Is there a simple way to delete a list element by value?

Question

I want to remove a value from a list if it exists in the list (which it may not).

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

The above gives the error:

ValueError: list.index(x): x not in list

So I have to do this:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

But is there not a simpler way to do this?

Answer 1

To remove the first occurrence of an element, use list.remove:

>>> xs = ['a', 'b', 'c', 'd']
>>> xs.remove('b')
>>> print(xs)
['a', 'c', 'd']

To remove all occurrences of an element, use a list comprehension:

>>> xs = ['a', 'b', 'c', 'd', 'b', 'b', 'b', 'b']
>>> xs = [x for x in xs if x != 'b']
>>> print(xs)
['a', 'c', 'd']

Answer 2

Usually Python will throw an Exception if you tell it to do something it can't so you'll have to do either:

if c in a:
    a.remove(c)

or:

try:
    a.remove(c)
except ValueError:
    pass

An Exception isn't necessarily a bad thing as long as it's one you're expecting and handle properly.

Answer 3

You can do

a=[1,2,3,4]
if 6 in a:
    a.remove(6)

but above need to search 6 in list a 2 times, so try except would be faster

try:
    a.remove(6)
except:
    pass

Answer 4

Consider:

a = [1,2,2,3,4,5]

To take out all occurrences, you could use the filter function in python. For example, it would look like:

a = list(filter(lambda x: x!= 2, a))

So, it would keep all elements of a != 2.

To just take out one of the items use

a.remove(2)

Answer 5

Here's how to do it inplace (without list comprehension):

def remove_all(seq, value):
    pos = 0
    for item in seq:
        if item != value:
           seq[pos] = item
           pos += 1
    del seq[pos:]

Answer 6

As stated by numerous other answers, list.remove() will work, but throw a ValueError if the item wasn't in the list. With python 3.4+, there's an interesting approach to handling this, using the suppress contextmanager:

from contextlib import suppress
with suppress(ValueError):
    a.remove('b')

Answer 7

If you know what value to delete, here's a simple way (as simple as I can think of, anyway):

a = [0, 1, 1, 0, 1, 2, 1, 3, 1, 4]
while a.count(1) > 0:
    a.remove(1)

You'll get [0, 0, 2, 3, 4]

Answer 8

In one line:

a.remove('b') if 'b' in a else None

sometimes it useful.

Even easier:

if 'b' in a: a.remove('b')

Answer 9

Another possibility is to use a set instead of a list, if a set is applicable in your application.

IE if your data is not ordered, and does not have duplicates, then

my_set=set([3,4,2])
my_set.discard(1)

is error-free.

Often a list is just a handy container for items that are actually unordered. There are questions asking how to remove all occurences of an element from a list. If you don't want dupes in the first place, once again a set is handy.

my_set.add(3)

doesn't change my_set from above.

Answer 10

If your elements are distinct, then a simple set difference will do.

c = [1,2,3,4,'x',8,6,7,'x',9,'x']
z = list(set(c) - set(['x']))
print z
[1, 2, 3, 4, 6, 7, 8, 9]

Answer 11

This example is fast and will delete all instances of a value from the list:

a = [1,2,3,1,2,3,4]
while True:
    try:
        a.remove(3)
    except:
        break
print a
>>> [1, 2, 1, 2, 4]

Answer 12

Finding a value in a list and then deleting that index (if it exists) is easier done by just using list's remove method:

>>> a = [1, 2, 3, 4]
>>> try:
...   a.remove(6)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 3, 4]
>>> try:
...   a.remove(3)
... except ValueError:
...   pass
... 
>>> print a
[1, 2, 4]

If you do this often, you can wrap it up in a function:

def remove_if_exists(L, value):
  try:
    L.remove(value)
  except ValueError:
    pass

Answer 13

We can also use .pop:

>>> lst = [23,34,54,45]
>>> remove_element = 23
>>> if remove_element in lst:
...     lst.pop(lst.index(remove_element))
... 
23
>>> lst
[34, 54, 45]
>>> 

Answer 14

Many of the answers here involve creating a new list. This involves copying all the data from the old list to the new list (except for the removed items). If your list is huge, you may not be able to afford it (or you should not want to).

