python-3.xlistdictionarylambdafunctional-programming

# List to dictionary using functional programming in python

I want to convert a list to dictionary using the rule `x if x odd; (x*2, x*3) if x is even` using functional programming in python.

Example `list([1, 2, 3])` to `dict({1:1, 4:2, 6:2, 3:3})`

I have a code that converts list to `{1: 1, (4, 2): (6, 2), 3: 3}`

``````from itertools import chain
print(
dict(
map(
lambda elem:
(lambda x: (x, x))(elem) if elem % 2 == 1
else (lambda x: chain(
((x*2, x), (x*3, x))
))(elem),
list([1, 2, 3])
)
)
)
``````

Could anyone figure it out?

The result should be the same as this code produces

``````d = dict()
for x in [1, 2, 3]:
if x % 2 == 1:
d[x] = x
else:
d[x*2] = x
d[x*3] = x
``````

Solution

• You can simply use dictionary comprehension:

``````lst = [1, 2, 3]

out = {k: v for v in lst for k in ((v,) if v % 2 else (v * 2, v * 3))}
print(out)
``````

Prints:

``````{1: 1, 4: 2, 6: 2, 3: 3}
``````

EDIT: Without for-loops or any libraries:

``````d = (
dict(map(lambda x: (x, x), filter(lambda x: x % 2, lst)))
| dict(map(lambda x: (x * 2, x), filter(lambda x: x % 2 == 0, lst)))
| dict(map(lambda x: (x * 3, x), filter(lambda x: x % 2 == 0, lst)))
)
print(d)
``````

Prints:

``````{1: 1, 3: 3, 4: 2, 6: 2}
``````