Given Tuple Matrix of length 2, map each column’s element value with next column and construct dictionary keys.
Input : test_list = [[(1, 4), (6, 3), (4, 7)], [(7, 3), (10, 14), (11, 22)]]
Output : {1: 7, 4: 3, 6: 10, 3: 14, 4: 11, 7: 22}
Explanation : 1 -> 7, 4 -> 3.., as in same column and indices.Input : test_list = [[(1, 4), (6, 3)], [(7, 3), (10, 14)]]
Output : {1: 7, 4: 3, 6: 10, 3: 14}
Explanation : Self explanatory column wise pairing.
Method #1 : Using loop
This is one of the ways in which this task can be performed. In this, we iterate for all the elements with their next column elements and construct dictionary key-value pair.
Python3
# Python3 code to demonstrate working of # Column Mapped Tuples to dictionary items# Using loop # initializing listtest_list = [[(5, 6), (7, 4), (1, 2)], [(7, 3), (10, 14), (11, 22)] ] # printing original listprint("The original list : " + str(test_list)) res = dict() # loop for tuple listsfor idx in range(len(test_list) - 1): for idx2 in range(len(test_list[idx])): # column wise dictionary pairing res[test_list[idx][idx2][0]] = test_list[idx + 1][idx2][0] res[test_list[idx][idx2][1]] = test_list[idx + 1][idx2][1] # printing result print("The constructed dictionary : " + str(res)) |
The original list : [[(5, 6), (7, 4), (1, 2)], [(7, 3), (10, 14), (11, 22)]]
The constructed dictionary : {5: 7, 6: 3, 7: 10, 4: 14, 1: 11, 2: 22}
Method #2 : Using dictionary comprehension + zip()
The combination of above functions provides one-liner to solve this problem. In this, we perform the task of zipping all the columns using zip() and dictionary comprehension is used to key-value pairs.
Python3
# Python3 code to demonstrate working of # Column Mapped Tuples to dictionary items# Using dictionary comprehension + zip() # initializing listtest_list = [[(5, 6), (7, 4), (1, 2)], [(7, 3), (10, 14), (11, 22)] ] # printing original listprint("The original list : " + str(test_list)) # nested dictionary comprehension to form pairing # paired using zip()res = {key[idx] : val[idx] for key, val in zip( *tuple(test_list)) for idx in range(len(key))} # printing result print("The constructed dictionary : " + str(res)) |
The original list : [[(5, 6), (7, 4), (1, 2)], [(7, 3), (10, 14), (11, 22)]]
The constructed dictionary : {5: 7, 6: 3, 7: 10, 4: 14, 1: 11, 2: 22}
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