Python | Convert column to separate elements in list of lists

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], [1], [2, 8], [6], [10, 5], [9]]

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], [1], [2, 8], [6], [10, 5], [9]]