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Harry Baker
Harry Baker

Reputation: 113

Python: Removing an Item From a List, But Returns a New List

Is there anyway to remove an item from a list in such a way that I return a new list without the removed element?

For instance, if I have a list ['a','b','c','d','e'] (called my_list), then I would like a method that does the following my_list.newRemove('a') = ['b','c','d','e'].

I want to do this in such a way that I could send the element to be removed and the new list as parameters of a function:

func(element, my_list.newRemove(element))

for example:

func('a', list.newRemove('a'))

which would pass 'a' AND ['b','c','d','e']

I know I can do this by just passing the element and the entire list to the function, and then creating a new subList in the function without the passed element. However, I feel like it would look better to do it the way I'm asking.

Is this just a fundamentally un-pythonic way of doing this? I learned to program with functional languages, and old habits die hard.

Edit: mistakenly wrote set rather than list

Edit 2: Here is an example of what I ultimately want to do:

myList = ['a','b','c','d']

for element in myList:
    function(element, myList.newRemove(element)

Where function does a calculation between the element and the rest of the elements in the list.

Upvotes: 2

Views: 14889

Answers (5)

hussein mohamed
hussein mohamed

Reputation: 1

My understanding of the question is the following if

>>> ls = [1,2,4,3,5,1]
>>> ls.sort()
is the equivalent to 
>>> sorted(ls)
[1,1,2,3,4,5]

with the only difference being that sorted preserves the original list unchanged, and returns a new list, while sort will change the original list and returns None

so what would be the equivalent of remove?

>>> ls = [1,2,4,3,5,1]
>>> ls.remove(1) # returns None
>>> print(ls)
[2,4,3,5,1]
>>> removed(ls, 1) ???

such a function is not built-in but can be created quite easily,

>>> removed = lambda ls, elem: [ x for i,x in enumerate(ls) if i != ls.index(elem) ]

What I see a lot of the answers neglected is that ls.remove() deletes only the first occurrence of an element, notice the second 1 is left unchanged, so actually this will be wrong,

>>> removed = lambda ls, elem: [ x for x in ls if x != elem ] # XX Wrong

Upvotes: 0

toto_tico
toto_tico

Reputation: 19027

The pythonic way of removing an element while copying is using comprehension lists: [x for x in my_list if x != element]. So there is no need of a function for this. For your example:

my_list = ['a','b','c','d']

for element in my_list:
    function(element, [x for x in my_list if x != element])

Upvotes: 1

r3v3r53
r3v3r53

Reputation: 172

Something like this?

def newListRemove(element, list): return element, filter(lambda x: x != element, list)

list = [1,2,3,4]
print newListRemove(2, list)
print list

returns

(2, [1, 3, 4])
[1, 2, 3, 4]

Upvotes: 1

Alok
Alok

Reputation: 42

def newListRemove(element, list):
  list.remove(element)
  return list

list = [1,2,3,4]
print(removeFromList(2, list))
print(list)

This should be what you need.

Upvotes: 0

Jacob Green
Jacob Green

Reputation: 61

I think this is what you mean, but I'm not sure.

def newRemove(element, x):
    x.remove(element)
    return element, x

print(newRemove("a", ["a", "b","c", "d"]))

which returns:

('a', ['b', 'c', 'd'])

hope this helps.

There is no built in function to do what you are asking, you'd have to make one.

And just a little pro-tip if you are trying to find a function that you can't remember what is called or don't know if it exists, to this: Type list. not the list you are trying to edit but literally list. and hit tab, which should then bring up all of the functions that can edit a list. Most of them are self explanatory, but it is easy to figure out what they do by trial and error.

Hope this helps for all your future coding endeavors

Upvotes: 0

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