Re: checking if an object IS in a list
On 20 juil, 07:17, Marc 'BlackJack' Rintsch [EMAIL PROTECTED] wrote: On Sat, 19 Jul 2008 13:13:40 -0700, nicolas.pourcelot wrote: On 18 juil, 17:52, Marc 'BlackJack' Rintsch [EMAIL PROTECTED] wrote: On Fri, 18 Jul 2008 07:39:38 -0700, nicolas.pourcelot wrote: So, I use something like this in 'sheet.objects.__setattr__(self, name, value)': if type(value) == Polygon: for edge in value.edges: if edge is_in sheet.objects.__dict__.itervalues(): object.__setattr__(self, self.__new_name(), edge) Ok, I suppose it's confused, but it's difficult to sum up. ;-) You are setting attributes with computed names? How do you access them? Always with `gettattr()` or via the `__dict__`? If the answer is yes, why don't you put the objects the into a dictionary instead of the extra redirection of an objects `__dict__`? Yes, I may subclass dict, and change its __getitem__ and __setitem__ methods, instead of changing objets __setattr__ and __getattr__... But I prefer sheet.objects.A = Point(0, 0) than sheet.objects[A] = Point(0, 0) But with computed names isn't the difference more like setattr(sheet.objects, name, Point(0, 0)) vs. sheet.objects[name] = Point(0, 0) and getattr(sheet.objects, name) vs. sheet.objects[name] Or do you really have ``sheet.objects.A`` in your code, in the hope that an attribute named 'A' exists? Oh and the `type()` test smells like you are implementing polymorphism in a way that should be replaced by OOP techniques. I wrote 'type' here by mistake, but I used 'isinstance' in my code. ;-) Doesn't change the code smell. OOP approach would be a method on the geometric objects that know what to do instead of a type test to decide what to do with each type of geometric object. Ciao, Marc 'BlackJack' Rintsch Thank you for your advises, it's very interesting to have an external point of view. No, I have not anything like 'sheet.objects.A' in the library code, but I use then the library in different programs where things like sheet.objects.A sometimes occur. However, since it is not so frequent, I may indeed subclass dict, and change __getitem__, __setitem__ and __delitem__, and then redirect __getattr__, __setattr__, __delattr__ to previous methods... This would not break library external API, and it may speed-up a little internal stuff. I'm not very expert in OOP ; imho it's largely a mean and not a goal. On one hand, 'a method on the geometric objects that know what to do' would require them to contain some code concerning the object manager... I don't like that very much. On the other hand, object manager should not rely on object's implementation... I will think about that. Thank you all of you for your answers. I'm sorry I may not have time to reply further, it was interesting :-) (even if it's a bit difficult for me to write clearly in english ;-)) -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
(1) You are searching through lists to find float objects by identity, not by value -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
On 20 juil, 23:18, John Machin [EMAIL PROTECTED] wrote: On Jul 21, 4:33 am, [EMAIL PROTECTED] wrote: (1) You are searching through lists to find float objects by identity, not by value You wrote I used short lists (a list of 20 floats) and the element checked was not in the list. (That was the case I usually deals with in my code.) :-D -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
On 18 juil, 17:52, Marc 'BlackJack' Rintsch [EMAIL PROTECTED] wrote: On Fri, 18 Jul 2008 07:39:38 -0700, nicolas.pourcelot wrote: So, I use something like this in 'sheet.objects.__setattr__(self, name, value)': if type(value) == Polygon: for edge in value.edges: if edge is_in sheet.objects.__dict__.itervalues(): object.__setattr__(self, self.__new_name(), edge) Ok, I suppose it's confused, but it's difficult to sum up. ;-) You are setting attributes with computed names? How do you access them? Always with `gettattr()` or via the `__dict__`? If the answer is yes, why don't you put the objects the into a dictionary instead of the extra redirection of an objects `__dict__`? Yes, I may subclass dict, and change its __getitem__ and __setitem__ methods, instead of changing objets __setattr__ and __getattr__... But I prefer sheet.objects.A = Point(0, 0) than sheet.objects[A] = Point(0, 0) Oh and the `type()` test smells like you are implementing polymorphism in a way that should be replaced by OOP techniques. I wrote 'type' here by mistake, but I used 'isinstance' in my code. ;-) If you make Point immutable you might be able to drop the must not be referenced twice requirement. Yes, but unfortunately I can't (or it would require complete redesign...) -- http://mail.python.org/mailman/listinfo/python-list
