Looping through and editing a list [duplicate] - python-3.x

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How to remove items from a list while iterating?
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Closed 10 months ago.
I am trying to get my code to go through each element in a list and decipher if it's the letter C. If it's a C remove it and print out the edited list.
Below is my code and it keeps looping.
letters = ["A", "B", "C", "D", "E", "C", "G"]
counter = 0
for letter in letters:
while letter != "":
if letter in letters == "C":
letters.remove("C")
counter += 1
print(letters)

Answer 1

You can do something like this:
With the pop argument
letters = ["A", "B", "C", "D", "E", "C", "G"]
counter = 0
i=0
while i < len(letters):
if i<len(letters) and letters[i] == "C":
letters.pop(i)
counter += 1
else:
i+= 1
print(letters) #['A', 'B', 'D', 'E', 'G']
You can remove the counter and obtain the number of "C" in the list by doing:
letters.count("C") at the begging
And you can use that to do:
With remove
letters = ["A", "B", "C", "D", "E", "C", "G"]
for _ in range(letters.count('C')):
letters.remove('C')
print(letters) #['A', 'B', 'D', 'E', 'G']
You can also use a list comprehension like that:
With list comprehension
letters = ["A", "B", "C", "D", "E", "C", "G"]
letters = [l for l in letters if l != 'C']
print(letters) #['A', 'B', 'D', 'E', 'G']
Or, you could also use filter with a lambda function:
With filter
letters = ["A", "B", "C", "D", "E", "C", "G"]
letters = [*filter(lambda l: l!='C', letters)]
print(letters) #['A', 'B', 'D', 'E', 'G']