Iterate over list of tuples and unpack first elements only

Question

Assume the following list: foo = [(1, 2, 3, 4), (5, 6, 7, 8)]

Is there a way to iterate over the list and unpack the first two elements of the inner tuple only?

This is a usual pattern: {a: b for a, b, _, _ in foo} , but this breaks if foo is modified (program change) and the tuple now contains 5 elements instead of 4 (the the list comprehension would need to be modified accordingly). I really like to name the elements instead of calling {f[0]: f[1] for f in foo} , so ideally, there would be some sort of "absorb all not unpacked variable", so one could call {a: b for a, b, absorb_rest in foo} . If that's possible, it wouldn't matter how many elements are contained in the tuple (as long as there are at least 2).

Answer 1

You can use extended iterable unpacking , where you extract the first two elements of the tuple, and ignore the rest of the elements. Note that this only works for python3

{a:b for a, b, *c in foo}                                                                                                                                                             

Answer 2

You can use extended iterable unpacking , keeping in this way the first two values from the iterable and ignoring the rest:

{a: b for a, b, *_ in foo}
# {1: 2, 5: 6}

Answer 3

Try this :

dict_ = {a:b for a,b, *_ in foo}

Output :

{1: 2, 5: 6}

If foo is changed to [(1, 2, 3, 4,9,0), (5, 6, 7, 8, 11, 16)] by program later, dict_ still remains : {1: 2, 5: 6}

Answer 4

use lambda

foo = [(1, 2, 3, 4), (5, 6, 7, 8)]
output = list(map(lambda x:{x[0]:x[1]}, foo))
print(output)

output

[{1: 2}, {5: 6}]