Remove a sublist from a list in which the numbers alreadys exist in another sublist

Question

Given a list t :

[[0, 4], [0, 4, 5, 7], [1, 2, 3, 6], [1, 3], [2, 6], [5, 7]]

I want to remove sublists in t , eg [0,4] , both numbers in which already exist in [0,4,5,7] , I want to keep the larger groups and remove smaller groups. And so in the end [[0, 4, 5, 7], [1,2,3,6]] will be my final results.

I have tried the following code but don't know what is wrong:

V=0
while V<60:
    for o in t:
        for p in t:
            if o!= p:
                for s in range(len(o)):
                    w=[]
                    if o[s] in p:
                        w.append(o[s])
                        #print w
                    if w==o:
                        t.remove(w)
    V=V+1
print t

Answer 1

You could use sets for this:

lists = [[0, 4], [0, 4, 5, 7], [1, 2, 3, 6], [1, 3], [2, 6], [5, 7]]

sets = map(set, lists)
ret = []
for l in lists:
  if not any(set(l) < s for s in sets):
    ret.append(l)
print ret

Here, set(l) < s checks that the list l is a proper subset of s .

Or, if you like conciseness:

sets = map(set, lists)
ret = [l for l in lists if not any(set(l) < s for s in sets)]
print ret

Answer 2

l = [[0, 4], [0, 4, 5, 7], [1, 2, 3, 6], [1, 3], [2, 6], [5, 7]]
final = l[:]

for m in l:
    for n in l:
        if set(m).issubset(set(n)) and m != n:
            final.remove(m)
            break   
print final

output: 
[[0, 4, 5, 7], [1, 2, 3, 6]]