First post from a Python newbiw

[Steve Turner]

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

[Marc 'BlackJack' Rintsch]

Re: First post from a Python newbiw

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

[Steve Turner]

Re: First post from a Python newbiw

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

[Christoph Zwerschke]

Re: First post from a Python newbiw

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

[Steve Turner]

Re: First post from a Python newbiw

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

[Marc 'BlackJack' Rintsch]

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 +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

[Terry Reedy]

Re: First post from a Python newbiw


"Christoph Zwerschke" <cito@online.de wrote in message
news:fqf4do$gq7 $1@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.

[Jeff Schwab]

Re: First post from a Python newbiw

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.

[Arnaud Delobelle]

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

[Christoph Zwerschke]

Re: First post from a Python newbiw

Arnaud Delobelle schrieb:

It's a FAQ:
http://www.python.org/doc/faq/progra...mensional-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