python remove element from nested list if it exists in another list +

Question

lets say I have a list:

A = ['x', 'y', 'z']

and another one - nested:

B = [['x', 'a', 'b', 'c'], ['y', 'c'], ['x', 'a', 'c', 'z']]

how can I remove 'z' from the last sublist in list B based on the knowledge that 'z' is in A and also every first element [0] of the sublist B is in list A ?

so basically i need to detele all elements from such a nested list where element is in A and where it not stands in the first position of that nested list ?

I am trying this but get stacked:

for i in B:
    for j in i[1:]:
        if j in A:
            del j

but I am missing something.

@IanAuld: from the question: and where it not stands in the first position of that nested list — Martijn Pieters
– Martijn Pieters, Commented May 29, 2016 at 19:55
Can you add expected output? I don't quite get where it not stands in the first position of that nested list, you seem to be ignoring all first elements — Padraic Cunningham
– Padraic Cunningham, Commented May 29, 2016 at 19:59

Kasravnd · Accepted Answer · 2016-05-29 19:58:04Z

1

You can use a nested list comprehension:

>>> [sub[:1]+[i for i in sub[1:] if not(sub[0] in A and i in A)] for sub in B]
[['x', 'a', 'b', 'c'], ['y', 'c'], ['x', 'a', 'c']]

Here you'll preserve the items from sublists (from second item to end) that doesn't met your conditions for deleting.

answered May 29, 2016 at 19:58

Kasravnd

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2 Comments

Martijn Pieters Over a year ago

Yeah, you could, but here I wouldn't. not (sub[0] in A and i in A) is not exactly rolling of the tongue.

Kasravnd Over a year ago

@MartijnPieter Maybe, but sometimes it's cool to use double negative elimination in scripts :-).

Martijn Pieters · Accepted Answer · 2016-05-29 20:00:30Z

1

Use a list comprehension to update a list in-place:

for sublist in B:
    if sublist[0] in A:
        sublist[1:] = [v for v in sublist[1:] if v not in A]

j in your loop is only another reference to a value in a sublist. del j removes that one reference, but the list still contains that reference. You can remove values from the list with del listobj[index] or listobj.pop(value) (watch for subtleties in what those mean), but deleting from a list while iterating over it will result in items being skipped if you are not careful.

By assigning to a slice, you replace those elements in the list itself, in-place.

Note that you probably want to make A a set; membership testing is far faster when using a set:

Aset = set(A)

for sublist in B:
    if sublist[0] in Aset:
        sublist[1:] = [v for v in sublist[1:] if v not in Aset]

Demo:

>>> A = ['x', 'y', 'z']
>>> B = [['x', 'a', 'b', 'c'], ['y', 'c'], ['x', 'a', 'c', 'z']]
>>> Aset = set(A)
>>> for sublist in B:
...     if sublist[0] in Aset:
...         sublist[1:] = [v for v in sublist[1:] if v not in Aset]
...
>>> B
[['x', 'a', 'b', 'c'], ['y', 'c'], ['x', 'a', 'c']]

edited May 29, 2016 at 20:00

answered May 29, 2016 at 19:55

Martijn Pieters

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1 Comment

JanM Over a year ago

thank you, much appreciated. A was already a set but then I modified it again to list(A) because I thought playing with both list would be somehow easy.

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