Python - Scoring Matrix using Dictionary
Last Updated :
14 Mar, 2023
Sometimes, while working with Python records, we can have a problem in which we need to resolve scoring in Python Matrix records. This means mapping of each key of dictionary with its value to aggregate score of each row. This kind of problem can have applications in gaming and web development domains. Let's discuss certain ways in which this task can be performed.
Input : test_list = [['gfg', 'best'], ['geeks'], ['is', 'for']] Output : [18, 15, 12] Input : test_list = [['gfg', 'geeks', 'CS']] Output : [20]
Method #1 : Using loop This is one of the way in which this task can be performed. In this, we iterate for the matrix elements and perform the values substitutions using dictionary and perform row summations.
Python3
# Python3 code to demonstrate working of
# Scoring Matrix using Dictionary
# Using loop
# initializing list
test_list = [['gfg', 'is', 'best'], ['gfg', 'is', 'for', 'geeks']]
# printing original list
print("The original list is : " + str(test_list))
# initializing test dict
test_dict = {'gfg' : 5, 'is' : 10, 'best' : 13, 'for' : 2, 'geeks' : 15}
# Scoring Matrix using Dictionary
# Using loop
res = []
for sub in test_list:
sum = 0
for val in sub:
if val in test_dict:
sum += test_dict[val]
res.append(sum)
# printing result
print("The Row scores : " + str(res))
Output : The original list is : [['gfg', 'is', 'best'], ['gfg', 'is', 'for', 'geeks']]
The Row scores : [28, 32]
Time complexity: O(M^N) as the number of combinations generated is M choose N.
Auxiliary space: O(L) as the size of the resultant list is L.
Method #2 : Using list comprehension + sum() This is yet another way to solve this problem. In this, we perform the summation using sum() and list comprehension is used for iterations and score assignments.
Python3
# Python3 code to demonstrate working of
# Scoring Matrix using Dictionary
# Using list comprehension + sum()
# initializing list
test_list = [['gfg', 'is', 'best'], ['gfg', 'is', 'for', 'geeks']]
# printing original list
print("The original list is : " + str(test_list))
# initializing test dict
test_dict = {'gfg' : 5, 'is' : 10, 'best' : 13, 'for' : 2, 'geeks' : 15}
# Scoring Matrix using Dictionary
# Using list comprehension + sum()
res = [sum(test_dict[word] if word.lower() in test_dict else 0 for word in sub) for sub in test_list]
# printing result
print("The Row scores : " + str(res))
Output : The original list is : [['gfg', 'is', 'best'], ['gfg', 'is', 'for', 'geeks']]
The Row scores : [28, 32]
Time complexity: O(M^N) as the number of combinations generated is M choose N.
Auxiliary space: O(L) as the size of the resultant list is L.
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