36

I'm trying to merge three dictionaries, which all have the same keys, and either lists of values, or single values.

one={'a': [1, 2], 'c': [5, 6], 'b': [3, 4]}
two={'a': [2.4, 3.4], 'c': [5.6, 7.6], 'b': [3.5, 4.5]}
three={'a': 1.2, 'c': 3.4, 'b': 2.3}

What I need is for all the items in the values to be added to one list.

result={'a': [1, 2, 2.4, 3.4, 1.2], 'c': [5, 6, 5.6, 7.6, 2.3], 'b': [3, 4, 3.5, 4.5, 3.4]}

I have tried several things, but most put the values into nested lists. E.g.

out=dict((k, [one[k], two.get(k), three.get(k)]) for k in one)
{'a': [[1, 2], [2.4, 3.4], 1.2], 'c': [[5, 6], [5.6, 7.6], 3.4], 'b': [[3, 4], [3.5, 4.5], 2.3]}

I tried updating it by looping through the values:

out.update((k, [x for x in v]) for k,v in out.iteritems())

but the results was exactly the same. I have tried to simply add the lists, but because the third dictionary has only a float, I couldn't do it. 

check=dict((k, [one[k]+two[k]+three[k]]) for k in one)

So I tried to first add the lists in values of one and two, and then append the value of three. Adding the lists worked well, but then when I tried to append the float from the third dictionary, suddenly the whole value went to 'None'

check=dict((k, [one[k]+two[k]]) for k in one)
{'a': [[1, 2, 2.4, 3.4]], 'c': [[5, 6, 5.6, 7.6]], 'b': [[3, 4, 3.5, 4.5]]}
new=dict((k, v.append(three[k])) for k,v in check.items())
{'a': None, 'c': None, 'b': None}
branwen85
  • 1,606
  • 5
  • 20
  • 25

6 Answers6

39

As a one-liner, with a dictionary comprehension:

new = {key: value + two[key] + [three[key]] for key, value in one.iteritems()}

This creates new lists, concatenating the list from one with the corresponding list from two, putting the single value in three into a temporary list to make concatenating easier.

Or with a for loop updating one in-place:

for key, value in one.iteritems():
    value.extend(two[key])
    value.append(three[key])

This uses list.extend() to update original list in-place with the list from two, and list.append() to add the single value from three.

Where you went wrong:

  • your first attempt creates a new list with the values from one, two and three nested within rather than concatenating the existing lists. Your attempt to clean that up just copied those nested lists across.

  • Your second attempt didn't work because the value in three is not a list so could not be concatenated. I created a new list just for that one value.

  • Your last attempt should not have used list.append() in a generator expression, because you store the return value of that method, which is always None (its change is stored in v directly and the list doesn't need returning again).

Demo of the first approach:

>>> one={'a': [1, 2], 'c': [5, 6], 'b': [3, 4]}
>>> two={'a': [2.4, 3.4], 'c': [5.6, 7.6], 'b': [3.5, 4.5]}
>>> three={'a': 1.2, 'c': 3.4, 'b': 2.3}
>>> {key: value + two[key] + [three[key]] for key, value in one.iteritems()}
{'a': [1, 2, 2.4, 3.4, 1.2], 'c': [5, 6, 5.6, 7.6, 3.4], 'b': [3, 4, 3.5, 4.5, 2.3]}
Martijn Pieters
  • 1,048,767
  • 296
  • 4,058
  • 3,343
25

Arbitrary dictionary number and keys

The issues with your attempt are covered by @MartijnPieters' solution.

For a generalised solution, consider using itertools.chain to chain multiple dictionaries. You can also use a defaultdict for the more general case where you do not find the same keys in each dictionary.

from collections import defaultdict
from itertools import chain
from operator import methodcaller

# dictionaries with non-equal keys, values all lists for simplicity
one = {'a': [1, 2], 'c': [5, 6], 'b': [3, 4], 'e': [6.2]}
two = {'a': [2.4, 3.4], 'c': [5.6, 7.6], 'b': [3.5, 4.5], 'f': [1.3]}
three = {'a': [1.2], 'c': [3.4], 'b': [2.3], 'e': [3.1]}

# initialise defaultdict of lists
dd = defaultdict(list)

# iterate dictionary items
dict_items = map(methodcaller('items'), (one, two, three))
for k, v in chain.from_iterable(dict_items):
    dd[k].extend(v)

print(dd)

