python program to remove duplicate elements from nested lists

Question

I am new to stack overflow so please excuse me if I am not clear.

aim: to write a program that would remove the elements that are present in preceding nested lists

here is the list: gang=[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 12, 15], [6, 12], [5, 10, 15]]

desired output: gang=[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 15], [5, 10]]

attempt:

for i in range(1,len(gang)):
    elem=gang[i]
    for j in range(len(elem)):
        for hey in range(i):
            if elem[j] in gang[hey]:
                elem.pop(j)

explanation of the attempt:

1)to traverse through list gang, I used a for loop (range is 1,len(gang) bcs 0th element wont be changed as it is the first one in the list)

2)I have declared a variable, 'elem' which equates to gang[i] i.e. the elements of gang from gang[1] to gang[-1]

3)Now, to traverse through the elements of the elements i.e. the elements of each nested list, I have used another for loop

4)the last for loop (for hey in range(i)) is used so that I can confirm if elem[j] i.e. element of nested list, exists in preceding nested lists, and if the condition is satisfied elem[j] would be removed

expected output:

gang=[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 15], [5, 10]]

output:

if elem[j] in gang[hey]:
IndexError: list index out of range

question: why is this error showing? potential fix for it? any better way to achieve aim?

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check this https://stackoverflow.com/questions/15037226/python-remove-duplicate-items-from-nested-list — Chathuranga, Mar 13 '21 at 05:47
Shouldn't your output be this: [[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 9, 15], [], [5]]? — pakpe, Mar 13 '21 at 06:02

pakpe · Accepted Answer · 2021-03-13T06:00:36.607

Here it is. You create a visited list, iterate backward through each sublist, and remove an element if it is already visited:

gang=[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 12, 15], [6, 12], [5, 10, 15]]
visited = []
for lst in gang:
    for i in range(len(lst)-1, -1, -1):
        if lst[i] not in visited:
            visited.append(lst[i])
        else:
            lst.pop(i)
print(gang)
#[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 9, 15], [], [5]]

score 0 · Answer 2 · answered Mar 13 '21 at 06:01

0

The library itertools is very helpful in this case

import itertools
gang=[[1], [7], [11], [13], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 12, 15], [6, 12], [5, 10, 15]]
gang.sort()#Sorting the list
gang = list(k for k,_ in itertools.groupby(gang))#This removes repetition
print (gang)

Check it on a repl here https://repl.it/join/keztvujw-siddharthagraw2

answered Mar 13 '21 at 06:01

Siddharth Agrawal

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This code returns [[1], [2, 4, 6, 8, 10, 12, 14], [3, 6, 9, 12, 15], [5, 10, 15], [6, 12], [7], [11], [13]]. This is the same as the input list except that the sublists are shuffled. – pakpe Mar 13 '21 at 06:06
Oh my bad. Itertools removes same objects but not sub items. Let me write a code myself to do this task – Siddharth Agrawal Mar 13 '21 at 06:08

python program to remove duplicate elements from nested lists

2 Answers2