How do you eliminate items from nested lists

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Given a list of ranked ballots and a list of candidates to eliminate, I need to return a new list of ranked ballots where all the candidates in the list of candidates to eliminate have been removed.

This is the code I've tried so far

import doctest 

def eliminate_candidate(ballots, to_eliminate):
    (list, list) -> list

    >>> eliminate_candidate([['NDP', 'LIBERAL'], ['GREEN', 'NDP'], \
    ['NDP', 'BLOC']], ['NDP', 'LIBERAL'])
    [[], ['GREEN'], ['BLOC']]
    >>> eliminate_candidate([], [])
    >>> eliminate_candidate([[]], [])
    new_list = []

    # Copy ballots onto new list
    for item in ballots:

    for item in new_list:
        for element in item:
            if element in to_eliminate:

    return new_list

My first doctest is failing, giving me this output instead:

[['LIBERAL'], ['GREEN'], ['BLOC']]

This is the function required. It searches through the sublists:

def eliminate_candidate(ballots, to_eliminate):
    return [[party for party in ballot if party not in to_eliminate] for ballot in ballots]

ballots = [['NDP', 'LIBERAL'], ['GREEN', 'NDP'], ['NDP', 'BLOC']]
to_eliminate = ['NDP', 'LIBERAL']

print(eliminate_candidate(ballots, to_eliminate))


[[], ['GREEN'], ['BLOC']]

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Using sets becomes pretty easy!

ballots = [['NDP', 'LIBERAL'], ['GREEN', 'NDP'], ['NDP', 'BLOC']]
to_eliminate = ['NDP', 'LIBERAL']

result = [list(set(x) - set(to_eliminate)) for x in ballots]
[[], ['GREEN'], ['BLOC']]


result = [list(set(x).difference(to_eliminate)) for x in ballots]
[[], ['GREEN'], ['BLOC']]

Python, Problem can be removing all occurrences of nested list. Let's discuss certain ways in which this problem can be solved. Method #1 : Using list comprehension. The� 2.1 Select the list you will remove values from in the Find values in box; 2.2 Select the list you will delete values based on in the According to box; 2.3 select the Single cell option in the Based on section; 2.4 Click the OK button. See screenshot: 3. Then a dialog box pops up to tell you how many cells have been selected, please click the

ballots = [['NDP', 'LIBERAL'], ['GREEN', 'NDP'], ['NDP', 'BLOC']]
to_eliminate = ['NDP', 'LIBERAL']

res = [[element for element in b if element not in to_eliminate] for b in ballots]


[[], ['GREEN'], ['BLOC']]

Remove elements from deep nested list, list[[All, All, 2;;]] (* or *) Drop[list, None, None, 1]. {{{1812.}, {4262.}, {1272.}, {4048. }, {3961.}, {7739.}, {3173.}, {1201.}, {6856.}, {11860.}, {1324.}� There are several ways to remove items from a nested dictionary. Remove an Item by Key If you know the key of the item you want, you can use pop () method. It removes the key and returns its value.

How to remove a certain column from a nested list in Python, Looking at your example that should work but remove unexpected elements. Unless you really did not try that exact example :) No python on my phone so cant � In this case you can do it in two ways: - If "Cultivar" wont repeat in your gallery you can use it in the remove method for identify which register you want to delete. Something like this Remove(PP;{Cultivar: Gallery1.Selected.Cultivar})

11. Lists — How to Think Like a Computer Scientist: Learning with , Although a list can contain another list, the nested list still counts as a single We can also remove elements from a list by assigning an empty list to them:. Create a dictionary, using the List items as keys. This will automatically remove any duplicates because dictionaries cannot have duplicate keys.

Remove nested lists from a list if nested list contains a certain value , This is the simple way to comprehensive list as shown below: regions = [i for i in regions if i and (-1 not in i)]. The output is: [[9, 10, 7, 8], [7, 6, 10]� Remove items from a Nested List If you know the index of the item you want, you can use pop () method. It modifies the list and returns the removed item. L = ['a', ['bb', 'cc', 'dd'], 'e'] x = L.pop (1) print(L) # Prints ['a', ['bb', 'dd'], 'e'] # removed item print(x) # Prints cc

  • Don't use list as the name of a variable!
  • Your output seems to be different to the requirement stated by the OP.
  • But that his output is not as the stated requirement.