Convert a list comprehension to functional programming

Question

I have a list of dictionaries

lst = [{'a': (1, 2, 3), 'b': (2, 3)},
       {'c': (3, 6), 'd': (4, 8), 'e': (5, 10)},
       {'d': (6, 12), 'e': (7, 14)}]

For each key in each dictionary, I want to keep only the first element of the values. So the desired output is

[{'a': 1, 'b': 2}, {'c': 3, 'd': 4, 'e': 5}, {'d': 6, 'e': 7}]

I can get it using a list comprehension like

[{key: val[0] for key, val in dct.items()} for dct in lst]

However, I want to know if it's possible to get the same output using map, itemgetter, itertools, functools etc. What I have so far:

map(dict.values, lst)

But I don't know how to go from here.

Answer 1

For nested iterations, I don't think we can do it without the help of lambda expressions:

from operator import itemgetter, methodcaller

list(map(
     lambda items: dict(zip(
             map(itemgetter(0), items),
             map(itemgetter(0), map(itemgetter(1), items))
         )), map(methodcaller('items'), lst)))
# [{'a': 1, 'b': 2}, {'c': 3, 'd': 4, 'e': 5}, {'d': 6, 'e': 7}]

I have to say it's very ugly.

Update: I found a way to avoid lambda:

1. First, according to the comment area, we simplify the above expression (the outermost list is omitted here to reduce the difficulty of understanding):

func = lambda d: dict(zip(d, map(itemgetter(0), d.values())))
map(func, lst)

2. It is easy to observe that dict can be moved outside lambda. We just need to add another map:

func = lambda d: zip(d, map(itemgetter(0), d.values()))
map(dict, map(func, lst))

3. Similarly, we can move the zip outside lambda:

func = lambda d: map(itemgetter(0), d.values())
map(dict, map(zip, lst, map(func, lst)))

4. This seems to be the end, and there seems to be no way to convert lambda into a combination of multiple built-in functions, but there are still ways, let's first try to move d.values outside lambda. Here, since the element type of the list is determined, we directly use dict.values instead of operator.methodcaller:

func = lambda values: map(itemgetter(0), values)
map(dict, map(zip, lst, map(func, map(dict.values, lst))))

5. The answer is ready to come out. We can eliminate lambda by using functools.partial:

map(dict, map(zip, lst, map(partial(map, itemgetter(0)), map(dict.values, lst))))

Test:

>>> from operator import itemgetter
>>> from functools import partial
>>> lst = [{'a': (1, 2, 3), 'b': (2, 3)},
...        {'c': (3, 6), 'd': (4, 8), 'e': (5, 10)},
...        {'d': (6, 12), 'e': (7, 14)}]
>>> map(dict, map(zip, lst, map(partial(map, itemgetter(0)), map(dict.values, lst))))
<map object at 0x000002A0542CBB20>
>>> list(_)
[{'a': 1, 'b': 2}, {'c': 3, 'd': 4, 'e': 5}, {'d': 6, 'e': 7}]