CODEWITHSUNDEEP
pythondictionary
istern
istern

Reputation: 393

Accessing nested tupels in dictionary values

I'm posting this question after not finding the answer in the many existing issues related to this topic.
I have a dictionary similar to this one:

foo = {'a': [(1, 0.5), (2, 0.3)], 'b': [(3, 0.4), (4, 0.1)]}

and would like to access the items of the nested tupels, in order to create the following dictionary out of it:

foo = {'a': [1, 2], 'b': [3, 4]}

How do I go about doing this, preferably in a pythonic way?

Upvotes: 1

Views: 87

Answers (5)

timgeb
timgeb

Reputation: 78650

I think these two solutions are easiest to read:

>>> from operator import itemgetter
>>> {k:map(itemgetter(0), foo[k]) for k in foo}
{'a': [1, 2], 'b': [3, 4]}

or import-less:

>>> {k:[x[0] for x in foo[k]] for k in foo}
{'a': [1, 2], 'b': [3, 4]}

Upvotes: 1

Iron Fist
Iron Fist

Reputation: 10951

You can do it with defaultdict from collections module:

>>> from collections import defaultdict
>>> d = defaultdict(list)
>>> d = defaultdict(list)
>>> for k in foo:
        for v in foo[k]:
            d[k].append(v[0])
>>> d
defaultdict(<class 'list'>, {'a': [1, 2], 'b': [3, 4]})
>>> d['a']
[1, 2]

Upvotes: 0

vk1011
vk1011

Reputation: 7179

{k:list(zip(x,y)[0]) for k,(x,y) in foo.iteritems()}

This gives:

{'a': [1, 2], 'b': [3, 4]}

Upvotes: 0

Anton Protopopov
Anton Protopopov

Reputation: 31662

Using dict comprehension: 

In [214]: {key:list(map(lambda x: x[0], foo[key])) for key in foo}
Out[214]: {'a': [1, 2], 'b': [3, 4]}

Upvotes: 0

Paweł Kordowski
Paweł Kordowski

Reputation: 2768

what about?

{x: map(lambda z: z[0], y) for x, y in foo.iteritems()}

Upvotes: 0

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