python – How to merge multiple dicts with same key or different key?

python – How to merge multiple dicts with same key or different key?

Heres 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)

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))

python – How to merge multiple dicts with same key or different key?

Here is one approach you can use which would work even if both dictonaries dont 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]}