Python program to Convert Matrix to Dictionary Value List

Given Matrix, the task is to write a Python program to map each column’s values as customized keys from another list.

Input : test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]], map_list = [4, 5, 6]
Output : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}
Explanation : 4 is mapped with all the 0th index of lists, 4, 1 ,3, 10.

Input : test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1]], map_list = [4, 5, 6]
Output : {4: [4, 1, 3], 5: [5, 3, 8], 6: [6, 5, 1]}
Explanation : 4 is mapped with all the 0th index of lists, 4, 1 ,3.

Method 1 : Using dictionary comprehension + zip()

In this, mapping of columns with custom list index elements is done using zip(), dictionary comprehension is done to assign extracted keys to mapping values.

Python3

# Python3 code to demonstrate working of
# Convert Matrix to Dictionary Value List
# Using dictionary comprehension + zip()
from collections import defaultdict

# initializing list
test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]

# printing original list
print("The original list is : " + str(test_list))

# initializing map list
map_list = [4, 5, 6]

# mapping column using zip(), dictionary comprehension for key
# converts to list of dictionary
temp = [{key : val for key,
         val in zip(map_list, idx)} for idx in test_list]

# convert to dictionary value list
res = defaultdict(list)
{res[key].append(sub[key]) for sub in temp for key in sub}

# printing result
print("Converted Dictionary : " + str(dict(res)))

Output

The original list is : [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]
Converted Dictionary : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}

Time Complexity: O(n*m)
Auxiliary Space: O(k)

Method 2 : Using dict() + list comprehension + zip()

In this, the task of mapping values to dictionary keys and conversion is done using dict() and zip() and dictionary comprehension. Rest functionalities are similar to the above method.

Python3

# Python3 code to demonstrate working of
# Convert Matrix to Dictionary Value List
# Using dict() + list comprehension + zip()
from collections import defaultdict

# initializing list
test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]

# printing original list
print("The original list is : " + str(test_list))

# initializing map list
map_list = [4, 5, 6]

# mapping column using zip() and conversion using dict()
# converts to list of dictionary
temp = [dict(zip(map_list, sub)) for sub in test_list]

# convert to dictionary value list
res = defaultdict(list)
{res[key].append(sub[key]) for sub in temp for key in sub}

# printing result
print("Converted Dictionary : " + str(dict(res)))

Output

The original list is : [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]
Converted Dictionary : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}

Method 3: Using loop

First initializes an empty dictionary res. Then, it loops over each row of the matrix test_list. For each row, it loops over each element of the row using a range function. For each element, it checks whether the corresponding key already exists in the res dictionary. If not, it creates a new key with an empty list as its value. Finally, it appends the element to the corresponding list. After the loops are completed, the resulting dictionary is printed.

Python3

# Python3 code to demonstrate working of
# Convert Matrix to Dictionary Value List
# Using loops

# initializing list
test_list = [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]

# printing original list
print("The original list is : " + str(test_list))

# initializing map list
map_list = [4, 5, 6]

# convert to dictionary value list
res = {}

# iterate over each row of the matrix
for row in test_list:
    # iterate over each element of the row
    for i in range(len(row)):
        # if the key doesn't exist in the result dictionary, create an empty list
        if map_list[i] not in res:
            res[map_list[i]] = []
        # append the value to the corresponding list
        res[map_list[i]].append(row[i])

# printing result
print("Converted Dictionary : " + str(res))

Output

The original list is : [[4, 5, 6], [1, 3, 5], [3, 8, 1], [10, 3, 5]]
Converted Dictionary : {4: [4, 1, 3, 10], 5: [5, 3, 8, 3], 6: [6, 5, 1, 5]}

Time complexity: O(m*n), because it performs the same number of iterations as the original code.
Auxiliary space: O(m*n) as well, because it creates a dictionary with m * n keys and a list of m * n elements