- Given a dictionary with a values list, our task is to extract the sum of values, found using mappings.
Input : test_dict = {4 : [‘a’, ‘v’, ‘b’, ‘e’],
1 : [‘g’, ‘f’, ‘g’],
3 : [‘e’, ‘v’]}, map_vals = {‘a’ : 3, ‘g’ : 8, ‘f’ : 10, ‘b’ : 4, ‘e’ : 7, ‘v’ : 2}
Output : {4: 16, 1: 26, 3: 9}
Explanation : “g” has 8, “f” has 10 as magnitude, hence 1 is mapped by 8 + 8 + 10 = 26. ( sum of list mapped ).
Input : test_dict = {4 : [‘a’, ‘v’, ‘b’, ‘e’],
1 : [‘g’, ‘f’, ‘g’]}, map_vals = {‘a’ : 3, ‘g’ : 8, ‘f’ : 10, ‘b’ : 4, ‘e’ : 7, ‘v’ : 2}
Output : {4: 16, 1: 26}
Explanation : “g” has 8, “f” has 10 as magnitude, hence 1 is mapped by 8 + 8 + 10 = 26. ( sum of list mapped ).
Method 1 : Using loop + items()
In this, each key is iterated in the dictionary and each value of each dict. is iterated and summed using mapped sum dictionary initialized.
Python3
# Python3 code to demonstrate working of# Dictionary Values Mapped Summation# Using loop + items()# initializing dictionarytest_dict = {4 : ['a', 'v', 'b', 'e'], 1 : ['g', 'f', 'g'], 3 : ['e', 'v']} # printing original dictionaryprint("The original dictionary is : " + str(test_dict))# values mapped map_vals = {'a' : 3, 'g' : 8, 'f' : 10, 'b' : 4, 'e' : 7, 'v' : 2}res = dict()# items() getting keys and valuesfor key, vals in test_dict.items(): sum = 0 for val in vals: # summing with mappings sum += map_vals[val] res[key] = sum# printing resultprint("The extracted values sum : " + str(res)) |
Output:
The original dictionary is : {4: [‘a’, ‘v’, ‘b’, ‘e’], 1: [‘g’, ‘f’, ‘g’], 3: [‘e’, ‘v’]}
The extracted values sum : {4: 16, 1: 26, 3: 9}
Time complexity: O(nm), where n is the number of keys in the dictionary and m is the average length of the value lists.
Auxiliary space: O(n), where n is the number of keys in the dictionary because we are creating a new dictionary to store the results.
Method #2 : Using dictionary comprehension + sum()
In this, we perform the task of getting sum of each value using sum(). The dictionary comprehension is used for getting keys and assigning to values sum corresponding in shorthand.
Python3
# Python3 code to demonstrate working of# Dictionary Values Mapped Summation# Using dictionary comprehension + sum()# initializing dictionarytest_dict = {4 : ['a', 'v', 'b', 'e'], 1 : ['g', 'f', 'g'], 3 : ['e', 'v']} # printing original dictionaryprint("The original dictionary is : " + str(test_dict))# values mapped map_vals = {'a' : 3, 'g' : 8, 'f' : 10, 'b' : 4, 'e' : 7, 'v' : 2}# sum() gets sum of each mapped values res = {key : sum(map_vals[val] for val in vals) for key, vals in test_dict.items()}# printing resultprint("The extracted values sum : " + str(res)) |
Output:
The original dictionary is : {4: [‘a’, ‘v’, ‘b’, ‘e’], 1: [‘g’, ‘f’, ‘g’], 3: [‘e’, ‘v’]}
The extracted values sum : {4: 16, 1: 26, 3: 9}
Method #3: Using list comprehension + lambda function
- First, we initialize a dictionary called test_dict with some key-value pairs where each value is a list of characters.
We then print out the original dictionary using the print statement and the str() function to convert the dictionary to a string.
We initialize another dictionary called map_vals with some key-value pairs where each key is a character and each value is an integer.
We create a new dictionary called res by extracting the values from the original dictionary and applying the mappings from the map_vals dictionary.
Finally, we print out the result dictionary using the print statement and the str() function to convert the dictionary to a string.
Python3
# initializing dictionarytest_dict = {4 : ['a', 'v', 'b', 'e'], 1 : ['g', 'f', 'g'], 3 : ['e', 'v']} # printing original dictionaryprint("The original dictionary is : " + str(test_dict))# values mapped map_vals = {'a' : 3, 'g' : 8, 'f' : 10, 'b' : 4, 'e' : 7, 'v' : 2}# extracting values and applying mappingsres = {key: sum(map(lambda val: map_vals[val], vals)) for key, vals in test_dict.items()}# printing resultprint("The extracted values sum : " + str(res)) |
The original dictionary is : {4: ['a', 'v', 'b', 'e'], 1: ['g', 'f', 'g'], 3: ['e', 'v']}
The extracted values sum : {4: 16, 1: 26, 3: 9}
Time complexity: O(n^2), where n is the number of elements in the original dictionary, due to the nested loops used to extract the values and apply the mappings.
Auxiliary space: O(n), where n is the number of elements in the original dictionary, due to the use of a dictionary to store the results.
Method 4 : using the reduce() function from the functools module.
Import the reduce function from the functools module.
Create a dictionary named test_dict with integer keys and list values.
Create a dictionary named map_vals with characters as keys and integer values.
Use the reduce() function with a lambda function to map and sum the values for each key in the test_dict dictionary.
The lambda function takes two arguments x and y. x is the accumulated sum and y is the current value in the list of values for each key in test_dict.
map_vals[y] maps each value in the list to its corresponding integer value in map_vals.
The reduce() function takes a third argument, which is the initial value of the accumulated sum. In this case, it is set to 0.
The result is a new dictionary named res, where each key is mapped to the sum of its corresponding values.
Finally, the program prints the resulting dictionary using the print() function.
Python3
# import required modulefrom functools import reduce# initializing dictionary and map_valstest_dict = {4 : ['a', 'v', 'b', 'e'], 1 : ['g', 'f', 'g'], 3 : ['e', 'v']}map_vals = {'a' : 3, 'g' : 8, 'f' : 10, 'b' : 4, 'e' : 7, 'v' : 2}# using reduce() to map and sum values for each keyres = {key: reduce(lambda x, y: x + map_vals[y], vals, 0) for key, vals in test_dict.items()}# printing resultprint("The extracted values sum : " + str(res)) |
The extracted values sum : {4: 16, 1: 26, 3: 9}
Time complexity of O(n*m). where n is the number of keys in the dictionary and m is the maximum number of values in the dictionary.
Auxiliary space complexity of O(n), where n is the number of keys in the dictionary and m is the maximum number of values in the dictionary.