checking if an object IS in a list
Hi, I want to test if an object IS in a list (identity and not equality test). I can if course write something like this : test = False myobject = MyCustomClass(*args, **kw) for element in mylist: if element is myobject: test = True break and I can even write a isinlist(elt, mylist) function. But most of the time, when I need some basic feature in python, I discover later it is in fact already implemented. ;-) So, is there already something like that in python ? I tried to write : 'element is in mylist' but this appeared to be incorrect syntax... All objects involved all have an '__eq__' method. Thanks, N. P. PS: Btw, how is set element comparison implemented ? My first impression was that 'a' and 'b' members are considered equal if and only if hash(a) == hash(b), but I was obviously wrong : class A(object): ... def __eq__(self,y): ... return False ... def __hash__(self): ... return 5 ... a=A();b=A() a==b False hash(b)==hash(a) True b in set([a]) False S=set([a]) S.difference([b]) set([__main__.A object at 0xb7a91dac]) So there is some equality check also, maybe only if '__eq__' is implemented ? -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
On 18 juil, 11:30, Peter Otten [EMAIL PROTECTED] wrote: [EMAIL PROTECTED] wrote: Hi, I want to test if an object IS in a list (identity and not equality test). I can if course write something like this : test = False myobject = MyCustomClass(*args, **kw) for element in mylist: if element is myobject: test = True break and I can even write a isinlist(elt, mylist) function. But most of the time, when I need some basic feature in python, I discover later it is in fact already implemented. ;-) So, is there already something like that in python ? I tried to write : 'element is in mylist' but this appeared to be incorrect syntax... There is no is in operator in Python, but you can write your test more concisely as any(myobject is element for element in mylist) Thanks a lot However, any() is only available if python version is = 2.5, but I may define a any() function on initialisation, if python version 2.5 I think something like id(myobject) in (id(element) for element in mylist) would also work, also it's not so readable, and maybe not so fast (?)... An is in operator would be nice... PS: Btw, how is set element comparison implemented ? My first impression was that 'a' and 'b' members are considered equal if and only if hash(a) == hash(b), but I was obviously wrong : class A(object): ... def __eq__(self,y): ... return False ... def __hash__(self): ... return 5 ... a=A();b=A() a==b False hash(b)==hash(a) True b in set([a]) False S=set([a]) S.difference([b]) set([__main__.A object at 0xb7a91dac]) So there is some equality check also, maybe only if '__eq__' is implemented ? In general equality is determined by __eq__() or __cmp__(). By default object equality checks for identity. Some containers (like the built-in set and dict) assume that a==b implies hash(a) == hash(b). Peter So, precisely, you mean that if hash(a) != hash(b), a and b are considered distinct, and else [ie. if hash(a) == hash(b)], a and b are the same if and only if a == b ? -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
In fact, 'any(myobject is element for element in mylist)' is 2 times slower than using a for loop, and 'id(myobject) in (id(element) for element in mylist)' is 2.4 times slower. -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
On 18 juil, 12:26, Peter Otten [EMAIL PROTECTED] wrote: [EMAIL PROTECTED] wrote: I think something like id(myobject) in (id(element) for element in mylist) would also work, also it's not so readable, and maybe not so fast (?)... An is in operator would be nice... And rarely used. Probably even less than the (also missing) in, | in, you-name-it operators... Maybe, but imho myobject is in mylist is highly readable, when myobject in mylist is not. -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
