Related
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.
I have Dictionary is below, My product output is below. I need to create a new dictionary with two types Out_1, Out_2
product = {'Product1': {'index': '1', '1': 'Book', '2': 'Pencil', '3': 'Pen','value': '1'},
'Product2': {'index': '2', '1': 'Marker', '2': 'MYSQL', '3': 'Scale','value': '0'}}
If value inside product is 0 then extract the keys
Expected Output
Out_1 = {'Product2': {1:'Marker': '2': 'Compass', '3': 'Scale', 'value': 0}}
Out_2 = {'Product2':['Marker','Compass','Scale', '0']}
Psuedo code is below. i tried to create but not able to create as above
Out_1 = {}
Out_2 = {i:[]}
for i,j in product.items():
for a,b in j.items():
if a['value'] == 0:
Out_2.append(i)
I am getting indices error, How to get Out_1, Out_2
You can use dict comprehensions for this.
out_1 = {k: v for k, v in product.items() if v['value']=='0'}
out_2 = {k: list(v.values()) for k, v in product.items() if v['value']=='0'}
Hi do you really need this index variable? If not yes why would not you use list of dicts instead of dict of dicts. However here is what you wanted:
products = {'Product1': {'index': '1', '1': 'Book', '2': 'Pencil', '3': 'Pen','value': '1'},
'Product2': {'index': '2', '1': 'Marker', '2': 'MYSQL', '3': 'Scale','value': '0'}}
for k,product in products.items():
product.pop('index', None)
if product['value'] == '0':
products[k] = list(product.values())
print(products)
>>> {'Product1': {'1': 'Book', '2': 'Pencil', '3': 'Pen', 'value': '1'}, 'Product2': ['Marker', 'MYSQL', 'Scale', '0']}
I was not assigning it to any other variables like out1/2 in case you have more than 2 products
Here it is:
Code:
product = {'Product1': {'index': '1', '1': 'Book', '2': 'Pencil', '3': 'Pen','value': '1'},
'Product2': {'index': '2', '1': 'Marker', '2': 'MYSQL', '3': 'Scale','value': '0'}}
output_1 = {}
output_2 = {}
for key,val in product.items():
if (val['value'] == '0'):
output_1[key]=val
output_2[key]=val.values()
print(output_1)
print(output_2)
Output:
{'Product2': {'1': 'Marker', 'index': '2', '3': 'Scale', '2': 'MYSQL', 'value': '0'}}
{'Product2': ['Marker', '2', 'Scale', 'MYSQL', '0']}
I have two files which are a and b. I want to import certain information from data b to data a with the unique id from every response.
data
a= [{'id':'abc23','name':'aa','age':'22',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}},
{'id':'abc25','name':'bb','age':'32',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}},
{'id':'abc60','name':'cc','age':'24',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}}]
b=[{'id':'abc23','read':'2','speak':'abc','write':'2'},
{'id':'abc25','read':'3','speak':'def','write':'3'},
{'id':'abc60','read':'5','speak':'dgf','write':'1'}]
Code that I used to import from b to a :
from pprint import pprint
for dest in a:
for source in b:
if source['id'] == dest['id']:
dest['data'].update(source)
pprint(a)
Output from the code that i used :
[{ 'age': '22',
'data': {'id': 'abc23', 'read': '2', 'speak': 'abc', 'write': '2'},
'id': 'abc23',
'name': 'aa',
'responses': {'a': 1, 'b': 2}},
{ 'age': '32',
'data': {'id': 'abc25', 'read': '3', 'speak': 'def', 'write': '3'},
'id': 'abc25',
'name': 'bb',
'responses': {'a': 1, 'b': 2}},
{ 'age': '24',
'data': {'id': 'abc60', 'read': '5', 'speak': 'dgf', 'write': '1'},
'id': 'abc60',
'name': 'cc',
'responses': {'a': 1, 'b': 2}}]
But... This is the output that I want:
[{'age': '22',
'data': {'read': '2', 'speak': 'abc'},
'id': 'abc23',
'name': 'aa',
'responses': {'a': 1, 'b': 2}},
{'age': '32',
'data': {'read': '3', 'speak': 'def'},
'id': 'abc25',
'name': 'bb',
'responses': {'a': 1, 'b': 2}},
{'age': '24',
'data': {'read': '5', 'speak': 'dgf'},
'id': 'abc60',
'name': 'cc',
'responses': {'a': 1, 'b': 2}}]
It can't work the way you want with your code.
You do
dest['data'].update(source)
where source is
{'id':'abc23','read':'2','speak':'abc','write':'2'}
and dest['data'] is {'read':'','speak':''}.
When you update it will add all key-value pairs to dest['data'] and preserve the ones that won't be overwritten.
from pprint import pprint
for dest in a:
for source in b:
if source['id'] == dest['id']:
dest['data'] = {k: v for k, v in source.items() if k in dest.get('data', {})}
pprint(a)
This one will look for all the fields that are 'updateable' for each case. You might want to hardcode it, depending on your use case.
This is one approach by changing b to a dict for easy lookup.
