Pablo Torres schreef: > Hi guys! > > I am working on Conway's Game of Life right now and I've run into a > little problem. > I represent dead cells with 0s and live ones with 1s. Check this out: > > >>> grid = [[0] * 3] * 3 > >>> grid > [[0, 0, 0], [0, 0, 0], [0, 0, 0]] > >>> grid[0][0] = 1 > [[1, 0, 0], [1, 0, 0], [1, 0, 0]] > > Now, that's not what I want. I just want the first element of the > first sublist to be a 1, not the first of every one. Here's what I > tried and didn't work: > > 0. grid = [[0] * 3][:] * 3 then the same grid[0][0] = 1 statement > 1. grid = [[0][:] * 3] * 3 followed by the same assignment > 2. (grid[0])[0] = 1 with the original initialization of the grid > > So, that means that it isn't a problem with two different variables > refering to the same list (0. and 1. prove that). What I don't > understand is why 2. doesn't work either. I'm baby-feeding my > instructions to Python and the mistake is still there. Any ideas? > > Hope you can help. Thanks in advance,
The multiplication operator doesn't make new copies; all elements reference the same object, as you can see with id(): >>> lst = [0] * 3 >>> lst [0, 0, 0] >>> id(lst[0]), id(lst[1]), id(lst[2]) (9788716, 9788716, 9788716) As a consequence, if you modify one element, you change them all (since it's all the same object). Solution: make sure to create independent lists. >>> grid = [[0] * 3 for i in range(3)] >>> grid [[0, 0, 0], [0, 0, 0], [0, 0, 0]] >>> grid[0][0] = 1 -- If I have been able to see further, it was only because I stood on the shoulders of giants. -- Isaac Newton Roel Schroeven -- http://mail.python.org/mailman/listinfo/python-list
