Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 766449
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 14, 20262026-05-14T17:03:00+00:00 2026-05-14T17:03:00+00:00

As per the title, I have a nested lists like so (the nested list

  • 0

As per the title, I have a nested lists like so (the nested list is a fixed length):

        # ID,  Name, Value
list1 = [[ 1, "foo",    10],
         [ 2, "bar",  None],
         [ 3, "fizz",   57],
         [ 4, "buzz", None]]

I’d like to return a list (the number of items equal to the length of a sub-list from list1), where the sub-lists are the indices of rows without None as their Xth item, i.e.:

[[non-None ID indices], [non-None Name indices], [non-None Value indices]]

Using list1 as an example, the result should be:

[[0, 1, 2, 3], [0, 1, 2, 3], [0, 2]]

My current implementation is:

indices = [[] for _ in range(len(list1[0]))]
for i, row in enumerate(list1):
    for j in range(len(row)):
        if not isinstance(row[j], types.NoneType):
            indices[j].append(i)

…which works, but can be slow (the lengths of the lists are in the hundreds of thousands).

Is there a better/more efficient way to do this?

EDIT:

I’ve refactored the above for loops into nested list comprehensions (similar to SilentGhost’s answer). The following line gives the same result as the my original implementation, but runs approximately 10x faster.

[[i for i in range(len(list1)) if list1[i][j] is not None] for j in range(len(log[0]))]
  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-14T17:03:01+00:00Added an answer on May 14, 2026 at 5:03 pm
    >>> [[i for i, j in enumerate(c) if j is not None] for c in zip(*list1)]
    [[0, 1, 2, 3], [0, 1, 2, 3], [0, 2]]
    

    in python-2.x you could use itertools.izip instead of zip to avoid generating intermediate list.

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Ask A Question

Stats

  • Questions 486k
  • Answers 486k
  • Best Answers 0
  • User 1
  • Popular
  • Answers
  • Editorial Team

    How to approach applying for a job at a company ...

    • 7 Answers
  • Editorial Team

    What is a programmer’s life like?

    • 5 Answers
  • Editorial Team

    How to handle personal stress caused by utterly incompetent and ...

    • 5 Answers
  • Editorial Team
    Editorial Team added an answer I have done something similar. But I would like it… May 16, 2026 at 8:00 am
  • Editorial Team
    Editorial Team added an answer 1) ...but the problem is that it appears to be… May 16, 2026 at 8:00 am
  • Editorial Team
    Editorial Team added an answer The most likely reason that it gets removed from cache… May 16, 2026 at 8:00 am

Trending Tags

analytics british company computer developers django employee employer english facebook french google interview javascript language life php programmer programs salary

Top Members

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.