How to merge multiple dicts with same key or different key?
PythonPython 3.xDictionaryMergePython Problem Overview
I have multiple dicts/key-value pairs like this:
d1 = {key1: x1, key2: y1}
d2 = {key1: x2, key2: y2}
I want the result to be a new dict (in most efficient way, if possible):
d = {key1: (x1, x2), key2: (y1, y2)}
Actually, I want result d to be:
d = {key1: (x1.x1attrib, x2.x2attrib), key2: (y1.y1attrib, y2.y2attrib)}
If somebody shows me how to get the first result, I can figure out the rest.
Python Solutions
Solution 1 - Python
Here's a general solution that will handle an arbitrary amount of dictionaries, with cases when keys are in only some of the dictionaries:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(list)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key].append(value)
print(dd)
Shows:
defaultdict(<type 'list'>, {1: [2, 6], 3: [4, 7]})
Also, to get your .attrib
, just change append(value)
to append(value.attrib)
Solution 2 - Python
assuming all keys are always present in all dicts:
ds = [d1, d2]
d = {}
for k in d1.iterkeys():
d[k] = tuple(d[k] for d in ds)
Note: In Python 3.x use below code:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = tuple(d[k] for d in ds)
and if the dic contain numpy arrays:
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = np.concatenate(list(d[k] for d in ds))
Solution 3 - Python
Here is one approach you can use which would work even if both dictonaries don't have same keys:
d1 = {'a':'test','b':'btest','d':'dreg'}
d2 = {'a':'cool','b':'main','c':'clear'}
d = {}
for key in set(d1.keys() + d2.keys()):
try:
d.setdefault(key,[]).append(d1[key])
except KeyError:
pass
try:
d.setdefault(key,[]).append(d2[key])
except KeyError:
pass
print d
This would generate below input:
{'a': ['test', 'cool'], 'c': ['clear'], 'b': ['btest', 'main'], 'd': ['dreg']}
Solution 4 - Python
dict1 = {'m': 2, 'n': 4}
dict2 = {'n': 3, 'm': 1}
Making sure that the keys are in the same order:
dict2_sorted = {i:dict2[i] for i in dict1.keys()}
keys = dict1.keys()
values = zip(dict1.values(), dict2_sorted.values())
dictionary = dict(zip(keys, values))
gives:
{'m': (2, 1), 'n': (4, 3)}
Solution 5 - Python
If you only have d1 and d2,
from collections import defaultdict
d = defaultdict(list)
for a, b in d1.items() + d2.items():
d[a].append(b)
Solution 6 - Python
This function merges two dicts even if the keys in the two dictionaries are different:
def combine_dict(d1, d2):
return {
k: tuple(d[k] for d in (d1, d2) if k in d)
for k in set(d1.keys()) | set(d2.keys())
}
Example:
d1 = {
'a': 1,
'b': 2,
}
d2` = {
'b': 'boat',
'c': 'car',
}
combine_dict(d1, d2)
# Returns: {
# 'a': (1,),
# 'b': (2, 'boat'),
# 'c': ('car',)
# }
Solution 7 - Python
Python 3.x Update
From Eli Bendersky answer:
Python 3 removed dict.iteritems use dict.items instead. See Python wiki: https://wiki.python.org/moin/Python3.0
from collections import defaultdict
dd = defaultdict(list)
for d in (d1, d2):
for key, value in d.items():
dd[key].append(value)
Solution 8 - Python
Assume that you have the list of ALL keys (you can get this list by iterating through all dictionaries and get their keys). Let's name it listKeys
. Also:
listValues
is the list of ALL values for a single key that you want to merge.allDicts
: all dictionaries that you want to merge.
result = {}
for k in listKeys:
listValues = [] #we will convert it to tuple later, if you want.
for d in allDicts:
try:
fileList.append(d[k]) #try to append more values to a single key
except:
pass
if listValues: #if it is not empty
result[k] = typle(listValues) #convert to tuple, add to new dictionary with key k
Solution 9 - Python
Assuming there are two dictionaries with exact same keys, below is the most succinct way of doing it (python3 should be used for both the solution).