On 18 juil, 13:13, Peter Otten [EMAIL PROTECTED] wrote: [EMAIL PROTECTED] wrote: In fact, 'any(myobject is element for element in mylist)' is 2 times slower than using a for loop, and 'id(myobject) in (id(element) for element in mylist)' is 2.4 times slower. This is not a meaningful statement unless you at least qualify with the number of item that are actually checked. For sufficently long sequences both any() and the for loop take roughly the same amount of time over here. Sorry. I used short lists (a list of 20 floats) and the element checked was not in the list. (That was the case I usually deals with in my code.) -- http://mail.python.org/mailman/listinfo/python-list
Re: checking if an object IS in a list
What is your (concrete) use case, by the way? I try to make it simple (there is almost 25000 lines of code...) I have a sheet with geometrical objects (points, lines, polygons, etc.) The sheet have an object manager. So, to simplify : sheet.objects.A = Point(0, 0) sheet.objects.B = Point(0, 2) sheet.objects.C = Middle(A, B) Then we have : sheet.objects.A == sheet.objects.B True since have and B have the same coordinates. But of course A and B objects are not same python objects. In certain cases, some geometrical objects are automatically referenced in the sheet, without being defined by the user. (Edges for polygons, for example...) But they must not be referenced twice. So if the edge of the polygon is already referenced (because the polygon uses an already referenced object for its construction...), it must not be referenced again. However, if there is an object, which accidentally have the same coordinates, it must be referenced with a different name. So, I use something like this in 'sheet.objects.__setattr__(self, name, value)': if type(value) == Polygon: for edge in value.edges: if edge is_in sheet.objects.__dict__.itervalues(): object.__setattr__(self, self.__new_name(), edge) Ok, I suppose it's confused, but it's difficult to sum up. ;-) Another possibility :-) from itertools import imap id(x) in imap(id, items) I didn't know itertools. Thanks :-) -- http://mail.python.org/mailman/listinfo/python-list
Re: wxpython and wxtextctrl
Ok, as I guessed, it was Boa installation which changed the wxpython version used. I removed Boa... Thanks ! Nicolas Greg Krohn a écrit : Nicolas Pourcelot wrote: Hello, my script worked well until today : when I tried to launch it, I got the following : frame = MyFrame(None,-1,Geometrie,size=wx.Size(600,400)) File /home/nico/Desktop/wxGeometrie/version 0.73/geometrie.py, line 74, in __init__ self.commande.Bind(wx.EVT_CHAR, self.EvtChar) AttributeError: wxTextCtrl instance has no attribute 'Bind' (self.commande is a wxTextCtrl instance) I don't understand : why did wxTextCtrl lost its Bind attribute ?? As there's not so much changes on my computer since yesterday, I suppose this is due to Boa package installation on my Ubuntu Hoary ? Does Boa installation changes wxpython version ? Is wxTextCtrl attribute .Bind() obsolete ?? Thanks, Nicolas control.Bind is relativly new. The wxTextCtrl notation (vs wx.TextCtrl) is the old way (but it IS kept around for backwards compatablility). My guess is that your code is for for a newer version of wxPython than what you actually have. Try printing the version from in your code: import wxPyhon print wxPython.__version__ self.commande.Bind(wx.EVT_CHAR, self.EvtChar) -greg -- http://mail.python.org/mailman/listinfo/python-list
wxpython and wxtextctrl
Hello, my script worked well until today : when I tried to launch it, I got the following : frame = MyFrame(None,-1,Geometrie,size=wx.Size(600,400)) File /home/nico/Desktop/wxGeometrie/version 0.73/geometrie.py, line 74, in __init__ self.commande.Bind(wx.EVT_CHAR, self.EvtChar) AttributeError: wxTextCtrl instance has no attribute 'Bind' (self.commande is a wxTextCtrl instance) I don't understand : why did wxTextCtrl lost its Bind attribute ?? As there's not so much changes on my computer since yesterday, I suppose this is due to Boa package installation on my Ubuntu Hoary ? Does Boa installation changes wxpython version ? Is wxTextCtrl attribute .Bind() obsolete ?? Thanks, Nicolas -- http://mail.python.org/mailman/listinfo/python-list
exporting from Tkinter Canvas object in PNG
Hello, I'm new to this mailing list and quite to Pyhon too. I would like to know how to export the contain of the Canvas object (Tkinter) in a PNG file ? Thanks :) Nicolas Pourcelot -- http://mail.python.org/mailman/listinfo/python-list
