How to remove list element from list of list in python

Question

I tried to remove the third and the fourth list from the list of list in python.

My list of list is below:

List =  [
            ['101', 'Dashboard', '1', '1'],
            ['102', 'Potential Cstomer', '1', '1'],
            ['102-01', 'Potential Cstomer', '1', '1'],
            ['102-02', 'Potential Cstomer Activity', '1', '1']
        ]

After remove the third and fourth element of list, I would like to be like this:

NewList =  [
            ['101', 'Dashboard'],
            ['102', 'Potential Cstomer'],
            ['102-01', 'Potential Cstomer'],
            ['102-02', 'Potential Customer Activity']
        ]

I tried my code like below but it did not make any change.

    NewList     = [list(element) for element in List if element[0] or element[1]]

    print NewList

How should I change my current code to achieve my expected result? Thanks.

Moses Koledoye · Accepted Answer · 2017-10-11 23:21:08Z

6

Slice each nested list in a list comprehension. The slice notation starts at index 0 and stops at 1 i.e. [0, 2):

NewList = [element[:2] for element in List]

When the start index is not specified, it is taken as None which is the same as start index of the list when None appears before the first :.

Same as:

NewList = [element[slice(None, 2)] for element in List] # More verbose

In Python 3, you could use extended unpacking to achieve the same thing applying the 'splat' operator *:

NewList = [elements for *elements, _, _ in List]

edited Oct 11, 2017 at 23:21

answered Oct 11, 2017 at 23:15

Moses Koledoye

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1 Comment

Houy Narun Over a year ago

Thank you very for your helpful and detail answer. It works perfect. I've learned something from that. Cheer :)

jwqz · Accepted Answer · 2017-10-11 23:17:41Z

1

How about this:

 for s in List:
    del s[3]
    del s[2]

That deletes in place.

answered Oct 11, 2017 at 23:17

jwqz

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1 Comment

ShadowRanger Over a year ago

del s[2:] would clear all but the first two entries more efficiently. You could use del s[2:4] if you only want to remove indices 2 and 3. Note that this solution doesn't copy in any way, so the data is dropped (Moses's solution shallow copies, so the top two levels of the new list are independent).

Ajax1234 · Accepted Answer · 2017-10-11 23:23:43Z

1

This solution uses negative indexing to allow for arbitrary length sublists, along as the original condition of two trailing digits is maintained.

List =  [
        ['101', 'Dashboard', '1', '1'],
        ['102', 'Potential Cstomer', '1', '1'],
        ['102-01', 'Potential Cstomer', '1', '1'],
        ['102-02', 'Potential Cstomer Activity', '1', '1']
    ]
new_final_list = [i[:-2] for i in List]
for i in new_final_list:
   print(i)

Output:

['101', 'Dashboard'], 
['102', 'Potential Cstomer']
['102-01', 'Potential Cstomer']
['102-02', 'Potential Cstomer Activity']

edited Oct 11, 2017 at 23:23

answered Oct 11, 2017 at 23:19

Ajax1234

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2 Comments

ShadowRanger Over a year ago

Minor note on behavior: This removes the last two elements, rather than preserving the first two. When there are a uniform four elements in each sublist, the two operations are equivalent, but if the elements are uneven length, or you performed the operation twice, the behavior is very different. Preserving the first two is safer and idempotent.

Ajax1234 Over a year ago

@ShadowRanger thank you for pointing that out. While other solutions are more than satisfactory performance-wise, I thought I would include this solution to allow for arbitrary length sublists.

K. Winfred · Accepted Answer · 2018-11-14 19:46:36Z

-1

Please reference code below:

Names = [["Tom", 32, 12], ["John", 54, 16], ["James", 52, 15]]
Names_new = []

for i in Names:
    # print(i + )
    del i[0]
    Names_new.append(i)

print(Names_new)

answered Nov 14, 2018 at 19:46

K. Winfred

1

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