比较不同大小的元组并在python中重新排列/替换元组元素

Question

Hello I am new to python, looking for way of comparing two of different size tuples say for example tuple a=(3,4) and b=(1,3,4) compare a and b and rearrange/replace tuple b as per tuple a (ie) b=(3,4,1) and it should work all different length tuples.

This is sample input list:

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

This is my code just iterating over list of tuples:

for i, (a, b) in enumerate(l):
print a, b
print len(a), len(b)

I want output such that when tuple a of length x=2 compared with tuple b of length y=3 should rearrange/Replace tuple b as per positions of elements in tuple a and with all elements of tuple b in it. In other words compare tuple of shorter length with tuple of larger length and rerrange/replace the tuple of larger length tuple wrt shorter length tuple, with all element of larger tuple in it.

I want Output as:

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

I have seen concept of sorting in python it sorts the tuples by key parameter but i couldn't find any help for comparing the tuples elements on its position element and rearranging it or replacing the whole tuple.

I would like to have suggestions on it for resolving the issue.

Answer 1

I actually figured out the solution for above problem. Thanks any way all for comments and feed back :)

Input:

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

Solution Code:

for i, (a, b) in enumerate(l):
# print a, b
# print len(a), len(b)
x= tuple(set(a).symmetric_difference(set(b)))
if(len(a)>len(b)):
    new_a=b+x
    l[i] = (new_a,b)
else:
    new_b=a+x
    l[i] = (a, new_b)

Output:

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