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I have the following two toy dicts

d1 = {

 'a': [2,4,5,6,8,10],

 'b': [1,2,5,6,9,12],

 'c': [0,4,5,8,10,21]

 }

d2 = {

 'a': [12,15],

 'b': [14,16],

 'c': [23,35]

  }

and I would like get a unique dictionary where I stack the second dictionary values after the first ones, within the same square brackets.

I tried the following code

d_comb = {key:[d1[key], d2[key]] for key in d1}

but the output I obtain has two lists within a list for each key, i.e.

{'a': [[2, 4, 5, 6, 8, 10], [12, 15]],

 'b': [[1, 2, 5, 6, 9, 12], [14, 16]],

 'c': [[0, 4, 5, 8, 10, 21], [23, 35]]}

whereas I would like to obtain

{'a': [2, 4, 5, 6, 8, 10, 12, 15],

 'b': [1, 2, 5, 6, 9, 12, 14, 16],

 'c': [0, 4, 5, 8, 10, 21, 23, 35]}

How can I do that with a line or two of code?

You almost had it, instead use + to append both lists:

{key: d1[key] + d2[key] for key in d1}



{'a': [2, 4, 5, 6, 8, 10, 12, 15],

 'b': [1, 2, 5, 6, 9, 12, 14, 16],

 'c': [0, 4, 5, 8, 10, 21, 23, 35]}

if not all the keys from d2 are in d1, then the simplest thing is using set union and dict.get:

combined_keys = d1.keys() | d2.keys()

d_comb = {key: d1.get(key, ) + d2.get(key, ) for key in combined_keys}

You could use extended iterable unpacking:

d1 = {

 'a': [2,4,5,6,8,10],

 'b': [1,2,5,6,9,12],

 'c': [0,4,5,8,10,21]

 }

d2 = {

 'a': [12,15],

 'b': [14,16],

 'c': [23,35]

  }



d_comb = {key:[*d1[key], *d2[key]] for key in d1}



print(d_comb)

Output

{'c': [0, 4, 5, 8, 10, 21, 23, 35], 'b': [1, 2, 5, 6, 9, 12, 14, 16], 'a': [2, 4, 5, 6, 8, 10, 12, 15]}

The code will work irrespective of whether d1 or d2 have the same set of keys. I have added a key 'e' in d1 and 'd' in d2.

d1 = {'a': [2, 4, 5, 6, 8, 10], 'b': [1, 2, 5, 6, 9, 12], 'c': [0, 4, 5, 8, 10, 21], 'e':[0,0,0]}

d2 = {'a': [12, 15], 'b': [14, 16], 'c': [23, 35], 'd': [13, 3]}



d2_keys_not_in_d1 = d2.keys() - d1.keys()

d1_keys_not_in_d2 = d1.keys() - d2.keys()

common_keys = d2.keys() & d1.keys()



for i in common_keys:

    d[i]=d1[i]+d2[i]

for i in d1_keys_not_in_d2:

    d[i]=d1[i]

for i in d2_keys_not_in_d1:

    d[i]=d2[i]

d

{'a': [2, 4, 5, 6, 8, 10, 12, 15],

 'b': [1, 2, 5, 6, 9, 12, 14, 16],

 'c': [0, 4, 5, 8, 10, 21, 23, 35],

 'd': [13, 3],

 'e': [0, 0, 0]}

You can use itertools.chain to efficiently construct a single list from input lists:

from itertools import chain

d_comb = {key: list(chain(d1[key], d2[key])) for key in d1}

For the more general case covering an arbitrary number of dictionaries and keys which are not equal across dictionaries, see Merging dictionary value lists in python.