Merging dictionary value lists in python
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]})
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}}
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
andthree
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 alwaysNone
(its change is stored inv
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]}
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
.