Python Forum

Dear members of python-forum,

I encounter a problem of list definition (i think). I would like to define def(S,T), where:
(i) S and T are two sets (lists in Python) and such that S is a subset of T.
(ii) def(S,T) gives the list of lists K such that S \subset K \subset T. (K is "between" S and T).

For example if S=[2] and T=[2,3,4], i want to get all the lists between [2] and [2,3,4], that is, L=[[2],[2,3],[2,4],[2,3,4]].

In the program below, i define subset(T,S) as being the elements in T which are not in S. In my example, setminus(T,S)=[3,4].


Notice that:
union(A,B) is the union of the lists A and B. (the union of sets).
interzz(A,B) is the intersection of lists A and B (the intersection of sets).
doublonsL removes the lists appearing several times in the list of lists.

The program works as follows: from the infimum S=[2] (for example) and the supremum T=[2,3,4], we compute the elements setminus(T,S) which are in T but not in S. We start from L=[[2]]. The elements 3 and 4 are in T but not in S. We take elements in L and we add lists with these elements setminus(T,S). For example:

[2]
[2],[2,3]
[2],[2,4],[2,3],[2,3,4].

My program, instead of giving what i want, gives Boolean([2],[2,3,4])=
[[2, 3, 4]]. That is, [2], [2,3] and [2,4] are missing.

I guess that my program is ugly but, actually, i just need something that works. Your suggestions would probably be helpful; thank you in advance.

def doublons(LM):# This def is OK.
    LP=[]
    for k in LM:
        if k not in LP:
            LP.append(k)
    return(LP)


def doublonsL(LM): # This def is OK.
    LP=[]
    for listtt in LM:
        if listtt not in LP:
            LP.append(listtt)
    return(LP)

def union(A,B):# This def is OK. 
    uni=A
    for k in B:
        uni.append(k)
    return doublons(sorted(uni))  


def interzz(lst1, lst2): # This def is OK.
    lst3 = [value for value in lst1 if value in lst2] 
    return doublons(sorted(lst3)) 


def setminus(S,T):# This def is OK. 
    for k in T:
        S.remove(k)
    return doublons(S)


def issubset(S,T):# This def is OK. 
    if interzz(S,T)==S:
        return "true"
    else:
        return "false"


def Boolean(SQ,TQ): # The problem comes from this def.
    if issubset(SQ,TQ)=="false": # We check that S is a subset of T.
        print("error subset")
    ZED=setminus(TQ,SQ) # All the elements of TQ which are not in SQ.
    LILI=[SQ]
    MOLI=LILI[:] 
    for k in ZED:
        for PPP in MOLI:
            LILI.append(union(PPP,[k]))
        LILI=MOLI[:]
        MOLI=LILI[:]
    for kj in range(0,len(LILI)):
        LILI[kj]=sorted(doublons(LILI[kj]))
    return doublonsL(LILI) 

Wishing you a nice week,
Hassediagram.

python has lists x = [1,2,3]
and sets x = {1,2,3}
plus all set operations see: https://docs.python.org/3.7/library/stdt...-frozenset

aslo see itertools for permutations, combinations and product

Also, Python has the boolean singletons True and False, so you don't need "true" and "false". If issubset returned those, you could just use if issubset(SQ, TQ):. Of course, since all issubset does is check interzz(), you could just replace that with if interzz(SQ, TQ) == SQ:.

(Feb-25-2019, 03:40 PM)Larz60+ Wrote: [ -> ]python has lists x = [1,2,3]
and sets x = {1,2,3}
plus all set operations see: https://docs.python.org/3.7/library/stdt...-frozenset

aslo see itertools for permutations, combinations and product

Thank you Larz60 for your answer; i meant sets in the mathematical sense, which i translate into lists in Python.

As Larz said, this functionality is all covered in Python. But if you need to recreate it for homework, to get the subsets, start with a list of the empty set. Loop through the items in the set, adding to the list of subsets every subset plus the item:

>>> subsets = [[]]
>>> fullset = [1, 2, 3]
>>> for item in fullset:
...    subsets = subsets + [subset + [item] for subset in subsets]
...
>>> subsets
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

You should be able to modify that to get the list of subsets you want.

(Feb-25-2019, 04:14 PM)ichabod801 Wrote: [ -> ]As Larz said, this functionality is all covered in Python. But if you need to recreate it for homework, to get the subsets, start with a list of the empty set. Loop through the items in the set, adding to the list of subsets every subset plus the item:

>>> subsets = [[]]
>>> fullset = [1, 2, 3]
>>> for item in fullset:
...    subsets = subsets + [subset + [item] for subset in subsets]
...
>>> subsets
[[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

You should be able to modify that to get the list of subsets you want.

Thank you ichabod801, this works ! I will use your trick.
But why didn't it work in my program ?

Thank you also for your suggestion about the intersections.

(Feb-25-2019, 04:23 PM)Hassediagram Wrote: [ -> ]But why didn't it work in my program ?

I don't know. Your code is really hard to read and very complicated, I just figured there must be a simpler solution. Always keep it simple when programming.

(Feb-25-2019, 04:34 PM)ichabod801 Wrote: [ -> ]

(Feb-25-2019, 04:23 PM)Hassediagram Wrote: [ -> ]But why didn't it work in my program ?

I don't know. Your code is really hard to read and very complicated, I just figured there must be a simpler solution. Always keep it simple when programming.

Yes that is true; ok i got my problem solved, thank you very much ichabod801 !