Removing elements that in another array

Question

I have two arrays in python. I want to remove all the elements that present in first array. Here an example of arrays:

array1 =[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11]
array2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 2, 4, 6, 8, 10, 3, 6, 9, 4, 8, 5, 10, 6, 7, 8, 9, 10, 11]

The result would be:

array2 = []

Here what I've done:

for x in array1:
    for y in array2:
        if x==y:
            array2.remove(x)

print (array2)

But I'm seeing [10] as result. Why? And how to fix this?

Answer 1

filtered_arr = [i for i in array2 if i not in array1]

Answer 2

I'm guessing the issue has to do with how you are looping and removing items in the list, making the program skip over the third 10 .

You can use a quick list comprehension to solve the issue:

array3 = [i for i in array2 if i not in array1]

This is basically a simpler way of typing:

array3 = []
for i in array2:
    if i not in array1:
        array3.append(i)

---

Additionally, you probably wouldn't want to use sets. For example:

array1 = [1, 2, 3]
array2 = [1, 2, 3, 3, 4, 4, 5, 6]

array3 = list(set(array2) - set(array1))

array3 will only contain [4, 5, 6] , rather than [4, 4, 5, 6] , since set s cannot contain duplicates.

Answer 3

def diff(first, second):
    return [item for item in first if item not in second]

Answer 4

Consider the following example of unexpected behavior when modifying a list while iterating over it:

a = [10, 10]
for x in a:
  if x == 10:
    a.remove(x)
    print 'item removed'

After the first removal, the second 10 element becomes the first element of the list and x has already pointed to the first element (position 0). Thus the second iteration will never occur.