Ex:
a= [{'id':'abc23','name':'aa','age':'22',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}},
{'id':'abc25','name':'bb','age':'32',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}},
{'id':'abc60','name':'cc','age':'24',
'data':{'read':'','speak':''},
'responses':{'a':1,'b':2}}]
b=[{'id':'abc23','read':'2','speak':'abc','write':'2'},
{'id':'abc25','read':'3','speak':'def','write':'3'},
{'id':'abc60','read':'5','speak':'dgf','write':'1'}]
b = {i.pop('id'): i for i in b} #Convert to dict key = ID & value = `read`, `speak`, `write`
for i in a:
i['data'].update(b[i['id']]) #Update list
print(a)
Output:
[{'age': '22',
'data': {'read': '2', 'speak': 'abc', 'write': '2'},
'id': 'abc23',
'name': 'aa',
'responses': {'a': 1, 'b': 2}},
{'age': '32',
'data': {'read': '3', 'speak': 'def', 'write': '3'},
'id': 'abc25',
'name': 'bb',
'responses': {'a': 1, 'b': 2}},
{'age': '24',
'data': {'read': '5', 'speak': 'dgf', 'write': '1'},
'id': 'abc60',
'name': 'cc',
'responses': {'a': 1, 'b': 2}}]
2 dictionary d1,d2, create a new dictionary with same keys.
d1 = {'product': '8', 'order': '8', 'tracking': '3'}
d2 = {'order': 1, 'product': 1,'customer':'5'}
dict3 = { k: [ d1[k], d2[k] ] if k in d2 else [d1[k]] for k in d1 }
dict3
{'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3']}
How to pass else [d2[k]] for k in d2 to get the expected out
My Expected out
{'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3'],'customer':['5']}
Disclaimer. I have done with defaultdict. Please give answer in dict comprehension only
You could use a nested ternary ... if ... else (... if ... else ...), but what if there are three dictionaries, or four?
Better use a nested list comprehension and iterate over the different dictionaries.
>>> d1 = {'product': '8', 'order': '8', 'tracking': '3'}
>>> d2 = {'order': 1, 'product': 1,'customer':'5'}
>>> {k: [d[k] for d in (d1, d2) if k in d] for k in set(d1) | set(d2)}
{'customer': ['5'], 'order': ['8', 1], 'product': ['8', 1], 'tracking': ['3']}
You have to iterate over both the dictionaries to include all the keys in new constructed dict.
You can achieve this by using defaultdict
from collections import defaultdict
res = defaultdict(list)
for key, value in d1.items():
res[key].append(value)
for key, value in d2.items():
res[key].append(value)
Output:
>>> dict(res)
>>> {'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3'], 'customer': ['5']}
Using a defaultdict without a comprehension is a much, much better way to go, but as requested:
d1 = {'product': '8', 'order': '8', 'tracking': '3'}
d2 = {'order': 1, 'product': 1,'customer':'5'}
d3 = {
k: [d1[k], d2[k]]
if (k in d1 and k in d2)
else [d1[k]]
if k in d1
else [d2[k]]
for k in list(d1.keys()) + list(d2.keys())
}
d3 is now:
{'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3'], 'customer': ['5']}
>>> d1 = {'product': '8', 'order': '8', 'tracking': '3'}
>>> d2 = {'order': 1, 'product': 1, 'customer': '5'}
>>> dict3 = {k: [d1[k], d2[k]] if k in d1 and k in d2 else [d1[k]] if k in d1 else [d2[k]] for list in [d1, d2] for k in list}
>>> dict3
{'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3'],'customer':['5']}
d1 = {'product': '8', 'order': '8', 'tracking': '3'}
d2 = {'order': 1, 'product': 1, 'customer': '5'}
list_ = []
for i in [d1, d2]:
list_.append(i)
list_
[{'product': '8', 'order': '8', 'tracking': '3'},
{'order': 1, 'product': 1, 'customer': '5'}]
dict_={}
for d in list_:
for k,v in d.items():
dict_.setdefault(k,[]).append(v)
dict_
{'product': ['8', 1], 'order': ['8', 1], 'tracking': ['3'], 'customer': ['5']}
Comprehension
combined_key = {key for d in list_ for key in d}
combined_key
{'customer', 'order', 'product', 'tracking'}
super_dict = {key:[d[key] for d in list_ if key in d] for key in combined_key}
super_dict
{'customer': ['5'], 'tracking': ['3'], 'order': ['8', 1], 'product': ['8', 1]}
This question already has answers here:
List of dicts to/from dict of lists
(14 answers)
Closed 4 years ago.
so given a list of dictionaries:
my_dict= [{'a': '4', 'b': '5', 'c': '1', 'd': '3'},
{'a': '1', 'b': '8', 'c': '1', 'd': '2'},
{'a': '7', 'b': '4', 'c': '1', 'd': '5'}]
and a list of keys in the dictionary for example [ 'a', 'b']
I am trying to make a list of dictionaries for both 'a' and 'b' for their respective values i.e the final product will resemble
new_dict = ['a':['4', '1', '7'], 'b':['5', '8', '4']]
any help will be appreciated
Using collections
Demo:
import collections
d = collections.defaultdict(list)
my_dict= [{'a': '4', 'b': '5', 'c': '1', 'd': '3'}, {'a': '1', 'b': '8', 'c': '1', 'd': '2'}, {'a': '7', 'b': '4', 'c': '1', 'd': '5'}]
for i in my_dict:
for k,v in i.items():
d[k].append(v)
print( d )
Output:
defaultdict(<type 'list'>, {'a': ['4', '1', '7'], 'c': ['1', '1', '1'], 'b': ['5', '8', '4'], 'd': ['3', '2', '5']})