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# Python | Convert column to separate elements in list of lists

• Last Updated : 11 Jun, 2021

There are instances in which we might require to extract a particular column of a Matrix and assign its each value as separate entity in list and this generally has a utility in Machine Learning domain. Let’s discuss certain ways in which this action can be performed.
Method #1 : Using list slicing and list comprehension
The functionality of list slicing and comprehension can be clubbed to perform the particular task of extracting a column from a list and then it can be added as new element using list comprehension.

## Python3

 `# Python3 code to demonstrate ``# column to separate elements in list of lists``# using list slicing and list comprehension` `# initializing list of list``test_list ``=` `[[``1``, ``3``, ``4``],``             ``[``6``, ``2``, ``8``],``             ``[``9``, ``10``, ``5``]]` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))` `# using list slicing and list comprehension``# column to separate elements in list of lists``res ``=` `[i ``for` `nest_list ``in` `[[j[``1` `: ], [j[``0``]]]``         ``for` `j ``in` `test_list] ``for` `i ``in` `nest_list]` `# printing result``print` `(``"The list after column shift is : "` `+` `str``(res))`

Output

```The original list is : [[1, 3, 4], [6, 2, 8], [9, 10, 5]]
The list after column shift is : [[3, 4], , [2, 8], , [10, 5], ]
```

Method #2 : Using itertools.chain() + list comprehension + list slicing
The above method can be improved by inducing the concept of element chaining and reduce the overhead of the list comprehension and reducing the time taken to execute this particular task.

## Python3

 `# Python3 code to demonstrate ``# column to separate elements in list of lists``# using itertools.chain()+ list comprehension + list slicing``from` `itertools ``import` `chain` `# initializing list of list``test_list ``=` `[[``1``, ``3``, ``4``],``             ``[``6``, ``2``, ``8``],``             ``[``9``, ``10``, ``5``]]` `# printing original list``print` `(``"The original list is : "` `+` `str``(test_list))` `# using itertools.chain() + list comprehension + list slicing``# column to separate elements in list of lists``res ``=` `list``(chain(``*``[``list``((sub[``1``: ], [sub[``0``]]))``                      ``for` `sub ``in` `test_list]))` `# printing result``print` `(``"The list after column shift is : "` `+` `str``(res))`

Output

```The original list is : [[1, 3, 4], [6, 2, 8], [9, 10, 5]]
The list after column shift is : [[3, 4], , [2, 8], , [10, 5], ]
```

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