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 8579709
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: June 11, 20262026-06-11T20:40:03+00:00 2026-06-11T20:40:03+00:00

I have two nested lists with strings ( list_a and list_b ), details below:

  • 0

I have two nested lists with strings (list_a and list_b), details below:

list_a = [
('shop1', 'stand1', 'shelf1', 'fruit1'),
('shop1', 'stand1', 'shelf2', 'fruit2'),
('shop1', 'stand1', 'shelf3', 'fruit3'),
('shop1', 'stand2', 'shelf1', 'fruit1'),
('shop1', 'stand2', 'shelf2', 'fruit2'),
('shop1', 'stand2', 'shelf3', 'fruit3'),
('shop2', 'stand3', 'shelf1', 'fruit1'),
('shop2', 'stand3', 'shelf2', 'fruit2'),
('shop2', 'stand3', 'shelf3', 'fruit3')
]
list_b = [
('shop1', 'stand1', 'shelf1', 'fruit1'),
('shop1', 'stand1', 'shelf2', 'fruit2'),
('shop1', 'stand1', 'shelf2', 'fruit2'),
('shop1', 'stand1', 'shelf3', 'fruit3'),
('shop1', 'stand1', 'shelf3', 'fruit3'),
('shop1', 'stand1', 'shelf3', 'fruit3'),
('shop1', 'stand2', 'shelf1', 'fruit1'),
('shop1', 'stand2', 'shelf1', 'fruit1'),
('shop1', 'stand2', 'shelf1', 'fruit1'),
('shop1', 'stand2', 'shelf2', 'fruit2'),
('shop1', 'stand2', 'shelf2', 'fruit2'),
('shop1', 'stand2', 'shelf2', 'fruit2'),
('shop1', 'stand2', 'shelf3', 'fruit3'),
('shop2', 'stand3', 'shelf1', 'fruit1'),
('shop2', 'stand3', 'shelf1', 'fruit1'),
('shop2', 'stand3', 'shelf2', 'fruit2'),
('shop2', 'stand3', 'shelf3', 'fruit3'),
('shop2', 'stand3', 'shelf3', 'fruit3'),
('shop2', 'stand3', 'shelf3', 'fruit3')
]

and I would like to find identical rows from list_b in list_a, count “duplicated” rows and merge list_a with one additional column (number of occurrences) as a new list, like this below:

result_list = [
('shop1', 'stand1', 'shelf1', 'fruit1', 1),
('shop1', 'stand1', 'shelf2', 'fruit2', 2),
('shop1', 'stand1', 'shelf3', 'fruit3', 3),
('shop1', 'stand2', 'shelf1', 'fruit1', 3),
('shop1', 'stand2', 'shelf2', 'fruit2', 3),
('shop1', 'stand2', 'shelf3', 'fruit3', 1),
('shop2', 'stand3', 'shelf1', 'fruit1', 2),
('shop2', 'stand3', 'shelf2', 'fruit2', 1),
('shop2', 'stand3', 'shelf3', 'fruit3', 3)
]

Is there any quick and efficient way to do this?

  • 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
    2026-06-11T20:40:04+00:00Added an answer on June 11, 2026 at 8:40 pm
    dict_a = {row: 0 for row in list_a}
    for row in list_b:
        if row in dict_a:
            dict_a[row] += 1
    
    result = [row + (dict_a[row],) for row in list_a]
    

    On Python 2.6 use dict((row, 0) for row in list_a) instead of the dictionary comprehension.

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

Sidebar

Related Questions

I have two nested lists of different sizes: A = [[1, 7, 3, 5],
I have two sortable lists, one being nested, with a mouse over effect on
Let's say I have two lists of strings: a = ['####/boo', '####/baa', '####/bee', '####/bii',
I have two tables: Campaigns, Campaign_statistics. I need to output campaigns list with nested
I have two nested packages that I am calling via fast_cgi. From the first
I have two nested loop in XSL like this, at this moment I use
I have two masterpages. A main.Master and a search.Master. The search.Master is a nested
I have a navigation bar that has two dropdowns (as nested ul's). I'm trying
I have two classes, one nested in the other. Public Class Operation Public Property
So I have some nested lists (only one level deep) and I'm running into

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.