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# Python – Columns to Dictionary Conversion in Matrix

Given a Matrix, Convert to Dictionary with elements in 1st row being keys, and subsequent rows acting as values list.

Input : test_list = [[4, 5, 7], [10, 8, 4], [19, 4, 6], [9, 3, 6]]
Output : {4: [10, 19, 9], 5: [8, 4, 3], 7: [4, 6, 6]}
Explanation : All columns mapped with 1st row elements. Eg. 4 -> 10, 19, 9.

Input : test_list = [[4, 5, 7], [10, 8, 4], [19, 4, 6], [9, 3, 7]]
Output : {4: [10, 19, 9], 5: [8, 4, 3], 7: [4, 6, 7]}
Explanation : All columns mapped with 1st row elements. Eg. 7 -> 4, 6, 7.

Method #1 : Using list comprehension + dictionary comprehension

This is one of the ways in which this task can be performed. In this, list comprehension is responsible for construction of values and mapping and dictionary conversion is done using dictionary comprehension.

## Python3

 `# Python3 code to demonstrate working of ``# Columns to Dictionary Conversion in Matrix``# Using dictionary comprehension + list comprehension`` ` `# initializing list``test_list ``=` `[[``4``, ``5``, ``7``], [``10``, ``8``, ``3``], [``19``, ``4``, ``6``], [``9``, ``3``, ``6``]]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# dictionary comprehension performing re making of result ``# dictionary``res ``=` `{test_list[``0``][idx]: [test_list[ele][idx]``      ``for` `ele ``in` `range``(``1``, ``len``(test_list))]``      ``for` `idx ``in` `range``(``len``(test_list[``0``]))}`` ` `# printing result ``print``(``"Reformed dictionary : "` `+` `str``(res))`

Output

```The original list : [[4, 5, 7], [10, 8, 3], [19, 4, 6], [9, 3, 6]]
Reformed dictionary : {4: [10, 19, 9], 5: [8, 4, 3], 7: [3, 6, 6]}
```

Method #2 : Using dictionary comprehension + zip()

This is yet another way to solve this problem. In this, we map all column elements with each other using zip() and dictionary comprehension is used to perform remaking of dictionary.

## Python3

 `# Python3 code to demonstrate working of ``# Columns to Dictionary Conversion in Matrix``# Using dictionary comprehension + zip()`` ` `# initializing list``test_list ``=` `[[``4``, ``5``, ``7``], [``10``, ``8``, ``3``], [``19``, ``4``, ``6``], [``9``, ``3``, ``6``]]`` ` `# printing original list``print``(``"The original list : "` `+` `str``(test_list))`` ` `# appropriate slicing before zip function ``res ``=` `{ele[``0``]: ``list``(ele[``1``:]) ``for` `ele ``in` `zip``(``*``test_list)}`` ` `# printing result ``print``(``"Reformed dictionary : "` `+` `str``(res))`

Output

```The original list : [[4, 5, 7], [10, 8, 3], [19, 4, 6], [9, 3, 6]]
Reformed dictionary : {4: [10, 19, 9], 5: [8, 4, 3], 7: [3, 6, 6]}
```