Given a Python list and an element (value). How to remove the element (value) from the given list?
Here’s an example of what you want to accomplish:
[1, 2, 99, 4, 99]
[1, 2, 4, 99]
An alternative would return the list with the element (value)
99 removed in all occurrences:
[1, 2, 4]
We’ll first look at the first problem variant in Method 1 and then examine the second in Method 2.
Method 1: list.remove()
>>> lst = [1, 2, 99, 4, 99] >>> lst.remove(99) >>> lst [1, 2, 4, 99]
In the first line of the example, you create the list
lst. You then remove the integer element 99 from the list—but only its first occurrence. The result is the list with only four elements
[1, 2, 4, 99].
You can call this method on each list object in Python. Here’s the syntax:
|Object you want to remove from the list. Only the first occurrence of the element is removed.|
list.remove(element) has return value
None. It operates on an existing list and, therefore, doesn’t return a new list with the removed element
You can also check out our full blog tutorial on the Python
Removed Element Does Not Exist
Let’s have a look at another example where this fails next!
To remove an element from the list, use the
list.remove(element) method you’ve already seen previously:
>>> lst = ["Alice", 3, "alice", "Ann", 42] >>> lst.remove("Ann") >>> lst ['Alice', 3, 'alice', 42]
The method goes from left to right and removes the first occurrence of the element that’s equal to the one to be removed.
If you’re trying to remove element
x from the list but
x does not exist in the list, Python raises a
>>> lst = ['Alice', 'Bob', 'Ann'] >>> lst.remove('Frank') Traceback (most recent call last): File "<pyshell#19>", line 1, in <module> lst.remove('Frank') ValueError: list.remove(x): x not in list
To fix this, you can enclose it in a try/except environment like so:
lst = ['Alice', 'Bob', 'Ann'] try: lst.remove('Frank') except: pass print('done!')
This code goes through and returns
'done!' without any error message.
Method 2: Python List remove() All
list.remove(x) method only removes the first occurrence of element
x from the list.
But what if you want to remove all occurrences of element
x from a given list?
The answer is to use a simple loop:
lst = ['Ann', 'Ann', 'Ann', 'Alice', 'Ann', 'Bob'] x = 'Ann' while x in lst: lst.remove(x) print(lst) # ['Alice', 'Bob']
You simply call the
remove() method again and again until element
x is not in the list anymore.
For comprehensibility, I want to show you one method that is not optimal—but that’s recommended all over the web (e.g., here).
Method 3: List Comprehension: Remove Elements Conditionally
Using list comprehension is cheating because this method does not really remove elements from a list object. It merely creates a new list with some elements that meet your condition.
List comprehension is a compact way of creating lists. The simple formula is:
[ expression + context ]
- Expression: What to do with each list element?
- Context: What list elements to select? It consists of an arbitrary number of for and if statements.
[x for x in range(3)] creates the list
[0, 1, 2].
You can also define a condition such as “all values unlike a given element 1”, e.g.,
x!=1, in the context part by using an
This leads us to a way to remove all elements that are equal to a certain value such as our value
99 in a given list.
>>> lst = [1, 2, 99, 4, 99] >>> lst_new = [x for x in lst if x != 99] >>> lst_new [1, 2, 4]
While you iterate over the whole list
lst, you include only the elements that are not equal to the element
99. Thus, the resulting list has all elements
However, the original list
lst has not changed!
>>> lst [1, 2, 99, 4, 99]
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