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Python – Flatten Nested Dictionary to Matrix

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  • Last Updated : 19 Jan, 2022
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Sometimes, while working with data, we can have a problem in which we need to convert nested dictionary into Matrix, each nesting comprising of different row in matrix. This can have application in many data domains. Lets discuss certain ways in which this task can be performed.
Method #1 : Using loop + zip() + map() 
The combination of above functions can be used to perform this task. In this, we use brute force to flatten the dictionary keys and then use map() and zip() to align them as rows of matrix.
 

Python3




# Python3 code to demonstrate working of
# Flatten Nested Dictionary to Matrix
# using zip() + loop + map()
   
# initializing dictionary
test_dict = {'Gfg1' : {'CS':1, 'GATE' : 2},
             'Gfg2' : {'CS':2, 'GATE' : 3},
             'Gfg3' : {'CS':4, 'GATE' : 5}}
   
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# Flatten Nested Dictionary to Matrix
# using zip() + loop + map()
temp = list(test_dict.values())
sub = set()
for ele in temp:
    for idx in ele:
        sub.add(idx)
res = []
res.append(sub)
for key, val in test_dict.items():
    temp2 = []
    for idx in sub:
        temp2.append(val.get(idx, 0))
    res.append(temp2)
 
res = [[idx for idx, val in test_dict.items()]] + list(map(list, zip(*res)))
   
# printing result 
print("The Grouped dictionary list is : " + str(res))

Output : 

The original dictionary is : {‘Gfg3’: {‘GATE’: 5, ‘CS’: 4}, ‘Gfg1’: {‘GATE’: 2, ‘CS’: 1}, ‘Gfg2’: {‘GATE’: 3, ‘CS’: 2}} 
The Grouped dictionary list is : [[‘Gfg3’, ‘Gfg1’, ‘Gfg2’], [‘GATE’, 5, 2, 3], [‘CS’, 4, 1, 2]] 

 

 
Method #2 : Using union() + list comprehension 
The combination of above methods can be used to perform this task. In this, we compute the union using union() rather than the nested loops. The result is compiled using list comprehension.
 

Python3




# Python3 code to demonstrate working of
# Flatten Nested Dictionary to Matrix
# using union() + list comprehension
   
# initializing dictionary
test_dict = {'Gfg1' : {'CS':1, 'GATE' : 2},
             'Gfg2' : {'CS':2, 'GATE' : 3},
             'Gfg3' : {'CS':4, 'GATE' : 5}}
   
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# Flatten Nested Dictionary to Matrix
# using union() + list comprehension
temp = set().union(*test_dict.values())
res = [list(test_dict.keys())]
res += [[key] + [sub.get(key, 0) for sub in test_dict.values()] for key in temp]
   
# printing result 
print("The Grouped dictionary list is : " + str(res))

Output : 

The original dictionary is : {‘Gfg3’: {‘GATE’: 5, ‘CS’: 4}, ‘Gfg1’: {‘GATE’: 2, ‘CS’: 1}, ‘Gfg2’: {‘GATE’: 3, ‘CS’: 2}} 
The Grouped dictionary list is : [[‘Gfg3’, ‘Gfg1’, ‘Gfg2’], [‘GATE’, 5, 2, 3], [‘CS’, 4, 1, 2]] 

 


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