如何在没有循环的情况下从列表中删除两个元素？

Question

I have a list

A = [A, B, #, C]

Using a.remove('#') I can remove the hashtag, but how can I remove both the hashtag and the element before without a loop considering I don't know what the element before the hashtag is?

Answer 1

Use list.index to find the position of the hashtag and then use two slices to remove the two elements.

pos = A.index('#')
A = A[:pos - 1] + A[pos + 1:]

A[:pos - 1] gives everything up to, and not including, the element before the hashtag. A[pos + 1:] gives everything after the hashtag, then you can combine these two into one list

Alternatively refer to @usr2564301's solution in the comments which removes it in-place (without having to reassign A): A[pos-1:pos+1] = [] , using the same pos from above

Note : This assumes that the hashtag is not the first element in the list, otherwise unexpected behavior will occur (refer to @sertsedat's comment below)

Answer 2

You basically need to know the index of the hashtag.

A = ['A', 'B', '#', 'C']
hashtag_index = A.index('#')
# here I am removing the hashtag
A.pop(hashtag_index)
# here the element before
if hashtag_index != 0:
    A.pop(hashtag_index - 1)

print(A)
# ['A', 'C']

You can also use del in one line (doesn't work when index == 0)

del A[hashtag_index - 1:hashtag_index+1]

Answer 3

If you are sure that the element is in the list, you can use index to get the position in the list and then a couple of pop to remove the items (assuming that if the hashtag is the first of the list, you just want to remove that):

a= ['A', 'B', '#', 'C']
index = a.index('#')
# if the element is not in the list, you will get a ValueError!
a.pop(index)
if index > 0:
    a.pop(index-1)

print(a)

# output
['A', 'C']

If you are not sure that the element is in the list, you should then check first if it is, in order to avoid getting a ValueError :

a= ['A', 'B', '#', 'C']
if '#' in a:
    index = a.index('#')   
    a.pop(index)
    if index > 0:
        a.pop(index-1)

print(a)

# output
['A', 'C']

Answer 4

one way to do it:

A = ['A', 'B', '#', 'C']

print(A) #output = ['A', 'B', '#', 'C']

index = A.index("#")
A.pop(index)
A.pop(index-1)

print(A) #output = ['A', 'C']