Re: Checking if a variable is a dictionary
Guillermo a écrit : Mamma mia! My head just exploded. I've seen the light. So you only need to ·want· to have a protocol? That's amazing... Far beyond the claim that Python is easy. You define protocols in writing basically! Even my grandma could have her own Python protocol. Okay, so I think I know where's the catch now -- you must rely on the fact that the protocol is implemented, Indeed. But even with static declarative typing à la Java, you must rely on the fact that the protocol is *correctly* implemented (what about having file.close sending mails to all your contacts and then rebooting the computer ?-). IOW, at some point, you have to trust the coder anyway. there's no way to enforce it if you're expecting a parrot-like object. Yes there is: You'd try to call the speak() method That's it. and deal with the error if there's no such method? Either catch the AttributeError and raise a TypeError or ValueError (or some more specific one) instead, or just let the AttributeError propagate - whichever makes more sens given the context. I'd say the default would be to catch the AttributeError and raise a TypeError instead (that's at least what len() does when passed an unsized object). -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
if type(a) is dict: print a is a dictionnary! class MyDict(dict): pass a = MyDict() type(a) is dict = False isinstance(a, dict) = True So the question you need to answer is whether you want to determine whether an object is exactly of type dict, or whether it you are willing to accept types derived from dict also. -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On Mon, 10 Mar 2008 14:29:48 +, Andrew Koenig wrote: So the question you need to answer is whether you want to determine whether an object is exactly of type dict, or whether it you are willing to accept types derived from dict also. Or other mappings that don't inherit from dict but behave just like dicts, such as UserDict. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
A protocol is just an interface that an object agrees to implement. In your case, you would state that every object stored in your special dict must implement the to_tagged_value method with certain agreeable semantics. Hm... I've searched about the implementation of protocols and now (I believe) I know how to implement the iterable protocol, for instance, but have no clue about how to define my own... I'm surely not thinking the right way, but I don't seem to be able to wrap my head around the implementation details of custom protocols... Is it just a matter of extending the different object classes (dict, list, tuple...)? Where do you put the interface/protocol code? :-? Regards, Guillermo -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On Sun, 09 Mar 2008 05:20:41 -0700, Guillermo wrote: A protocol is just an interface that an object agrees to implement. In your case, you would state that every object stored in your special dict must implement the to_tagged_value method with certain agreeable semantics. Hm... I've searched about the implementation of protocols and now (I believe) I know how to implement the iterable protocol, for instance, but have no clue about how to define my own... I'm surely not thinking the right way, but I don't seem to be able to wrap my head around the implementation details of custom protocols... Is it just a matter of extending the different object classes (dict, list, tuple...)? Where do you put the interface/protocol code? :-? A protocol is a convention that you insist other objects must follow, not code. To implement that convention, naturally you need to write code, but the convention makes the protocol, not the code. Some examples: To be considered a writable file, your object must include a write() method and a close() method. It doesn't matter what those methods do exactly, so long as they behave reasonably. For instance, the close() method might do nothing, and the write() method might send an SMS. To be a random-access writable file, it must also include a seek() method. To be a sequence, your object must obey the sequence protocol, which says it has a __getitem__ method which takes integer arguments starting at 0 and increasing, until it raises IndexError. You don't have to write any code to create a protocol, you just have to tell people what it is. Here's my parrot protocol: To be a parrot, your object must have a method speak() which takes an integer argument and returns the case-insensitive string spam repeated that many times. I'm done. I have a protocol. But if I want to actually use it, then I need an object that obeys it, and if I can't wait for somebody else to write it, I need to write one myself. So here it is: class Viking(object): def __init__(self, case='upper'): if case == 'upper': self.case = str.upper else: self.case = str.lower def speak(self, n): return self.case(spam*n) And now Viking instances are also parrots, or rather, they obey the parrot protocol: v = Viking() v.speak(3) 'SPAMSPAMSPAM' I hope this helps. -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Mamma mia! My head just exploded. I've seen the light. So you only need to ·want· to have a protocol? That's amazing... Far beyond the claim that Python is easy. You define protocols in writing basically! Even my grandma could have her own Python protocol. Okay, so I think I know where's the catch now -- you must rely on the fact that the protocol is implemented, there's no way to enforce it if you're expecting a parrot-like object. You'd try to call the speak() method and deal with the error if