So subclass B has no access to __not_here in A after all...
OK, in one of legacy Python I supported there are a lot of code floating
around like this. It works OK (in term of business logic and unit test).
That's probably due to luck :-)
It also uses a lot of __slot__ = ['attr_a', 'attr_b'...] in
On Mon, Jul 11, 2011 at 11:21 AM, Anthony Kong
wrote:
> Awesome, Thomas. The trick only works if there is only one leading
> underscore in the method names.
> The following example works as I expected for the derived class B.
> class A(object):
> def __init__(self):
> self.__not_here =
On Mon, Jul 11, 2011 at 10:53 AM, Anthony Kong
wrote:
> Thanks again for your input, Thomas.
> I normally prefer
> not_here = property(lambda self: self.__get_not_here(), lambda self, v:
> self.__set_not_here(v))
> than
> not_here = property(__get_not_here, __set_not_here)
> Because it allows me t
On 2:59 PM, Anthony Kong wrote:
> So the question: is it possible to use lambda expression at all for
> the setter? (As in the last, commented-out line)
>
> Python interpreter will throw an exception right there if I use the
> last line ('SyntaxError: lambda cannot contain assignment'). I'd use
>
>
> PS: are you sure the lambda self: self.__foo() trick works, with
> subclasses or otherwise? I haven't tested it, and I'm not saying it
> doesn't, but I have a feeling double-underscore name mangling might be a
> problem somewhere down the line?
>
>
Awesome, Thomas. The trick only works if there
# On 07/11/2011 06:53 PM, Anthony Kong wrote:
# But decorator! Of course! Thanks for reminding me this.
#
# In your example, where does '@not_here' come from? (Sorry, this syntax
# is new to me)
class A(object):
def __init__(self):
self.not_here = 1
@property
def not_here(se
Good point! Need to get my terminology right. Thanks
On Tue, Jul 12, 2011 at 2:43 AM, Ian Kelly wrote:
> On Mon, Jul 11, 2011 at 9:54 AM, Anthony Kong
> wrote:
> > Hi, all,
> > This question is in the same context of my two earlier questions. This
> > question was raised by some python beginner
Thanks again for your input, Thomas.
I normally prefer
not_here = property(lambda self: self.__get_not_here(), lambda self, v:
self.__set_not_here(v))
than
not_here = property(__get_not_here, __set_not_here)
Because it allows me to have a pair getter/setter (when there is a need for
it). Use o
On Mon, Jul 11, 2011 at 9:54 AM, Anthony Kong wrote:
> Hi, all,
> This question is in the same context of my two earlier questions. This
> question was raised by some python beginners, and I would like to check with
> the list to ensure I provide a correct answer.
> Here is a code snippet I used t
On 07/11/2011 05:54 PM, Anthony Kong wrote:
> Hi, all,
>
> This question is in the same context of my two earlier questions. This
> question was raised by some python beginners, and I would like to check
> with the list to ensure I provide a correct answer.
>
> Here is a code snippet I used to de
Hi, all,
This question is in the same context of my two earlier questions. This
question was raised by some python beginners, and I would like to check with
the list to ensure I provide a correct answer.
Here is a code snippet I used to demonstrate the keyword *property*:
class A(object):
On 6/6/2011 12:52 PM, Terry Reedy wrote:
def group(seq, n):
'Yield from seq successive disjoint slices of length n & the remainder'
if n<=0: raise ValueError('group size must be positive')
for i in range(0,len(seq), n):
yield seq[i:i+n]
for inn,out in (
(('',1), []), # no input, no output
#(('a
On 6/6/2011 1:29 PM, rusi wrote:
On Jun 5, 11:33 pm, Terry Reedy wrote:
Let me add something not said much here about designing functions: start
with both a clear and succinct definition *and* test cases. (I only
started writing tests first a year ago or so.)
I am still one year in the futu
Steven D'Aprano wrote:
For any non-trivial function, I usually start by writing the
documentation (a docstring and doctests) first. How else do you know what
the function is supposed to do if you don't have it documented?
Yes. In my early years I was no different than any other hacker in terms
Steven D'Aprano writes:
> On Mon, 06 Jun 2011 12:52:31 -0400, Terry Reedy wrote:
>
> > Let me add something not said much here about designing functions: start
> > with both a clear and succinct definition *and* test cases. (I only
> > started writing tests first a year ago or so.)
