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I have a list

A= [1,2,3,3,4,4,5,6,4,7]

I want to only keep one 3 and 4 if they are next to another identical one.

Thus A should become 

A= [1,2,3,4,5,6,4,7]

I tried to use set 

set[A]

{1, 2, 3, 4, 5, 6, 7}

But it removed the last 4 that I want to keep. It seems I should loop through the list and compare i and i+1. Wondering any faster and smarter way exists?

Kevin
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2 Answers2

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Using list comprehension:

[A[i] for i in range(len(A)) if (i==0) or A[i] != A[i-1]]
#[1, 2, 3, 4, 5, 6, 4, 7]

The logic is to iterate through each index in the list and keep the number if either of the following conditions hold:

  • i==0 (the first element)
  • A[i] != A[i-1]
pault
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Try this:

A = [1,2,3,3,4,4,5,6,4,7]
newlist = []
newlist.append(A[0])
for i, element in enumerate(A):
    if i > 0 and A[i - 1] != element:
        newlist.append(element)

Here, we loop through the list and check that every element is not equal to the element before it. It will return 

[1, 2, 3, 4, 5, 6, 4, 7]
Seth Rothschild
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