Argthtjtr

I am programming a simple software for practicing.

def check(num, lst):

    for i in lst:

        if num == i:

            return False

    else:

       lst.append(num)

       return True



timelst = 

for i1 in range(1,7):

    usednum = [i1]



    for i2 in range(1,7):

        if not check(i2, usednum):

            continue



        for i3 in range(1,7):

            if not check(i3, usednum):

                continue



            for i4 in range(1,7):

                if not check(i4, usednum):

                    continue



                for i5 in range(1,7):

                    if not check(i5, usednum):

                        continue



                    for i6 in range(1,7):

                        if not check(i6, usednum):

                            continue

                        else:



                            print(usednum)   #print the appending list before actual appending

                            timelst.append(usednum)

                            usednum.pop()

                            break



                    usednum.pop()

When running this, this is what I want the timelst to be like:

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

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

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

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

 ...

]

However this is what I actually get:

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

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

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

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

 ...

]

I am really confused about this and got stuck for a long time. I tried printing the 'usednum' list before appending it, and it returns perfectly what I want.

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

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

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

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

#printed lists while running the program

But the same problem still occurs everytime when I check the timelst after executing the program.

I used python 3.7 with spyder. There should not be a problem with my compiler, as I have tried running this on ipython and I still got the same result.

Can anyone please help me solving this problem? Thanks!

You can make your life a lot easier by using the itertools.permutations() function call:

import itertools



mylist = [1, 2, 3, 4, 5, 6]

for item in itertools.permutations(mylist):

    # Do something with item, which is a permutation of 'mylist'

    print(item)

Note, however, that this results in a big (6! ... that is six-factorial) list of results. If you only want combinations, take a look at itertools.combinations() instead.

The problem has been solved! I should have realised earlier that there had been an issue about the shallow and deep copy of the list I have appended. I have add

usednum = copy.deepcopy(usednum)

to the final loop, so it now looks like

for i6 in range(1, 7):

    if not check(i6, usednum):

        continue

    else:

        timelst.append(usednum)

        usednum = copy.deepcopy(usednum) #the new copy of the appended list, so now there's nothing to do with the appended one.

        print('i6',usednum.pop())

        break



    usednum.pop()

which can work properly now