CODEWITHSUNDEEP
pythonlist
user3457749
user3457749

Reputation: 1043

Remove list element without mutation

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 item in m (for example the item 'a'). However, when I use

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

the original list is mutated (the value 'a' is removed from the original list). Is there a way to get a new list sans-'a' without mutating the original? So I mean that m should still be [ 'a', 'b', 'c' ], and I will get a new list, which has to be [ 'b', 'c' ].

Upvotes: 90

Views: 101517

Answers (8)

GJebran
GJebran

Reputation: 11

The question is useful as I sometimes have a list that I use throughout my given script but I need to at a certain step to apply a logic on a subset of the list elements. In that case I found it useful to use the same list but only exclude the needed element for that individual step, without the need to create a totally new list with a different name. For this you can use either:

  1. list comprehension: say you have l=['a','b','c'] to exclude b, you can have [x for x in l if x!='b']
  2. set [only if order is unimortant]: list(set(l) - set(['b'])), pay attention here that you pass 'b' as list ['b']

Upvotes: 0

EHadavi
EHadavi

Reputation: 82

We can do it via built-in copy() function for list; However, should assign a new name for the copy;

m = ['a','b','c']
m_copy=m.copy()
m_copy.remove('a')

print (m)

['a', 'b', 'c']

print(m_copy)

['b', 'c']

Upvotes: 4

Oleg O
Oleg O

Reputation: 1065

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

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

This will remove duplicate elements from your original list though

Upvotes: 8

xmavericks
xmavericks

Reputation: 31

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']

Upvotes: 3

Amen
Amen

Reputation: 1633

Another approach to list comprehension is numpy:

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

Upvotes: 1

Andrew Johnson
Andrew Johnson

Reputation: 3186

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']

Upvotes: 1

请叫我小马哥
请叫我小马哥

Reputation: 1731

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)

Upvotes: 19

Krumelur
Krumelur

Reputation: 32577

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:]

Upvotes: 108

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