CODEWITHSUNDEEP
python
bb Yang
bb Yang

Reputation: 51

get difference between two lists of tuples

a=[(1,2),(2,3),(3,5),(4,1)]
b=[(1,3),(2,3),(3,5),(4,3)]

I want to compare each element in the list according to their index number, namely the first item to the first item, the second item to the second item, and so on..

if they match, do nothing, it they don't match, append to new list.

Upvotes: 2

Views: 1752

Answers (4)

Ajax1234
Ajax1234

Reputation: 71451

Using simple list comprehension:

a=[(1,2),(2,3),(3,5),(4,1)]
b=[(1,3),(2,3),(3,5),(4,3)]

new_list = [a[i] for i in range(len(a)) if a[i] != b[i]]

Upvotes: 0

timgeb
timgeb

Reputation: 78690

Vanilla:

>>> result = []
>>> for x, y in zip(a, b):
...     if x != y:
...         result.append(x)
...         result.append(y)
... 
>>> result
[(1, 2), (1, 3), (4, 1), (4, 3)]

Fun:

>>> sum(([x,y] for x,y in zip(a, b) if x != y), [])
[(1, 2), (1, 3), (4, 1), (4, 3)]

The general lesson you should learn here is that whenever you want to compare iterables element-wise, use the zip builtin. Everything after that is completely straight forward (in the vanilla version).

Upvotes: 2

Right leg
Right leg

Reputation: 16720

Here is a solution using a classical for loop.

l = []
for i in range(len(a)):
    if a[i] != b[i]:
        l.append(a[i])
        l.append(b[i])

Upvotes: 0

Mohd
Mohd

Reputation: 5613

If the lists are the same length like the example you provided, and you want to append the two elements that don't match you can use:

    a=[(1,2),(2,3),(3,5),(4,1)]
    b=[(1,3),(2,3),(3,5),(4,3)]
    c = []
    for i in xrange(len(a)):
        if a[i] != b[i]:
            c.append(a[i])
            c.append(b[i])
    print c

output:

[(1, 2), (1, 3), (4, 1), (4, 3)]

Upvotes: 0

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