Automatic redefining

[Hassediagram]

Dear Python-forum members,

I have registered to this forum because i have a small but resistant problem.
It is surely trivial, but i have not practiced Python for a long time and i don't remember the trick, and i would be grateful to benefit from someone's help.

(This def is to produce neighbors in a Hasse diagram).

Let "Collection" be a collection of vectors, initially empty.

From an initial vector that i call : "SET", i will produce other vectors that i will add to "Collection". Then i return "Collection".

To do this, I will add or remove elements of SET, and i add these vectors to "Collection".

My problem is that i don't want "SET" to change. When i write C=SET, i want to modify C, but i don't want "SET" to be modified when i modify C. However with this code, "SET" is modified when i modify C. (I can see the change with : print("SET before",SET); and : print("SET after",SET))

def PHa(SET):
    Collection=[]
    for k in range(1,n+1):
        if k not in SET:
            C=SET
            print("SET before",SET)
            C.append(k)
            print("SET after",SET)
            Collection.append(C)
    for k in range(1,n+1):
        if k in SET:
            C=SET
            C.remove(k)
            Collection.append(C)
    return Collection

Does someone know the trick to fix "SET" ?

Thank you very much in advance,
Hassediagram

[j.crater]

I suppose SET is a list. You should provide more details.
List is a mutable data type so changes to a list variable will remain. And assigning C to SET will point to same variable.
To copy the contents of SET to new variable C, you can do:

C=SET[:]

[ljmetzger]

n is not defined.

You probably want something like:

def PHa(SET, n):

or to define 'n' as a global outside the function (not recommended) and leave the 'PHa' line as is.

Lewis

[Hassediagram]

Dear Lewis, Yes I have defined n globally in the code before. Thank you for your remark.

---

Dear j.crater, Great ! It works. Thank you very much for your help; problem solved !

[Hassediagram]

Dear members of python-forum,

I have another problem. I think it is linked to the one i had last year (above).

I want to define a 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.

[Hassediagram]

Message deleted; i am opening a new topic.