Dictionary comprehension with conditional

Question

So I'm wondering if anyone can help me out with this issue I'm having.

Lets assume I have a dictionary:

d = {1: {2: 3}, 4: 5}

I want to create a dictionary of any contained dictionaries:

wanted_result = {2: 3}

what I am trying is this:

e = {inner_key: d[key][inner_key] for key in d.keys() for inner_key in d[key].keys() if isinstance(d[key], dict)}

However this is causing me to get an error saying that ints don't have keys, which I know, but I thought my conditional would exclude say 4 from my example from being included in the comprehension.

What should the result be if there is more than one nested dict? What if they are nested more deeply? — Tim Pietzcker, May 11 '12 at 21:10
say started at d = {1: {2: {3: 4}, 5: 6}, 7: 8}, would just assume to get {2: {3: 4}, 5: 6} . ie only gathering the dictionaries one layer deep — spdub, May 11 '12 at 21:21

score 10 · Answer 1 · answered May 11 '12 at 21:21

10

d = {1: {2: 3}, 4: 5, 6: {7: 8}}
s = {k: v for elem in d.values() if type(elem) is dict for k, v in elem.items()}
>> {2: 3, 7: 8}

answered May 11 '12 at 21:21

georg

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juliomalegria · Answer 2 · 2012-05-11T21:13:00.830

2

In this case, I would recommend you a for-loop and the update method:

d = {1: {2: 3}, 4: 5, 6: {7: 8}}
inner_dicts = {}
for val in d.values():
    if type(val) is dict:
        inner_dicts.update(val)
print inner_dicts
# {2: 3, 7: 8}

edited May 11 '12 at 21:13

answered May 11 '12 at 21:08

juliomalegria

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garnertb · Answer 3 · 2012-05-11T21:14:58.623

1

The following list comprehension will return each value that is a dict:

>>> d = {1: {2: 3}, 4: 5}
>>> [d[i] for i in d.keys() if isinstance(d[i],dict)]
[{2: 3}]

edited May 11 '12 at 21:14

answered May 11 '12 at 21:09

garnertb

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use `isinstance()`, it's much more pythonic. – Ashwini Chaudhary May 11 '12 at 21:13
didn't he explicitly asked for a "**dictionary** of any contained dictionaries"? – juliomalegria May 11 '12 at 21:15
Obviously that is what he asked for, I'm just wondering how he would use that for multiple keys that are dicts. – garnertb May 11 '12 at 21:16

Ashwini Chaudhary · Answer 4 · 2012-05-11T21:33:29.630

1

d = {1: {2: 3}, 4: 5,'j':{7:8}}
e={y:d[x][y] for x in d if isinstance(d[x],dict) for y in d[x]}
print(e)

{2: 3, 7: 8}

edited May 11 '12 at 21:33

answered May 11 '12 at 21:11

Ashwini Chaudhary

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didn't he explicitly asked for a "**dictionary** of any contained dictionaries"? – juliomalegria May 11 '12 at 21:14

score 1 · Accepted Answer · edited May 23 '17 at 12:11

My first idea was something like this:

d = {1: {2: 3}, 4: 5, 6: {7: 8}}
generator = (x.items() for x in d.itervalues() if type(x) == dict)
s = dict(sum( generator, list() )) 
# s = {2: 3, 7: 8}

But, in order to avoid letting sum() construct a large temporary list of all the items() you may use itertools.chain and iteritems() instead:

# chain "concatenates" iterables
from itertools import chain
d = {1: {2: 3}, 4: 5, 6: {7: 8}}
s = dict(chain( *(x.iteritems() for x in d.itervalues() if type(x) == dict) ))
# s = {2: 3, 7: 8}

You should be aware of edge cases. Especially if identical keys are in the part-dictionaries these will necessarily collapse somehow.

If you also want to support subclasses of dict to be collected from your d.values(), you may indeed want to use isinstance instead of type. See this question for details.

Why in earth did you choose `_` as your variable name? (http://stackoverflow.com/a/1739541/972208) — juliomalegria, May 11 '12 at 21:20
@julio.alegria I used this occasionally for temporary variables (be sure to not shadow another `x`). Wasn't aware of the special semantics in interactive mode though! — moooeeeep, May 11 '12 at 21:25

score 0 · Answer 6 · answered Jan 29 '13 at 08:29

0

Other way using lambda

(lambda x: {k:v for x in d.values() if type(x) == dict for k,v in x.items()})(d)

answered Jan 29 '13 at 08:29

Mirage

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Dictionary comprehension with conditional

6 Answers6