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Remove items from a list if it is in another list while keeping the duplicate - Python

How to remove items from a list if it is in another list while keeping the duplicate?

I've succeeded by doing this but is there a faster way?

x = [1,2,3,4,7,7,5,6,7,8,8]
y = [1,4,5,6,8,9]
z = []
for i in x:
  if i not in y:
   z.append(i)
print z

Correct output:

[2, 3, 7, 7, 7]

Alternatively, a list comprehension also works but are these the only way?

x = [1,2,3,4,7,7,5,6,7,8,8]
y = [1,4,5,6,8,9]
z = [i for i in x if not in y]

Although using set is a lot faster but it doesn't keep the duplicate:

x = [1,2,3,4,7,7,5,6,7,8,8]
y = [1,4,5,6,8,9]
print list(set(x) - set(y))

The set subtraction gave the output that loses the duplicate:

[2, 3, 7]
Solution

  • If order isn't important

    >>> x = [1,2,3,4,7,7,5,6,7,8,8]
>>> y = [1,4,5,6,8,9]
>>> from collections import Counter
>>> count=Counter(x)
>>> for i in y:
...     del count[i]
... 
>>> list(count.elements())
[2, 3, 7, 7, 7]