# How do I remove the first item from a list?

## Question:

How do I remove the first item from a list?

``````[0, 1, 2, 3]   →   [1, 2, 3]
``````

``````>>> x = [0, 1, 2, 3, 4]
>>> x.pop(0)
0
``````

More on this here.

Slicing:

``````x = [0,1,2,3,4]
x = x[1:]
``````

Which would actually return a subset of the original but not modify it.

With list slicing, see the Python tutorial about lists for more details:

``````>>> l = [0, 1, 2, 3, 4]
>>> l[1:]
[1, 2, 3, 4]
``````

you would just do this

``````l = [0, 1, 2, 3, 4]
l.pop(0)
``````

or `l = l[1:]`

Pros and Cons

Using pop you can retrieve the value

say `x = l.pop(0)`
`x` would be `0`

You can find a short collection of useful list functions here.

`list.pop(index)`

``````>>> l = ['a', 'b', 'c', 'd']
>>> l.pop(0)
'a'
>>> l
['b', 'c', 'd']
>>>
``````

`del list[index]`

``````>>> l = ['a', 'b', 'c', 'd']
>>> del l
>>> l
['b', 'c', 'd']
>>>
``````

These both modify your original list.

Others have suggested using slicing:

• Copies the list
• Can return a subset

Also, if you are performing many `pop(0)`, you should look at `collections.deque`

``````from collections import deque
>>> l = deque(['a', 'b', 'c', 'd'])
>>> l.popleft()
'a'
>>> l
deque(['b', 'c', 'd'])
``````
• Provides higher performance popping from left end of the list

Then just delete it:

``````x = [0, 1, 2, 3, 4]
del x
print x
# [1, 2, 3, 4]
``````

You can also use `list.remove(a)` to `pop` out the first element in the list.

``````>>>> a=[1,2,3,4,5]
>>>> a.remove(a)
>>>> print a
>>>> [2,3,4,5]
``````

You can use `list.reverse()` to reverse the list, then `list.pop()` to remove the last element, for example:

``````l = [0, 1, 2, 3, 4]
l.reverse()
print l
[4, 3, 2, 1, 0]

l.pop()
0
l.pop()
1
l.pop()
2
l.pop()
3
l.pop()
4
``````

If you are working with numpy you need to use the delete method:

``````import numpy as np

a = np.array([1, 2, 3, 4, 5])

a = np.delete(a, 0)

print(a) # [2 3 4 5]
``````

There is a data structure called `deque` or double ended queue which is faster and efficient than a list. You can use your list and convert it to deque and do the required transformations in it. You can also convert the deque back to list.

``````import collections
mylist = [0, 1, 2, 3, 4]

#make a deque from your list
de = collections.deque(mylist)

#you can remove from a deque from either left side or right side
de.popleft()
print(de)

#you can covert the deque back to list
mylist = list(de)
print(mylist)
``````

Deque also provides very useful functions like inserting elements to either side of the list or to any specific index. You can also rotate or reverse a deque. Give it a try!

### Unpacking assignment:

You could use unpacking assignment as mentioned in PEP 3132.

#### Solution:

You should try unpacking like the following:

``````>>> l = [0, 1, 2, 3, 4]
>>> _, *l = l
>>> l
[1, 2, 3, 4]
``````

#### Explanation:

As mentioned in PEP 3132:

This PEP proposes a change to iterable unpacking syntax, allowing to
specify a "catch-all" name which will be assigned a list of all items
not assigned to a "regular" name.

An example says more than a thousand words:

``````>>> a, *b, c = range(5)
>>> a
0
>>> c
4
>>> b
[1, 2, 3]
``````

This works for me, instead of using `pop` like below, which of course will be 0, because the result/return value of `pop`

``````>>> x = [0, 1, 2, 3].pop(0)
>>> x
0
``````

Using this below method will skip the first value:

``````>>> x = [0, 1, 2, 3][1:]
>>> x
[1, 2, 3]
``````
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