# Python | Product of kth column in List of Lists

• Last Updated : 14 May, 2020

Sometimes, while working with Python Matrix, we may have a problem in which we require to find the product of a particular column. This can have a possible application in day-day programming and competitive programming. Letâ€™s discuss certain ways in which this task can be performed.

Method #1 : Using loop + `zip()`
This task can be solved using the combination of above functions. In this, we pass in the zip() the list, to access all columns and explicit product function to get product of columns.

 `# Python3 code to demonstrate working of``# Column Product in List of Lists``# using loop + zip()`` ` `# getting Product``def` `prod(val) :``    ``res ``=` `1` `    ``for` `ele ``in` `val:``        ``res ``*``=` `ele``    ``return` `res ``     ` `# initialize list``test_list ``=` `[[``5``, ``6``, ``7``],``            ``[``9``, ``10``, ``2``], ``            ``[``10``, ``3``, ``4``]]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initialize K``K ``=` `2`` ` `# Column Product in List of Lists``# using loop + zip()``res ``=` `[prod(idx) ``for` `idx ``in` `zip``(``*``test_list)][K] `` ` `# printing result``print``(``"Product of Kth column is : "` `+` `str``(res))`
Output :
```The original list is : [[5, 6, 7], [9, 10, 2], [10, 3, 4]]
Product of Kth column is : 56
```

Method #2 : Using loop
This is brute force way to solve this problem. In this, we iterate through the matrix and take product of column by testing column number.

 `# Python3 code to demonstrate working of``# Column Product in List of Lists``# Using loop`` ` `# initialize list``test_list ``=` `[[``5``, ``6``, ``7``],``            ``[``9``, ``10``, ``2``], ``            ``[``10``, ``3``, ``4``]]`` ` `# printing original list``print``(``"The original list is : "` `+` `str``(test_list))`` ` `# initialize K``K ``=` `2`` ` `# Column Product in List of Lists``# Using loop``res ``=` `1``for` `sub ``in` `test_list:``    ``for` `idx ``in` `range``(``0``, ``len``(sub)):``        ``if` `idx ``=``=` `K:``            ``res ``=` `res ``*` `sub[idx]`` ` `# printing result``print``(``"Product of Kth column is : "` `+` `str``(res))`
Output :
```The original list is : [[5, 6, 7], [9, 10, 2], [10, 3, 4]]
Product of Kth column is : 56
```

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