# defaultdict(list,
#             {'a': [1, 2, 2.4, 3.4, 1.2],
#              'b': [3, 4, 3.5, 4.5, 2.3],
#              'c': [5, 6, 5.6, 7.6, 3.4],
#              'e': [6.2, 3.1],
#              'f': [1.3]})

Note defaultdict is a subclass of dict so there's generally no need to convert the result to a regular dict.

jpp
  • 159,742
  • 34
  • 281
  • 339
  • An alternative to using `defaultdict` is using [`dict.setdefault`](https://docs.python.org/3/library/stdtypes.html#dict.setdefault). `defaultdict` can come back to bite you if you forget that every time you'll access a key that's not there it will not throw an exception and return a list instead (or whatever your `defaultdict`'s default value factory method returns). – Dominykas Mostauskis Mar 23 '22 at 14:55
5

A robust solution. =)

def FullMergeDict(D1, D2):
  for key, value in D1.items():
    if key in D2:
      if type(value) is dict:
        FullMergeDict(D1[key], D2[key])
      else:
        if type(value) in (int, float, str):
          D1[key] = [value]
        if type(D2[key]) is list:
          D1[key].extend(D2[key])
        else:
          D1[key].append(D2[key])
  for key, value in D2.items():
    if key not in D1:
      D1[key] = value

if __name__ == '__main__':
  X = {
    'a': 'aaa',
    'c': [1,3,5,7],
    'd': 100,
    'e': {'k': 1, 'p': 'aa','t': [-1,-2]},
    'f': {'j':1}
  }
  Y = {
    'b': 'bbb',
    'd': 200,
    'e': {'k': 2, 'p': 'bb','o': [-4]},
    'c': [2,4,6],
    'g': {'v':2}
  }
  FullMergeDict(X, Y)
  exit(0)

Result:

  X = {
    'a': 'aaa',
    'b': 'bbb',
    'c': [1, 3, 5, 7, 2, 4, 6],
    'd': [100, 200],
    'e': {'k': [1, 2], 'o': [-4], 'p': ['aa', 'bb'], 't': [-1, -2]},
    'f': {'j': 1},
    'g': {'v': 2}}
Deoclecio Freire
  • 109
  • 2
  • 4
  • Excellent solution! I just used it to merge two dict of list in which the list values were tuples of tuple pairs. Oh, and the keys in the two dicts were also tuples. All tuples (keys and values) were coordinate pairs. Worked like a champ! – MarkS Aug 22 '21 at 12:51
2

If you have different keys and different types of values in dictionaries you can use the following approach:

from collections import defaultdict, Iterable

dct1 = {'a': [1, 2]}
dct2 = {'a': [3], 'b': [5, 6]}
dct3 = {'a': 4, 'c': 7}

result = defaultdict(list)
for dct in [dct1, dct2, dct3]:
    for k, v in dct.items():
        if isinstance(v, Iterable):
            result[k].extend(v)
        else:
            result[k].append(v)

print(result)
# defaultdict(<class 'list'>, {'a': [1, 2, 3, 4], 'b': [5, 6], 'c': [7]})
Mykola Zotko
  • 15,583
  • 3
  • 71
  • 73
1

One line solution (also handles for keys present in only one dict):

{ key:one.get(key,[])+two.get(key,[]) for key in set(list(one.keys())+list(two.keys())) }

Example 1:

one = {'a': [1, 2], 'c': [5, 6], 'b': [3, 4]}
two = {'a': [2.4, 3.4], 'c': [5.6, 7.6], 'd': [3.5, 4.5]}
{ key:one.get(key,[])+two.get(key,[]) for key in set(list(one.keys())+list(two.keys())) }

Output:

{'a': [1, 2, 2.4, 3.4], 'b': [3, 4], 'c': [5, 6, 5.6, 7.6], 'd': [3.5, 4.5]}

Example 2:

x={1:['a','b','c']}
y={1:['d','e','f'],2:['g']}
{ key:x.get(key,[])+y.get(key,[]) for key in set(list(x.keys())+list(y.keys())) }

Output:

{1: ['a', 'b', 'c', 'd', 'e', 'f'], 2: ['g']}

Shailesh Appukuttan
  • 770
  • 8
  • 19
0

See this help or not: 

>>> dic={}
>>> k=[]
>>> for i in 'abc':
    k=one[i]+two[i]
    k.append(three[i])
    dic[i]=k


>>> dic
{'c': [5, 6, 5.6, 7.6, 3.4], 'a': [1, 2, 2.4, 3.4, 1.2], 'b': [3, 4, 3.5, 4.5, 2.3]}
HBasiri
  • 343
  • 1
  • 6
  • 16