Remove list element without mutation

Question

Assume you have a list

>>> m = [\'a\',\'b\',\'c\']

I\'d like to make a new list n that has everything except for a given

Answer 1

Another approach to list comprehension is numpy:

>>> import numpy
>>> a = [1, 2, 3, 4]
>>> list(numpy.remove(a, a.index(3)))
[1, 2, 4]

Answer 2

I assume you mean that you want to create a new list without a given element, instead of changing the original list. One way is to use a list comprehension:

m = ['a', 'b', 'c']
n = [x for x in m if x != 'a']

n is now a copy of m, but without the 'a' element.

Another way would of course be to copy the list first

m = ['a', 'b', 'c']
n = m[:]
n.remove('a')

If removing a value by index, it is even simpler

n = m[:index] + m[index+1:]

Answer 3

You can create a new list without the offending element with a list-comprehension. This will preserve the value of the original list.

l = ['a', 'b', 'c']
[s for s in l if s != 'a']

Answer 4

We can do it without using in built remove function and also without creating new list variable

Code:

# List m
m = ['a', 'b', 'c']

# Updated list m, without creating new list variable
m = [x for x in m if x != a]

print(m)

output

>>> ['b', 'c']

Answer 5

There is a simple way to do that using built-in function :filter .

Here is ax example:

a = [1, 2, 3, 4]
b = filter(lambda x: x != 3, a)

Answer 6

If the order is unimportant, you can use set (besides, the removal seems to be fast in sets):

list(set(m) - set(['a']))