>
> For any no
On Mon, 06 Jun 2011 12:52:31 -0400, Terry Reedy wrote:
> Let me add something not said much here about designing functions: start
> with both a clear and succinct definition *and* test cases. (I only
> started writing tests first a year ago or so.)
For any non-trivial function, I usually start b
On Jun 5, 11:33 pm, Terry Reedy wrote:
> On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
>
> > writes:
>
> > f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
>
> f=lambda ... statements are inferior for practical purposes to the
> equivalent def f statements because the resultin
On 6/6/2011 9:42 AM, jyoun...@kc.rr.com wrote:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
Packing tail recursion into one line is bad for both understanding and
refactoring. Use better names and a docstring gives
def group(seq, n):
'Yield from seq successive disjoin
> f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
> Packing tail recursion into one line is bad for both understanding and
> refactoring. Use better names and a docstring gives
>
> def group(seq, n):
>'Yield from seq successive disjoint slices of length n plus the
> re
On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
writes:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f=lambda ... statements are inferior for practical purposes to the
equivalent def f statements because the resulting object is missing a
useful name attribute and a docstr
On 2:59 PM, Ian Kelly wrote:
> On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico wrote:
>> Python doesn't seem to have an inbuilt function to divide strings in
>> this way. At least, I can't find it (except the special case where n
>> is 1, which is simply 'list(string)'). Pike allows you to use the
writes:
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f("Hallo Welt", 3)
> ['Hal', 'lo ', 'Wel', 't']
>
> http://stackoverflow.com/questions/312443/how-do-you-split-a-list-into-evenly-s
> ized-chunks-in-python/312644
>
> It doesn't work with a huge list, but l
jyoun...@kc.rr.com wrote:
> I was surfing around looking for a way to split a list into equal
> sections.
non-recursive, same-unreadeable (worse?) one liner alternative:
def chunks(s, j):
return [''.join(filter(None,c))for c in map(None,*(s[i::j]for i in
range(j)))]
--
By ZeD
--
http:/
On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico wrote:
> Python doesn't seem to have an inbuilt function to divide strings in
> this way. At least, I can't find it (except the special case where n
> is 1, which is simply 'list(string)'). Pike allows you to use the
> division operator: "Hello, worl
jyoun...@kc.rr.com wrote:
> I was surfing around looking for a way to split a list into equal
> sections. I came upon this algorithm:
>
f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
f("Hallo Welt", 3)
> ['Hal', 'lo ', 'Wel', 't']
>
> http://stackoverflow.com/ques
On Sun, Jun 5, 2011 at 3:46 AM, wrote:
> It doesn't work with a huge list, but looks like it could be handy in certain
> circumstances. I'm trying to understand this code, but am totally lost. I
> know a little bit about lambda, as well as the ternary operator, but how
> does this part work:
>
I was surfing around looking for a way to split a list into equal sections. I
came upon this algorithm:
>>> f = lambda x, n, acc=[]: f(x[n:], n, acc+[(x[:n])]) if x else acc
>>> f("Hallo Welt", 3)
['Hal', 'lo ', 'Wel', 't']
http://stackoverflow.com/questions/312443/how-do-you-split-a-list
On Fri, Jun 11, 2010 at 10:11 PM, Ian Kelly wrote:
> On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis
> wrote:
>> Starting with an example.
>> In [23]: x = [1,2,3,4,4,4,5,5,3,2,2,]
>> In [24]: y = set(x)
>> In [25]: y
>> Out[25]: set([1, 2, 3, 4, 5])
>> In [26]: y2 = len(set(x))
>> In [27]: y2
>>
On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis wrote:
> Starting with an example.
> In [23]: x = [1,2,3,4,4,4,5,5,3,2,2,]
> In [24]: y = set(x)
> In [25]: y
> Out[25]: set([1, 2, 3, 4, 5])
> In [26]: y2 = len(set(x))
> In [27]: y2
> Out[27]: 5
>
> How would I do the above "y2 = len(set(x))" but ha
Starting with an example.
In [23]: x = [1,2,3,4,4,4,5,5,3,2,2,]
In [24]: y = set(x)
In [25]: y
Out[25]: set([1, 2, 3, 4, 5])
In [26]: y2 = len(set(x))
In [27]: y2
Out[27]: 5
How would I do the above "y2 = len(set(x))" but have len(set()) in a
dictionary. I know how to do ..
In [30]: d = dict(s=set
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