d1 = {'a': 1, 'b': 2, 'c':3}
d2 = {'a': 5, 'b': 6, 'c':7}
# get keys from one of the dictionary
ks = [k for k in d1.keys()]
print(ks)
['a', 'b', 'c']
# call values from each dictionary on available keys
d_merged = {k: (d1[k], d2[k]) for k in ks}
print(d_merged)
{'a': (1, 5), 'b': (2, 6), 'c': (3, 7)}
# to merge values as list
d_merged = {k: [d1[k], d2[k]] for k in ks}
print(d_merged)
{'a': [1, 5], 'b': [2, 6], 'c': [3, 7]}
If there are two dictionaries with some common keys, but a few different keys, a list of all the keys should be prepared.
d1 = {'a': 1, 'b': 2, 'c':3, 'd': 9}
d2 = {'a': 5, 'b': 6, 'c':7, 'e': 4}
# get keys from one of the dictionary
d1_ks = [k for k in d1.keys()]
d2_ks = [k for k in d2.keys()]
all_ks = set(d1_ks + d2_ks)
print(all_ks)
['a', 'b', 'c', 'd', 'e']
# call values from each dictionary on available keys
d_merged = {k: [d1.get(k), d2.get(k)] for k in all_ks}
print(d_merged)
{'d': [9, None], 'a': [1, 5], 'b': [2, 6], 'c': [3, 7], 'e': [None, 4]}
Solution 10 - Python
def merge(d1, d2, merge):
result = dict(d1)
for k,v in d2.iteritems():
if k in result:
result[k] = merge(result[k], v)
else:
result[k] = v
return result
d1 = {'a': 1, 'b': 2}
d2 = {'a': 1, 'b': 3, 'c': 2}
print merge(d1, d2, lambda x, y:(x,y))
{'a': (1, 1), 'c': 2, 'b': (2, 3)}
Solution 11 - Python
To supplement the two-list solutions, here is a solution for processing a single list.
A sample list (NetworkX-related; manually formatted here for readability):
ec_num_list = [((src, tgt), ec_num['ec_num']) for src, tgt, ec_num in G.edges(data=True)]
print('\nec_num_list:\n{}'.format(ec_num_list))
ec_num_list:
[((82, 433), '1.1.1.1'),
((82, 433), '1.1.1.2'),
((22, 182), '1.1.1.27'),
((22, 3785), '1.2.4.1'),
((22, 36), '6.4.1.1'),
((145, 36), '1.1.1.37'),
((36, 154), '2.3.3.1'),
((36, 154), '2.3.3.8'),
((36, 72), '4.1.1.32'),
...]
Note the duplicate values for the same edges (defined by the tuples). To collate those "values" to their corresponding "keys":
from collections import defaultdict
ec_num_collection = defaultdict(list)
for k, v in ec_num_list:
ec_num_collection[k].append(v)
print('\nec_num_collection:\n{}'.format(ec_num_collection.items()))
ec_num_collection:
[((82, 433), ['1.1.1.1', '1.1.1.2']), ## << grouped "values"
((22, 182), ['1.1.1.27']),
((22, 3785), ['1.2.4.1']),
((22, 36), ['6.4.1.1']),
((145, 36), ['1.1.1.37']),
((36, 154), ['2.3.3.1', '2.3.3.8']), ## << grouped "values"
((36, 72), ['4.1.1.32']),
...]
If needed, convert that list to dict:
ec_num_collection_dict = {k:v for k, v in zip(ec_num_collection, ec_num_collection)}
print('\nec_num_collection_dict:\n{}'.format(dict(ec_num_collection)))
ec_num_collection_dict:
{(82, 433): ['1.1.1.1', '1.1.1.2'],
(22, 182): ['1.1.1.27'],
(22, 3785): ['1.2.4.1'],
(22, 36): ['6.4.1.1'],
(145, 36): ['1.1.1.37'],
(36, 154): ['2.3.3.1', '2.3.3.8'],
(36, 72): ['4.1.1.32'],
...}
References
Solution 12 - Python
From blubb answer:
You can also directly form the tuple using values from each list
ds = [d1, d2]
d = {}
for k in d1.keys():
d[k] = (d1[k], d2[k])
This might be useful if you had a specific ordering for your tuples
ds = [d1, d2, d3, d4]
d = {}
for k in d1.keys():
d[k] = (d3[k], d1[k], d4[k], d2[k]) #if you wanted tuple in order of d3, d1, d4, d2
Solution 13 - Python
This library helped me, I had a dict list of nested keys with the same name but with different values, every other solution kept overriding those nested keys.
