Remove an item from a list contained in another item in python

[Yoriz]

You can step through the code at the following link
visualize code execution

[Gribouillis]

I have an even more avant-garde version that works for me. I don't know if this is documented somewhere

1 2	roots = [] roots.extend(s for s in S if not any(s.issubset(r) for r in roots))

[CompleteNewb]

Ahh, I understand your confusion now. It is clever, and I guess potentially confusing.

You are correct that roots starts out as an empty list the first time through the "for s in S:" loop.

roots = [], s = {2, 3, 4, 5}

Since roots is empty, s is not going to be a subset of any of the sets in roots and is appended to roots.

roots = [{2, 3, 4, 5}], s = {4, 5, 6}

Second time through the outer loop roots contains {2, 3, 4, 5}. s is not a subset of {2, 3, 4, 5}, so it is appended to roots.

roots = [{2, 3, 4, 5}, {4, 5, 6}] s = {2, 3}

Third time through the outer loop {2, 3} is a subset of {2, 3, 4, 5}, so it is not appended to roots.

Do follow how this goes now? Each time through the outer loop we test if s is a subset of any of the sets in root. If s is a subset of one of the sets in root, it is not appended to root. If s is not a subset of any set in root, s is appended to root.

Okay, thanks, but how would you write it if you don't want the "for r in roots" on the same line as the "any(s.issubset( r )"? I am trying to rewrite it that way, because I still have a hard time understanding the " line ??? of the for loop" and I just can't do it.

[Yoriz]

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items = [[2, 3, 4, 5], [2, 3], [7, 8], [2], [4, 5, 6]] sorted_items = sorted((set(item) for item in items), key=len, reverse=True)     roots = [] for item in sorted_items:     for root in roots:         if item.issubset(root):             break     else:         roots.append(item)   print(roots)

Output:
[{2, 3, 4, 5}, {4, 5, 6}, {8, 7}]

[deanhystad]

You can replace the "any" with a loop, but you need the "for r in roots".

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l = [[2,3,4,5],[2,3],[7,8],[2],[4,5,6]] S = sorted((set(x) for x in l), key=len, reverse=True)   roots = [] for s in S:     for r in roots:         if s.issubset(r):             break  # Exclude s from results     else:         roots.append(s)  # Include s in results   print(roots)

Echo...echo...

[Gribouillis]

Also

1	sorted_items = sorted(map(set, items), key=len, reverse=True)

would be prettier.

[CompleteNewb]

You can replace the "any" with a loop, but you need the "for r in roots".

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l = [[2,3,4,5],[2,3],[7,8],[2],[4,5,6]] S = sorted((set(x) for x in l), key=len, reverse=True)   roots = [] for s in S:     for r in roots:         if s.issubset(r):             break  # Exclude s from results     else:         roots.append(s)  # Include s in results   print(roots)

Echo...echo...

Well i tried rewriting it from my own comprehension and i came up with that which is in my mind the same thing, but it doesn't work??? What am i missing?

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ls = [[2,3,4,5],[2,3],[7,8],[2],[4,5,6]] l = [] roots = [] for s in ls:        l.append(set(s))   for s in l:     for r in roots:         if s.issubset(r):             break         else:             roots.append(s)   print(roots)

I got an empty list : roots = []

[deanhystad]

The else needs to be attached to for loop, not the if statement. Read about for..else.

[CompleteNewb]

Thank you very much everyone. Your help has been very helpful! I really appreciated it

[Gribouillis]

@CompleteNewb WARNING If you don't sort the list of sets by decreasing lengths, the algorithm may fail. It will keep sets that are subsets of other sets that come later in the list.