Flatten values of every key in python

Question

I have a dictionary like this:

migration_dict = {'30005': ['key42750','key43119', 'key44103', ['key333'],
['key444'], ['keyxx']], '30003': ['key43220', 'key42244','key42230',
['keyzz'], ['kehh']]}

How can I flatten the values of every key in order to have something like this:

migration_dict = {'30005': ['key42750','key43119', 'key44103', 'key333',
'key444', 'keyxx'], '30003': ['key43220', 'key42244','key42230',
'keyzz', 'kehh']}

Possible duplicate of [Flatten (an irregular) list of lists](https://stackoverflow.com/questions/2158395/flatten-an-irregular-list-of-lists) — Chris_Rands, Jun 09 '17 at 10:57

score 9 · Accepted Answer · answered Jun 09 '17 at 10:39

9

You can write a recursive function to flatten the value lists and use it in a dictionary comprehension to build a new dictionary:

def flatten(lst):
   for x in lst:
      if isinstance(x, list):
         for y in flatten(x): # yield from flatten(...) in Python 3
            yield y           #
      else:
         yield x

migration_dict = {k: list(flatten(v)) for k, v in dct.items()}
print(migration_dict)
# {'30005': ['key42750', 'key43119', 'key44103', 'key333', 'key444', 'keyxx'], '30003': ['key43220', 'key42244', 'key42230', 'keyzz', 'kehh']}

It handles any nesting depth in the dict value lists.

answered Jun 09 '17 at 10:39

Moses Koledoye

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Helpful, just for curiosity what if there is `interger` in our list ? like `l = [[1],2,3]` , I tried it but with `flatten()`, but it says `int object is not iteratble` – akash karothiya Jun 09 '17 at 10:44
@akashkarothiya Works the same, not sure how you're using the function. The ints are caught in the `else`, while the inner list(s) is processed recursively. Gives `[1, 2, 3]`. – Moses Koledoye Jun 09 '17 at 10:45
4

Just here to nitpick that this will run forever if any list has a reference to itself in it ;) (+1 ofc) – timgeb Jun 09 '17 at 10:48
@MosesKoledoye oh Yes, my bad thanks again, Is there any in-built function like `flatten` in python, here you wrote UDF ? – akash karothiya Jun 09 '17 at 10:48
@timegb Highly unlikely, but good observation. Patch: `if x is lst: raise ValueError('The list is self referencing')` – Moses Koledoye Jun 09 '17 at 10:52
@akashkarothiya I don't know of any that handles arbitrary depths. – Moses Koledoye Jun 09 '17 at 10:56
@MosesKoledoye ok, thanks again, i wish i could give more than +1, really helpful – akash karothiya Jun 09 '17 at 10:57
@timgeb With "runs forever" you mean "until it hits the recursion-limit"? :) – MSeifert Jun 09 '17 at 11:01
How to handle RuntimeError: maximum recursion depth exceeded while calling a Python object error? – e7lT2P Jun 09 '17 at 11:17

score 2 · Answer 2 · answered Jun 09 '17 at 10:38

for key in migration_dict:
    for i in migration_dict[key]:
        if type(i) == list:
            migration_dict[key].remove(i)
            for element in i:
                migration_dict[key].append(element)

Now this loop should do it. Note however, that this works only if the inner list doesn't have more lists within it. If it does then you might have to make a recursive function that flattens it out for you.

score 2 · Answer 3 · answered Jun 09 '17 at 10:50

If you don't mind using an 3rd party extension, you could use iteration_utilities.deepflatten¹ and a dict-comprehension:

>>> from iteration_utilities import deepflatten
>>> {key: list(deepflatten(value, ignore=str)) for key, value in migration_dict.items()}
{'30003': ['key43220', 'key42244', 'key42230', 'keyzz', 'kehh'],
 '30005': ['key42750', 'key43119', 'key44103', 'key333', 'key444', 'keyxx']}

This flattens all iterable items in your values (except strings).

¹ Disclaimer: I'm the author of that library

gold_cy · Answer 4 · 2017-06-09T10:52:06.273

Heres a oneliner:

Consider this dict, same structure:

m =  {'a': ['k1', 'k2', 'k3', ['k4'], ['k5'], ['k6']],
     'b': ['k1', 'k2', 'k3', ['k4'], ['k5'], ['k6']]}

import itertools

d = {k:list(itertools.chain.from_iterable(itertools.repeat(x,1) if isinstance(x, str) else x for x in v)) for k,v in m.iteritems()}

{'a': ['k1', 'k2', 'k3', 'k4', 'k5', 'k6'],
 'b': ['k1', 'k2', 'k3', 'k4', 'k5', 'k6']}

You could also use a 3rd party library more-iterools,

t = {k: list(more_itertools.collapse(v, base_type=str)) for k,v in m.iteritems()}

 {'a': ['k1', 'k2', 'k3', 'k4', 'k5', 'k6'],
 'b': ['k1', 'k2', 'k3', 'k4', 'k5', 'k6']}

score 0 · Answer 5 · answered Jun 09 '17 at 10:47

If its important to have flat values in your dictionary, maybe you could create a function that gets rid of all the (multi-dimensional)lists before adding them to the dictionary?

def flatten_values(value):
      if type(value) == list:
           for i in value:
                return i

Hope this helps, couldnt make this into a comment, because i don't have enough rep ;)

Flatten values of every key in python

5 Answers5