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 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
On Mon, Jul 11, 2011 at 9:54 AM, Anthony Kong anthony.hw.k...@gmail.com 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
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
Good point! Need to get my terminology right. Thanks
On Tue, Jul 12, 2011 at 2:43 AM, Ian Kelly ian.g.ke...@gmail.com wrote:
On Mon, Jul 11, 2011 at 9:54 AM, Anthony Kong anthony.hw.k...@gmail.com
wrote:
Hi, all,
This question is in the same context of my two earlier questions. This
# 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
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 is
On 2:59 PM, Anthony Kong wrote:
snip
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
On Mon, Jul 11, 2011 at 10:53 AM, Anthony Kong
anthony.hw.k...@gmail.com 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)
On Mon, Jul 11, 2011 at 11:21 AM, Anthony Kong
anthony.hw.k...@gmail.com 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):
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'...]
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
remainder'
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
On Jun 5, 11:33 pm, Terry Reedy tjre...@udel.edu wrote:
On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
jyoun...@kc.rr.com 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
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 by
Steven D'Aprano steve+comp.lang.pyt...@pearwood.info 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
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
On 6/6/2011 1:29 PM, rusi wrote:
On Jun 5, 11:33 pm, Terry Reedytjre...@udel.edu 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
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
jyoun...@kc.rr.com 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,
On 2:59 PM, Ian Kelly wrote:
On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico ros...@gmail.com 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
On 6/5/2011 5:31 AM, Alain Ketterlin wrote:
jyoun...@kc.rr.com 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
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']
On Sun, Jun 5, 2011 at 3:46 AM, jyoun...@kc.rr.com 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
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']
On Sat, Jun 4, 2011 at 12:09 PM, Chris Angelico ros...@gmail.com 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:
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
--
On Fri, Jun 11, 2010 at 10:11 PM, Ian Kelly ian.g.ke...@gmail.com wrote:
On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis vinc...@vincentdavis.net
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
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)
On Fri, Jun 11, 2010 at 9:31 PM, Vincent Davis vinc...@vincentdavis.net 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 =
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