List assignment, unexpected result

[Bengt Richter]

On Wed, 10 Jul 2002 22:33:02 +0100, "Steve Coates" <steve.coates at virgin.net> wrote:

>Thanks for the help. I think my favourite solution was
>
>grid = [['.' * 4] for i in range(4)]
>
I think you should take Dave Reed's advice and use

    grid = [['.',] * 4 for i in range(4)]

though you can leave out the the comma, since ['.',] is a list, not a one-element tuple

>but I'll need to read up on the syntax. I haven't seen this
>use of a for loop before.
>
As someone pointed out, it's a list comprehension, not a for loop.
Your "favorite solution" is a little different from Dave's suggestion
(as you've probably found out by now):

 >>> grid = [['.' * 4] for i in range(4)]
 >>> grid
 [['....'], ['....'], ['....'], ['....']]
 >>> grid [0][0] = '0'
 >>> grid
 [['0'], ['....'], ['....'], ['....']]

IOW, even trivial things are worth testing ;-)

vs. moving a right bracket to get Dave's version (w/o comma):
 >>> grid = [['.' * 4] for i in range(4)]
                     :
                     v
                 ,-<-'
                 v
 >>> grid = [['.'] * 4 for i in range(4)]
 >>> grid
 [['.', '.', '.', '.'], ['.', '.', '.', '.'], ['.', '.', '.', '.'], ['.', '.', '.', '.']]
 >>> grid [0][0] = '0'
 >>> grid
 [['0', '.', '.', '.'], ['.', '.', '.', '.'], ['.', '.', '.', '.'], ['.', '.', '.', '.']]

or to finish,

 >>> for i in range(4): grid[i][i] = i
 ...
 >>> grid
 [[0, '.', '.', '.'], ['.', 1, '.', '.'], ['.', '.', 2, '.'], ['.', '.', '.', 3]]

Regards,
Bengt Richter