2,3]
I get that the pop() function removes and returns the value of the element corresponding to the index. However, the following example is the reason why I'm confused:
>>> a = [1,2,3,4]
>>> def solution(array):
array.pop()
return array
>>> solution(a)
[1,2,3]
First, I get that the function that I've described returns [1,2,3]. However, why does it not return the pop() value? Shouldn't it return 4 since the pop() function inside the solution() has a pop() function which, in definition, returns the value of the popped element?? I thought this pop() function kind of acts like del and print function simultaneously.
When you call return array after array.pop() it will return the rest of the element expcept the pop. Because you are returning array not all the specific pop opereation. Return array.pop() instead of array
def solution(array):
return array.pop()
Or another way you can store the pop element and then you can return the pop variable.
def solution(array):
pop_op=array.pop()
return pop_op
cause you are not catching what pop() returns, the element is never printed, just in the python shell, you are returning the array modified, not the value returned by pop().
Edit:
Btw , you don't need to return the array, the original array is already modified.
I'm trying to return true if only all the previous elements are true up to the current position.
I have it set up with all function but I don't want to code it this way
def check(lightsOnOff, light):
for light in lights[:light]:
if not on:
return False
return True
count = count + 1
In general all is a useful construct to use, I can see why it looks wrong in this expression
all(list(lightsOnOff.values())[:light])
but the smelly part is actually the list(iterable)[:number] construction, which forces construction of the whole list then truncates it.
As an important aside, if lightsOnOff is a dict (not e.g. an OrderedDict) your code will be non-deterministic (see notes at bottom).
If you don't want to create a list and slice it, you can leverage itertools:
from itertools import islince
...
all(islice(lightsOnOff.values(), n))
As a frame challenge, if your dict has an order and you know the keys, you can simply rewrite it as:
all(lightsOnOff[k] for k in keys[:light])
and if your dict has keys that are ordered and e.g. integers, just use a list?
all(listOfLights[:light])
Provided you want to implement all yourself on an arbitrary list, you can do something like:
my_list = [1, 7, 2, 1, None, 2, 3]
up_to_ix = 5
def my_all(some_list, up_to_index):
for element in some_list[:up_to_index]:
if not element:
return False
return True
my_all(my_list, up_to_ix)
The function will loop through all elements in the list up to, but excluding the some_index and if it finds at least one Falsy value, will return False, otherwise True.
Is there a built-in python equivalent of std::find_if to find the first element of a list for which a given condition is true? In other words, something like the index() function of a list, but with an arbitrary unary predicate rather than just a test for equality.
I don't want to use list comprehension, because the specific predicate I have in mind is somewhat expensive to compute.
Using a tip from an answer to a related question, and borrowing from the answer taras posted, I came up with this:
>>> lst=[1,2,10,3,5,3,4]
>>> next(n for n in lst if n%5==0)
10
A slight modification will give you the index rather than the value:
>>> next(idx for idx,n in enumerate(lst) if n%5==0)
2
Now, if there was no match this will raise an exception StopIteration. You might want use a function that handles the exception and returns None if there was no match:
def first_match(iterable, predicate):
try:
return next(idx for idx,n in enumerate(iterable) if predicate(n))
except StopIteration:
return None
lst=[1,2,10,3,5,3,4]
print(first_match(lst, lambda x: x%5 == 0))
Note that this uses a generator expression, not a list comprehension. A list comprehension would apply the condition to every member of the list and produce a list of all matches. This applies it to each member until it finds a match and then stops, which is the minimum work to solve the problem.
Say, you have some predicate function pred and a list lst.
You can use itertools.dropwhile to get the first element in lst,
for which pred returns True with
itertools.dropwhile(lambda x: not pred(x), lst).next()
It skips all elements for which pred(x) is False and .next()
yields you the value for which pred(x) is True.
Edit:
A sample use to find the first element in lst divisible by 5
>>> import itertools
>>> lst = [1,2,10,3,5,3,4]
>>> pred = lambda x: x % 5 == 0
>>> itertools.dropwhile(lambda x: not pred(x), lst).next()
10
Given a list xs and a value item, how can I check whether xs contains item (i.e., if any of the elements of xs is equal to item)? Is there something like xs.contains(item)?
For performance considerations, see Fastest way to check if a value exists in a list.
Use:
if my_item in some_list:
...
Also, inverse operation:
if my_item not in some_list:
...
It works fine for lists, tuples, sets and dicts (check keys).
Note that this is an O(n) operation in lists and tuples, but an O(1) operation in sets and dicts.
In addition to what other have said, you may also be interested to know that what in does is to call the list.__contains__ method, that you can define on any class you write and can get extremely handy to use python at his full extent.
A dumb use may be:
>>> class ContainsEverything:
def __init__(self):
return None
def __contains__(self, *elem, **k):
return True
>>> a = ContainsEverything()
>>> 3 in a
True
>>> a in a
True
>>> False in a
True
>>> False not in a
False
>>>
I came up with this one liner recently for getting True if a list contains any number of occurrences of an item, or False if it contains no occurrences or nothing at all. Using next(...) gives this a default return value (False) and means it should run significantly faster than running the whole list comprehension.
list_does_contain = next((True for item in list_to_test if item == test_item), False)
The list method index will return -1 if the item is not present, and will return the index of the item in the list if it is present. Alternatively in an if statement you can do the following:
if myItem in list:
#do things
You can also check if an element is not in a list with the following if statement:
if myItem not in list:
#do things
There is also the list method:
[2, 51, 6, 8, 3].__contains__(8)
# Out[33]: True
[2, 51, 6, 3].__contains__(8)
# Out[33]: False
There is one another method that uses index. But I am not sure if this has any fault or not.
list = [5,4,3,1]
try:
list.index(2)
#code for when item is expected to be in the list
print("present")
except:
#code for when item is not expected to be in the list
print("not present")
Output:
not present
How do I get the last element of a list?
Which way is preferred?
alist[-1]
alist[len(alist) - 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.
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.
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.
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.
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 (i.e. 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
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'
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'
Another method:
some_list.reverse()
some_list[0]
Here is the solution for your query.
a=["first","second","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 second item in list
print(a[-1]) #prints the last item in the list.
print(a[-2]) #prints the second last item in the list.
Output:
>>> first
>>> second
>>> last
>>> second from last
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.
You can also use the code below, if you do not want to get IndexError when the list is empty.
next(reversed(some_list), None)
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.
Strange that nobody posted this yet:
>>> l = [1, 2, 3]
>>> *x, last_elem = l
>>> last_elem
3
>>>
Just unpack.
You can use ~ operator to get the ith element from end (indexed from 0).
lst=[1,3,5,7,9]
print(lst[~0])
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.
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
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))
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])
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].
If you use negative numbers, it will start giving you elements from last of the list
Example
lst=[1,3,5,7,9]
print(lst[-1])
Result
9
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.
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.
You could use it with next and iter with [::-1]:
>>> a = [1, 2, 3]
>>> next(iter(a[::-1]))
3
>>>
array=[1,2,3,4,5,6,7]
last_element= array[len(array)-1]
last_element
Another simple solution
Couldn't find any answer mentioning this. So adding.
You could try some_list[~0] also.
That's the tilde symbol