通过列表理解将元组分解为函数

Question

Let's say I have a list of tuples and a simple function that takes two arguments:

a_list = [(1,6), (6,8)]
add_it(a + b)

I now want to unpack the tuple into the function. I know I can do it by indices:

l = [add_it(tup[0], tup[1]) for tup in a_list]

But I kinda expected this would work, too:

l = [add_it(a, b) for tup in a_list for a, b in tup]

Because this works:

tup = (3,5)
a, b = tup
>>> 3 5

Can someone point me to what I am missing?

Answer 1

You should do:

l = [add_it(a, b) for a, b in a_list]
# or
l = [add_it(*tup) for tup in a_list]

because

a, b in tup cannot work as tup does not contain pairs, a_list does! Translated to your nested logic that would be:

l = [add_it(*(x for x in tup)) for tup in a_list]

But that is just unnecessarily wordy.

Answer 2

no need for a double loop. Just unpack in the loop arguments

l = [add_it(tup[0], tup[1]) for tup in a_list]

becomes

l = [add_it(a,b) for a,b in a_list]

note: that would work as well (unpacking arguments):

l = [add_it(*tup) for tup in a_list]

Answer 3

Because in your innermost for statement you are iterating over the tuple, so you cannot unpack it. You unpack the individual items of a sequence. If it were a tuple of pairs or a list of pairs, then it would work.

>>> a_list = [(1,6), (6,8)]
>>> b_list = [[(1,6)], [(6,8)]]
>>> def add_it(a, b): return a+b
...
>>> l = [add_it(a, b) for tup in a_list for a, b in tup]
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 1, in <listcomp>
TypeError: 'int' object is not iterable
>>> l = [add_it(a, b) for tup in b_list for a, b in tup]
>>> l
[7, 14]

And, as pointed out by others, it is overengineered, you only need one for expression in your comprehension since a_list is already a list of pairs:

>>> [add_it(a,b) for a,b in a_list]
[7, 14]