In these cases, it is much faster to alter the list in place. If you have to remove more than 1 element from the list it can be tricky. Suppose you loop over the list, and remove an item, then the list changes and a standard for-loop will not take this into account. The result of the loop may not be what you expected.

Example:

a = [0, 1, 2, 3, 4, 5]
for i in a:
    a.remove(i)  # Remove all items
print(a)

Out: [1, 3, 5]

A simple solution is to loop through the list in reverse order. In this case you get:

a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
    a.remove(i)  # Remove all items
print(a)

Out: []

Then, if you need to only remove elements having some specific values, you can simply put an if statement in the loop resulting in:

a = [0, 1, 2, 3, 4, 5]
for i in reversed(a):
    if i == 2 or i == 3:  # Remove all items having value 2 or 3.
        a.remove(i)
print(a)

Out: [0, 1, 4, 5]

Answer 15

With a for loop and a condition:

def cleaner(seq, value):    
    temp = []                      
    for number in seq:
        if number != value:
            temp.append(number)
    return temp

And if you want to remove some, but not all:

def cleaner(seq, value, occ):
    temp = []
    for number in seq:
        if number == value and occ:
            occ -= 1
            continue
        else:
            temp.append(number)
    return temp

Answer 16

 list1=[1,2,3,3,4,5,6,1,3,4,5]
 n=int(input('enter  number'))
 while n in list1:
    list1.remove(n)
 print(list1)

Answer 17

Overwrite the list by indexing everything except the elements you wish to remove

>>> s = [5,4,3,2,1]
>>> s[0:2] + s[3:]
[5, 4, 2, 1]

More generally,

>>> s = [5,4,3,2,1]
>>> i = s.index(3)
>>> s[:i] + s[i+1:]
[5, 4, 2, 1]

Answer 18

When nums is the list and c is the value to be removed:

To remove the first occurrence of c in the list, just do:

if c in nums:
    nums.remove(c)

To remove all occurrences of c from the list do:

while c in nums:
    nums.remove(c)

Adding the exception handling would be the best practice, but I mainly wanted to show how to remove all occurrences of an element from the list.

Answer 19

Say for example, we want to remove all 1's from x. This is how I would go about it:

x = [1, 2, 3, 1, 2, 3]

Now, this is a practical use of my method:

def Function(List, Unwanted):
    [List.remove(Unwanted) for Item in range(List.count(Unwanted))]
    return List
x = Function(x, 1)
print(x)

And this is my method in a single line:

[x.remove(1) for Item in range(x.count(1))]
print(x)

Both yield this as an output:

[2, 3, 2, 3, 2, 3]

Hope this helps. PS, pleas note that this was written in version 3.6.2, so you might need to adjust it for older versions.

Answer 20

arr = [1, 1, 3, 4, 5, 2, 4, 3]

# to remove first occurence of that element, suppose 3 in this example
arr.remove(3)

# to remove all occurences of that element, again suppose 3
# use something called list comprehension
new_arr = [element for element in arr if element!=3]

# if you want to delete a position use "pop" function, suppose 
# position 4 
# the pop function also returns a value
removed_element = arr.pop(4)

# u can also use "del" to delete a position
del arr[4]

Answer 21

you can use this for finding and removing specified Item from list

def find_and_delete_element(lst,item):
    for ele in lst:
        if ele == item:
            lst.remove(item)
    return lst

Answer 22

This removes all instances of "-v" from the array sys.argv, and does not complain if no instances were found:

while "-v" in sys.argv:
    sys.argv.remove('-v')

You can see the code in action, in a file called speechToText.py:

$ python speechToText.py -v
['speechToText.py']

$ python speechToText.py -x
['speechToText.py', '-x']

$ python speechToText.py -v -v
['speechToText.py']

$ python speechToText.py -v -v -x
['speechToText.py', '-x']

Answer 23

Maybe your solutions works with ints, but It Doesnt work for me with dictionarys.

