根据某些标准查找列表的共同元素？

Question

The input is in form of a list of tuples,

left_corners = [((380, 456), 1), ((1129, 456), 2), ((354, 328), 3), ((1137, 325), 4)]
right_corners = [((784, 456), 1), ((1535, 456), 2), ((778, 328), 3)]

The common elements would be the ones which have the second element of the tuple as 1, 2, 3 so I would want the final representation to look like -

[(((380, 456), 1),((784, 456), 1)), (((1129, 456), 2),((1535, 456), 2)), (((354, 328), 3), ((778, 328), 3))]

I can't think of a solution without a few loops (looping over both the elements and appending the elements which have the same second element in a common list). Is there a more pythonic way? I am open to other representations of this in numpy or something too.

Assumptions: We can assume both lists are sorted with the key being the second element of the list, but either left_corners or right_corners might have more elements than the other

Answer 1

If you use dictionaries, you could do something a bit more efficient

group_lists = {}
for corner in (left_corners+right_corners):
  group_lists[corner[1]] = group_lists.get(corner[1], [])
  group_lists[corner[1]].append(corner)

output = [group_lists[k] for k in group_lists]

And if you want your output to be a list of tuples instead of a list of lists you would just have to change that last line to

output = [tuple(group_lists[k]) for k in group_lists]

Answer 2

first, second = (left_corners, right_corners) if len(left_corners) > len(right_corners) else (right_corners, left_corners)

[(i, j) for i in first for j in second  if i[1] == j[1]]

I assumed that the second element (key to unify the tuples) can be anything not just 1, 2, 3

Another assumption is that the second element is sorted but can skip numbers for example, if this is not the case, then zip is the way to go.

left_corners = [(X, 1), (Y, 2), (Z, 3)]
right_corners = [(A, 1), (B, 3)]

Answer 3

Maybe (not sure, as the problem is not very clearly stated) this is what you want:

good_corners = [ [ t for t in corners if t[1] in {1,2,3}] for corners in (left_corners, right_corners) ]
result = list(zip(*good_corners))

Answer 4

How about a list comprehension?

a = [k for k in left_corners if k[1] in [1,2,3]] + [k for k in right_corners if k[1] in [1,2,3]]

output:

[((380, 456), 1), ((1129, 456), 2), ((354, 328), 3), ((784, 456), 1), ((1535, 456), 2), ((778, 328), 3)]

This is only valid, if the ordering is irrelevant. If it needs to have the order you can sort using a lambda function:

sorted(a, key=lambda x: x[1])

output:

[((380, 456), 1), ((784, 456), 1), ((1129, 456), 2), ((1535, 456), 2), ((354, 328), 3), ((778, 328), 3)]

Answer 5

This worked pretty well for me

combined_list = list(zip(left_corners, right_corners))