从列表部分匹配中删除列表 Python

Question

I am trying to remove a list from a list if it has a partial match

List 1

[['Person1', 'www.google.co.uk', '1'], ['Person2', 'www.amazon.co.uk', '2'],['Person3', 'www.google.co.uk', '2']]

List 2

[['Person1', 'www.google.co.uk', '1'], ['Person2', 'www.amazon.co.uk', '2'],['Person1', 'www.google.co.uk', '2']]

Desired Output:

['Person2', 'www.amazon.co.uk', '2']

At the moment I have the following code with will remove it as long as it is a full match:

[i for i in List1 if i in List2]

but as the number can vary i want to be a able to match on the first two entries so it will not matter what number is in the last column

Answer 1

You can do:

l1 = [['Person1', 'www.google.co.uk', '1'], ['Person2', 'www.amazon.co.uk', '2'],['Person3', 'www.google.co.uk', '2']]

l2 = [['Person1', 'www.google.co.uk', '2'], ['Person2', 'www.amazon.co.uk', '2'],['Person1', 'www.google.co.uk', '2']]

print ([ x for x in l1 if x[0:2] in [ l[0:2] for l in l2 ] ])

Output:

[['Person1', 'www.google.co.uk', '1'], ['Person2', 'www.amazon.co.uk', '2']]

Here you check for each sublist in l1 if values at index 0 and 1 are found in [ l[0:2] for l in l2 ] . The later is the same than l2 at the notable exception that it contains only two index by sublist instead of more. So you can use in .