In one hand, remove() has not worked for me. But maybe it works with basic Types. I guess the code bellow is also the way to remove items from objects list.

In the other hand, 'del' has not worked properly either. In my case, using python 3.6: when I try to delete an element from a list in a 'for' bucle with 'del' command, python changes the index in the process and bucle stops prematurely before time. It only works if You delete element by element in reversed order. In this way you dont change the pending elements array index when you are going through it

Then, Im used:

c = len(list)-1
for element in (reversed(list)):
    if condition(element):
        del list[c]
    c -= 1
print(list)

where 'list' is like [{'key1':value1'},{'key2':value2}, {'key3':value3}, ...]

Also You can do more pythonic using enumerate:

for i, element in enumerate(reversed(list)):
    if condition(element):
        del list[(i+1)*-1]
print(list)

Answer 24

this is my answer, just use while and for

def remove_all(data, value):
    i = j = 0
    while j < len(data):
        if data[j] == value:
            j += 1
            continue
        data[i] = data[j]
        i += 1
        j += 1
    for x in range(j - i):
        data.pop()

Answer 25

Benchmark of some of the simplest method:

import random
from copy import copy
sample = random.sample(range(100000), 10000)
remove = random.sample(range(100000), 1000)

%%timeit
sample1 = copy(sample)
remove1 = copy(remove)

for i in reversed(sample1):
    if i in remove1:
        sample1.remove(i)
# 271 ms ± 16 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove)

filtered = list(filter(lambda x: x not in remove1, sample1))
# 280 ms ± 18.9 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

filtered = [ele for ele in sample1 if ele not in remove1]
# 293 ms ± 72.1 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# remove all instances

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

for val in remove1:
    if val in sample1:
        sample1.remove(val)
# 558 ms ± 40.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence

%%timeit
sample1 = copy(sample)
remove1 = copy(remove) 

for val in remove1:
    try:
        sample1.remove(val)
    except:
        pass
# 609 ms ± 11.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
# only remove first occurrence

Answer 26

This is a less efficient solution, but it still works:

a = [ ] // that is your list

b // element(s) you need to delete

counter = a.count(b)

while counter > 0:
    if b in a:
       a.remove(b)
       counter -= 1

print(a)

Answer 27

This is a variation of the other answers that use:

• reversed()
• list comprehension
• list.remove(element)

As others have pointed out, reversed() is needed to ensure that the elements are deleted in reverse order, else, without it, when traversing forward, it will not process the record immediately to the right of the delete.

The difference is we use list comprehension to both iterate and delete the elements at the same time, so, no double traversal. We also use list comprehension to build the list of deleted elements.

a = [1,2,2,2,2,3,4,5]
deleted = [a.remove(n) or n for n in reversed(a) if n == 2]
print(deleted)
# [2, 2, 2, 2]
print(a)
# [1, 3, 4, 5]

For the values in the resultant list, we are using a.remove(n) or n. The left-hand expression, i.e. a.remove(n) always evaluates to None. So the or will always favor the right-hand expression, i.e. n. So, the resultant list will always be the list of deleted items. This is useful when the list contains more complex objects such as dictionaries.

materials = [{"n":"silver","t":"metal" },
             {"n":"iron"  ,"t":"metal" },
             {"n":"water" ,"t":"liquid"}]
deleted = [materials.remove(m) or m for m in reversed(materials) if m["t"] == "metal"]
print(deleted)
# [{'n': 'iron', 't': 'metal'}, {'n': 'silver', 't': 'metal'}]
print(materials)
# [{'n': 'water', 't': 'liquid'}]

The downside of this approach is it forces a temporary list to be built which, in most cases, you will discard shortly afterward. For example, in this example, the temporary list is discarded immediately after it was built:

a = [1,2,2,2,2,3,4,5]
[a.remove(n) or n for n in reversed(a) if n == 2]
print(a)
# [1, 3, 4, 5]

The saving grace of this solution is it implements the in-place deletion with 1 succinct line of python.