I finally decided to have a go with Python and am working through the
tutorial.
On my old BBC Computer I could do something like this:
DIM A(2,2)
to create a 3 by 3 array of data. Then I could set any point:
A(0,0) = foo
A(0,1) = bar
etc.
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??
--
Steve 9 1078
On Sun, 02 Mar 2008 14:15:09 +0000, 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
Marc 'BlackJack' Rintsch wrote:
: On Sun, 02 Mar 2008 14:15:09 +0000, 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
Marc 'BlackJack' Rintsch schrieb:
On Sun, 02 Mar 2008 14:15:09 +0000, 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
Christoph Zwerschke wrote:
: Marc 'BlackJack' Rintsch schrieb:
:: On Sun, 02 Mar 2008 14:15:09 +0000, 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
On Sun, 02 Mar 2008 21:58:31 +0100, Christoph Zwerschke wrote:
Marc 'BlackJack' Rintsch schrieb:
>On Sun, 02 Mar 2008 14:15:09 +0000, 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
"Christoph Zwerschke" <ci**@online.dewrote in message
news:fq**********@online.de...
| Marc 'BlackJack' Rintsch schrieb:
| On Sun, 02 Mar 2008 14:15:09 +0000, 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.
Christoph Zwerschke wrote:
Marc 'BlackJack' Rintsch schrieb:
>On Sun, 02 Mar 2008 14:15:09 +0000, 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.
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/progra...mensional-list
--
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