I'm trying to remove all instances of a value ("c") from a list.

letters = ["a", "b", "c", "c", "d"]
for i in letters:
    if i == "c":
        letters.remove(i)

print(letters)

The output is ["a", "b", "c", "d"] instead of the intended ["a", "b", "d"]

Why is it that only one "c" is removed, and not both/all?

CodePudding user response：

Filter is the right approach, when you have to filter the contents of a list.

list(filter(lambda x: x!='c', letters))
['a', 'b', 'd']

CodePudding user response：

You are updating a list and iterating it at the same time. When you remove the first c your new list will be ["a", "b", "c", "d"] but the index is stil in the 3rd position (where was the first c) so it moves directly to 4th position.

If you change your list to ["a", "b", "c", "d", "c"] you won't see the problem.

I suggest that you create a completely new list:

letters = ["a", "b", "c", "c", "d"]
new_letters = [i for i in letters if i != "c"]

print(new_letters)

CodePudding user response：

I recommend not deleting/adding elements inside a loop on the list's elements. Believe creating a new list and copying all the relevant elements to the new list will be better.

CodePudding user response：

Find duplictaes and then filter

letters = ["a", "b", "c", "c", "d"]


duplicates = []
for i in letters:
    count = letters.count(i)
    if count > 1 and i not in duplicates:
        duplicates.append(i)
#Filter    

print([x for x in letters if x not in duplicates])

output#

['a', 'b', 'd']

CodePudding user response：

Here is an (admittedly ugly but working) implementation with a for loop:

letters = ["a", "b", "c", "c", "d"]
for i in range(len(letters)-1, -1, -1):
    if letters[i] == "c":
       del(letters[i])

print(letters)

Explanation: counting backwards solves the index problem mentioned by Aymen

CodePudding user response：

You can remove the duplicates using collections.Counter,

In [1]: from collections import Counter

In [2]: [k for k, v in Counter(letters).items() if v == 1]
Out[2]: ['a', 'b', 'd']