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I need to place images so they are not on each other. I'm looking at matrix 3x3, so 9 places. Image is horizontal and it's taking 2 places. I tried this way (edited):

other_position = [e for e in range(9)]

def find_horizontal_position(avaiable_pos):
    myList = avaiable_pos
    print(avaiable_pos)
    try:
        possible_position = [0,1,3,4,6,7] # positions where image is not cut in half
        position = random.choice(possible_position)
        myList.remove(position)
        myList.remove(position + 1)
    except ValueError:
        return find_horizontal_position(avaiable_pos)
    return myList, position

for _ in range(2):
    other_position, position = find_horizontal_position(other_position)
    print(position)

Sometimes it generates output like this:

1
[0, 3, 4, 5, 6, 7, 8]
[0, 3, 4, 5, 6, 7, 8]
[0, 3, 4, 5, 6, 7, 8]
[0, 3, 4, 5, 6, 7, 8]
[3, 4, 5, 6, 7, 8]
4
[3, 6, 7, 8]

First image is placed on position 1, so 1 and 2 is taken out. Second image is on position 4, but by the way it takes away 0... Maybe there is a better way to do it?

1

In your except clause, incorporate a return statement:

return   find_horizontal_position(avaiable_pos)

otherwise what happens is that even though the position you attempted to place your image in is invalid, you will still execute the return statement

return myList, position

which in your example has position==0. Note that your approach might not find a valid solution even if one exists, I would suggest looping over all pairs

(img1leftposition, imp2leftposition), img1leftposition != imp2leftposition

where img1leftposition and img2leftposition are in possible_position (plenty of ways to do this, simple one using

for pair in zip(possible_position, possible_position):
    ...
  • My solution with additional return statement is still buggy. Now it correctly separates two images, but once in a while takes away one additional spot - I edited my post. Can you explain better your suggestion, give sample code? I don't know the number or rectangle images, it can be 1, 2 or 3. – Plajerity Apr 16 at 21:44
  • 0 was removed because when position==1, the first remove is succesful, so the try-statement only fails until you get to myList.remove(position + 1). – Juan Carlos Ramirez Apr 16 at 22:12
  • This is way I'm creating myList. It should take old variable (other position). – Plajerity Apr 16 at 22:34
  • errata: meant to say "because when position==0" – Juan Carlos Ramirez Apr 16 at 22:37
  • Once again, I created new variable and I'm removing positions from it. It should call old variable if it fails to match. – Plajerity Apr 16 at 22:40

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