https://pypi.org/project/deepmerge/
from deepmerge import always_merger
def process_parms(args):
temp_list = []
for x in args:
with open(x, 'r') as stream:
temp_list.append(yaml.safe_load(stream))
return always_merger.merge(*temp_list)
Solution 14 - Python
If keys are nested:
d1 = { 'key1': { 'nkey1': 'x1' }, 'key2': { 'nkey2': 'y1' } }
d2 = { 'key1': { 'nkey1': 'x2' }, 'key2': { 'nkey2': 'y2' } }
ds = [d1, d2]
d = {}
for k in d1.keys():
for k2 in d1[k].keys():
d.setdefault(k, {})
d[k].setdefault(k2, [])
d[k][k2] = tuple(d[k][k2] for d in ds)
yields:
{'key1': {'nkey1': ('x1', 'x2')}, 'key2': {'nkey2': ('y1', 'y2')}}
Solution 15 - Python
A better representation for two or more dicts with the same keys is a pandas Data Frame
IMO:
d1 = {"key1": "x1", "key2": "y1"}
d2 = {"key1": "x2", "key2": "y2"}
d3 = {"key1": "x3", "key2": "y3"}
d1_df = pd.DataFrame.from_dict(d1, orient='index')
d2_df = pd.DataFrame.from_dict(d2, orient='index')
d3_df = pd.DataFrame.from_dict(d3, orient='index')
fin_df = pd.concat([d1_df, d2_df, d3_df], axis=1).T.reset_index(drop=True)
fin_df
key1 key2
0 x1 y1
1 x2 y2
2 x3 y3
Solution 16 - Python
> Using below method we can merge two dictionaries having same keys.
def update_dict(dict1: dict, dict2: dict) -> dict:
output_dict = {}
for key in dict1.keys():
output_dict.update({key: []})
if type(dict1[key]) != str:
for value in dict1[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict1[key])
if type(dict2[key]) != str:
for value in dict2[key]:
output_dict[key].append(value)
else:
output_dict[key].append(dict2[key])
return output_dict
> Input: d1 = {key1: x1, key2: y1} d2 = {key1: x2, key2: y2}
> Output: {'key1': ['x1', 'x2'], 'key2': ['y1', 'y2']}
Solution 17 - Python
There is a great library funcy
doing what you need in a just one, short line.
from funcy import join_with
from pprint import pprint
d1 = {"key1": "x1", "key2": "y1"}
d2 = {"key1": "x2", "key2": "y2"}
list_of_dicts = [d1, d2]
merged_dict = join_with(tuple, list_of_dicts)
pprint(merged_dict)
Output:
{'key1': ('x1', 'x2'), 'key2': ('y1', 'y2')}
More info here: funcy -> join_with.
Solution 18 - Python
Modifying this answer to create a dictionary of tuples (what the OP asked for), instead of a dictionary of lists:
from collections import defaultdict
d1 = {1: 2, 3: 4}
d2 = {1: 6, 3: 7}
dd = defaultdict(tuple)
for d in (d1, d2): # you can list as many input dicts as you want here
for key, value in d.items():
dd[key] += (value,)
print(dd)
The above prints the following:
defaultdict(<class 'tuple'>, {1: (2, 6), 3: (4, 7)})
Solution 19 - Python
dicts = [dict1,dict2,dict3]
out = dict(zip(dicts[0].keys(),[[dic[list(dic.keys())[key]] for dic in dicts] for key in range(0,len(dicts[0]))]))
Solution 20 - Python
A compact possibility
d1={'a':1,'b':2}
d2={'c':3,'d':4}
context={**d1, **d2}
context
{'b': 2, 'c': 3, 'd': 4, 'a': 1}