Two dictionary is below
d1 = {'1': {'index': '1', 'sc': '4', 'st': '3'}, '2': {'index': '2', 'sc': '5', 'st': '5'}}
d2 = {'1': {'diff': 1}, '2': {'diff': 0}}
Code is below
z = {**d2, **d1} why this is not working
Tried below code also
def Merge(d1, d2):
return(d2.update(d1))
print(Merge(d1, d2))
Expetec out below
{'1': {'index': '1', 'sc': '4', 'st': '3', 'diff': 1},
'2': {'index': '2', 'sc': '5', 'st': '5', 'diff': 0}}
an alternate way using pandas
>>> import pandas as pd
>>> df = pd.DataFrame(d1)
>>> df2 = pd.DataFrame(d2)
>>> merged_dict = pd.concat([df,df2]).to_dict()
output
>>> merged_dict
{'1': {'index': '1', 'sc': '4', 'st': '3', 'diff': 1}, '2': {'index': '2', 'sc': '5', 'st': '5', 'diff': 0}}
generally, ** will capture any keyword arguments we pass to the function into a dictionary which that attributes arguments will reference. For example:
d1={'a':1,'b':2}
d2={'c':3,'d':4}
def merge(**di):
res = {}
for k, v in di.items():
try:
res[k].append(v)
except KeyError:
res[k] = [v]
return res
print(merge(**d1, **d2))
# {'a': [1], 'b': [2], 'c': [3], 'd': [4]}
However, if we pass in two dictionary with same keys:
d1 = {'1': {'index': '1', 'sc': '4', 'st': '3'}, '2': {'index': '2', 'sc': '5', 'st': '5'}}
d2 = {'1': {'diff': 1}, '2': {'diff': 0}}
def merge(**di):
res = {}
for k, v in di.items():
try:
res[k].append(v)
except KeyError:
res[k] = [v]
return res
print(merge(**d1, **d2))
# TypeError: merge() got multiple values for keyword argument '1'
This error is handled by continuing which keep the original one and skip the second dict key. Sorry I don't have a shorthand method for this.
d1 = {'1': {'index': '1', 'sc': '4', 'st': '3'}, '2': {'index': '2', 'sc': '5', 'st': '5'}}
d2 = {'1': {'diff': 1}, '2': {'diff': 0}}
def merge(*args):
res = {}
for di in args:
for k, v in di.items():
try:
res[k].update(v)
except KeyError:
res[k] = v
return res
print(merge(d1, d2))
# {'1': {'index': '1', 'sc': '4', 'st': '3', 'diff': 1}, '2': {'index': '2', 'sc': '5', 'st': '5', 'diff': 0}}
z = {**d2, **d1}
Overwrites everything in d2 with d1 values for keys '1', and '2'. It is tricky to merge dictionaries with the same keys so you don't overwrite key:value pairs within those keys.
The following will get you to the depth needed in both d1 and d2 to update d1 to your expected output:
d1['1']['diff'] = d2['1']['diff']
d1['2']['diff'] = d2['2']['diff']
print ('d1:', d1)
Output:
d1: {'1': {'index': '1', 'sc': '4', 'st': '3', 'diff': 1}, '2': {'index': '2', 'sc': '5', 'st': '5', 'diff': 0}}
>>> for key in d1:
... d1[key].update(d2[key])
>>> d1
{'1': {'index': '1', 'sc': '4', 'st': '3', 'diff': 1}, '2': {'index': '2', 'sc': '5', 'st': '5', 'diff': 0}}
Update:
If you want in another identifier d3.
d3 = d1.copy()
for key in d3:
d3[key].update(d2[key])
print(d3)
Dictionaries are mutable objects. update function just mutates/updates the object and return None. So, you need to create a copy (so you have another object) and change the new object, if you want initial data unaltered.
