Re: First post from a Python newbiw
Steve Turner wrote: I finally decided to have a go with Python and am working through the tutorial. Great! On my old BBC Computer [...] These were nice machines... In Python I thought I could do this with: a=[0,0,0] b=[a,a,a] b [[0, 0, 0], [0, 0, 0], [0, 0, 0]] b[1][1]='foo' b [[0, 'foo', 0], [0, 'foo', 0], [0, 'foo', 0]] I can understand why as b[1][1]='foo' is actually changing a[1] Apart from doing something like a=[0,0,0] b=[0,0,0] c=[0,0,0] d=[a,b,c] is there a better way of creating d?? It's a FAQ: http://www.python.org/doc/faq/programming/#how-do-i-create-a-multidimensional-list -- Arnaud -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Arnaud Delobelle schrieb: It's a FAQ: http://www.python.org/doc/faq/programming/#how-do-i-create-a-multidimensional-list Somewhere on my todo list I have read through the whole Python FAQ, but so far never got round doing it. Should probably set it to prio A. -- Christoph -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Marc 'BlackJack' Rintsch wrote: : On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: : :: Apart from doing something like :: a=[0,0,0] :: b=[0,0,0] :: c=[0,0,0] :: d=[a,b,c] :: :: is there a better way of creating d?? : : a = [[0] * 3 for dummy in xrange(3)] Thanks, Marc. -- Steve -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: Apart from doing something like a=[0,0,0] b=[0,0,0] c=[0,0,0] d=[a,b,c] is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Ciao, Marc 'BlackJack' Rintsch -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Marc 'BlackJack' Rintsch schrieb: On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: Apart from doing something like a=[0,0,0] b=[0,0,0] c=[0,0,0] d=[a,b,c] is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Why not simply [[0]*3]*3 ? -- Christoph -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Christoph Zwerschke wrote: : Marc 'BlackJack' Rintsch schrieb: :: On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: :: ::: Apart from doing something like ::: a=[0,0,0] ::: b=[0,0,0] ::: c=[0,0,0] ::: d=[a,b,c] ::: ::: is there a better way of creating d?? :: :: a = [[0] * 3 for dummy in xrange(3)] : : Why not simply [[0]*3]*3 ? I've just tried that and it gives the same as my earlier b=[a,a,a] -- Steve -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
On Sun, 02 Mar 2008 21:58:31 +0100, Christoph Zwerschke wrote: Marc 'BlackJack' Rintsch schrieb: On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: Apart from doing something like a=[0,0,0] b=[0,0,0] c=[0,0,0] d=[a,b,c] is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Why not simply [[0]*3]*3 ? Because: In [77]: a = [[0] * 3] * 3 In [78]: a Out[78]: [[0, 0, 0], [0, 0, 0], [0, 0, 0]] In [79]: a[0][0] = 42 In [80]: a Out[80]: [[42, 0, 0], [42, 0, 0], [42, 0, 0]] Ciao, Marc 'BlackJack' Rintsch -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Christoph Zwerschke [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] | Marc 'BlackJack' Rintsch schrieb: | On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: | | Apart from doing something like | a=[0,0,0] | b=[0,0,0] | c=[0,0,0] | d=[a,b,c] | | is there a better way of creating d?? | | a = [[0] * 3 for dummy in xrange(3)] | | Why not simply [[0]*3]*3 ? Because that is essentially the same as what the OP originally did, which does not work as he wanted. -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
Christoph Zwerschke wrote: Marc 'BlackJack' Rintsch schrieb: On Sun, 02 Mar 2008 14:15:09 +, Steve Turner wrote: Apart from doing something like a=[0,0,0] b=[0,0,0] c=[0,0,0] d=[a,b,c] is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Each element of a refers to a distinct array. Why not simply [[0]*3]*3 ? All three elements of the result refer to the same array. -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Each element of a refers to a distinct array. Why not simply [[0]*3]*3 ? All three elements of the result refer to the same array. ... whereas you reassign all three elements of [0]* 3. ((0,)*3,)*3 ((0, 0, 0), (0, 0, 0), (0, 0, 0)) You're safe in this one-- changing [0][0] won't change [1][0], 'cuz you can't! -- http://mail.python.org/mailman/listinfo/python-list
Re: First post from a Python newbiw
[EMAIL PROTECTED] wrote: is there a better way of creating d?? a = [[0] * 3 for dummy in xrange(3)] Each element of a refers to a distinct array. Why not simply [[0]*3]*3 ? All three elements of the result refer to the same array. ... whereas you reassign all three elements of [0]* 3. ((0,)*3,)*3 ((0, 0, 0), (0, 0, 0), (0, 0, 0)) You're safe in this one-- changing [0][0] won't change [1][0], 'cuz you can't! A technically correct solution. :) -- http://mail.python.org/mailman/listinfo/python-list
