Em terça-feira, 15 de setembro de 2015 21:47:10 UTC-3, Chris Angelico escreveu:
> On Wed, Sep 16, 2015 at 10:29 AM, Rafael David wrote:
> > Oooohhh ... I think I got it! I'm assigning a reference to peca and not the
> > value itself! Thank you very much MRAB and C Smith for the enlightenment :)
On Wed, Sep 16, 2015 at 10:29 AM, Rafael David wrote:
> Oooohhh ... I think I got it! I'm assigning a reference to peca and not the
> value itself! Thank you very much MRAB and C Smith for the enlightenment :)
Right! That's how Python's assignment always works. You may find, in
your case, that y
Em terça-feira, 15 de setembro de 2015 21:11:38 UTC-3, MRAB escreveu:
> On 2015-09-16 00:45, Rafael David wrote:
> > Hi guys,
> >
> > I'm newbie in Python (but not a newbie developer). I'm facing a
> > problem with a bidimensional list (list of lists) containing
> > dictionaries. I don't know if I
On 2015-09-16 00:45, Rafael David wrote:
Hi guys,
I'm newbie in Python (but not a newbie developer). I'm facing a
problem with a bidimensional list (list of lists) containing
dictionaries. I don't know if I didn't understand how lists and
dictionaries work in Python or if there is a mistake in m
>>tabuleiro[lin][col] = peca
use peca.copy() here or else a deep copy is made.
On Tue, Sep 15, 2015 at 4:45 PM, Rafael David wrote:
> Hi guys,
>
> I'm newbie in Python (but not a newbie developer). I'm facing a problem with
> a bidimensional list (list of lists) containing dictionaries. I don't
Hi guys,
I'm newbie in Python (but not a newbie developer). I'm facing a problem with a
bidimensional list (list of lists) containing dictionaries. I don't know if I
didn't understand how lists and dictionaries work in Python or if there is a
mistake in my code that I can't see. In the code I'm
On Feb 20, 8:12 am, "ssd" wrote:
> Hi,
>
> In the following code, (in Python 2.5)
> I was expecting to get in "b" variable the values b: [[0, 0], [0, 1],[0,
> 2], [0, 3],[0, 4], [1, 0],[1, 1], [1, 2], .]
> But I get only the last value [4,4], b: b: [[4, 4], [4, 4], [4, 4], ... ]
>
> My code:
On Fri, Feb 20, 2009 at 8:12 AM, ssd wrote:
>
> Hi,
>
> In the following code, (in Python 2.5)
> I was expecting to get in "b" variable the values b: [[0, 0], [0, 1],[0,
> 2], [0, 3],[0, 4], [1, 0],[1, 1], [1, 2], .]
> But I get only the last value [4,4], b: b: [[4, 4], [4, 4], [4, 4], ... ]
On Feb 20, 10:12 am, "ssd" wrote:
> Hi,
>
> In the following code, (in Python 2.5)
> I was expecting to get in "b" variable the values b: [[0, 0], [0, 1],[0,
> 2], [0, 3],[0, 4], [1, 0],[1, 1], [1, 2], .]
> But I get only the last value [4,4], b: b: [[4, 4], [4, 4], [4, 4], ... ]
>
> My code:
Hi,
In the following code, (in Python 2.5)
I was expecting to get in "b" variable the values b: [[0, 0], [0, 1],[0,
2], [0, 3],[0, 4], [1, 0],[1, 1], [1, 2], .]
But I get only the last value [4,4], b: b: [[4, 4], [4, 4], [4, 4], ... ]
My code:
a = ["",""]
b = []
for i in range (0,5):
On Jul 31, 2:51 pm, Alexnb <[EMAIL PROTECTED]> wrote:
> Lets say we have this list:
>
> funlist = ['a', 'b', 'c']
>
> and lets say I do this:
>
> if funlist[4]:
> print funlist[4]
>
> I will get the exception "list index out of range"
>
> How can I test if the list item is empty without getting
Alexnb:
> How can I test if the list item is empty without getting that exception?
In Python such list cell isn't empty, it's absent. So you can use
len(somelist) to see how much long the list is before accessing its
items. Often you can iterate on the list with a for, so you don't need
to care of
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