there's no such method? Thanks a lot! Guillermo -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On Sun, 09 Mar 2008 06:58:15 -0700, Guillermo wrote: Okay, so I think I know where's the catch now -- you must rely on the fact that the protocol is implemented, there's no way to enforce it if you're expecting a parrot-like object. You'd try to call the speak() method and deal with the error if there's no such method? That's right. That's called duck typing -- if all you want is something that quacks like a duck, then it doesn't matter if it actually is a duck or not. Or if you prefer: if it quacks like a duck and swims like a duck, then it's close enough to a duck as to make no difference. Sometimes though, you need to check for a parrot up front. So I'd so this: try: something.speak except AttributeError: # No speak() method, so it can't be a parrot. do_something_else() else: # It seems to follow the parrot protocol. yummy_goodness = something.speak(5) assert spam in yummy_goodness.lower() -- Steven -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On Mar 9, 9:23 am, Steven D'Aprano [EMAIL PROTECTED] cybersource.com.au wrote: On Sun, 09 Mar 2008 06:58:15 -0700, Guillermo wrote: Okay, so I think I know where's the catch now -- you must rely on the fact that the protocol is implemented, there's no way to enforce it if you're expecting a parrot-like object. You'd try to call the speak() method and deal with the error if there's no such method? That's right. That's called duck typing -- if all you want is something that quacks like a duck, then it doesn't matter if it actually is a duck or not. Or if you prefer: if it quacks like a duck and swims like a duck, then it's close enough to a duck as to make no difference. The syntax checker over here raised a warning on this one. So close to a duck as. (I'm using the so...as construct in my Propositional Calculus program.) try: something.speak except AttributeError: # No speak() method, so it can't be a parrot. do_something_else() It might be a dead parrot. HAR! else: # It seems to follow the parrot protocol. yummy_goodness = something.speak(5) assert spam in yummy_goodness.lower() P.S. Is 'resemblant' a word? So resemblant of a duck as. -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On 9 mar, 11:23, Steven D'Aprano [EMAIL PROTECTED] cybersource.com.au wrote: On Sun, 09 Mar 2008 06:58:15 -0700, Guillermo wrote: Okay, so I think I know where's the catch now -- you must rely on the fact that the protocol is implemented, there's no way to enforce it if you're expecting a parrot-like object. You'd try to call the speak() method and deal with the error if there's no such method? That's right. That's called duck typing -- if all you want is something that quacks like a duck, then it doesn't matter if it actually is a duck or not. In addition to duck typing, in some cases an explicit declaration may be useful: I implement the Parrot protocol. The zope.interface package does that: there is a way to define interfaces (a set of methods any implementation must provide), and classes can assert I implement this interface, and one can determine whether a class implements or not certain interface. See http://pypi.python.org/pypi/zope.interface Abstract Base Classes (ABC, for Python 3.0) provide a similar concept (but they're not the same thing). See http://www.python.org/dev/peps/pep-3119/ Of course there is some overlapping between all of these techniques - one should choose the best method in each case. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Guillermo a écrit : Hello, This is my first post here. I'm getting my feet wet with Python and I need to know how can I check whether a variable is of type dictionary. What makes you say you need to know this ? Except for a couple corner cases, you usually don't need to care about this. If you told us more about the actual problem (instead of asking about what you think is the solution), we might be of more help... Something like this: if isdict(a) then print a is a dictionary Should this work for subclasses of dicts ? Should this work for non-subclasses of dict implementing the dict protocol ? -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Sam a écrit : Hello if type(a) is dict: print a is a dictionnary! class MyDict(dict): pass a = MyDict() type(a) is dict = False -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Hello if type(a) is dict: print a is a dictionnary! ++ Sam -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
En Thu, 06 Mar 2008 10:10:47 -0200, Guillermo [EMAIL PROTECTED] escribi�: This is my first post here. I'm getting my feet wet with Python and I need to know how can I check whether a variable is of type dictionary. Something like this: if isdict(a) then print a is a dictionary if isinstance(a, dict): ... But note that it's more common in Python to try to use it in the intended way, and catch the possible errors. Look for duck typing in this group, and the difference between easier to ask forgiveness than permission and look before you leap. -- Gabriel Genellina -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Wow, I think I'm gonna like this forum. Thank you all for the prompt
answers!
What makes you say you need to know this ? Except for a couple corner
cases, you usually don't need to care about this. If you told us more
about the actual problem (instead of asking about what you think is the
solution), we might be of more help...
Good point.
I want to iterate recursively a dictionary whose elements might be
strings or nested tuples or dictionaries and then convert values to a
tagged format according to some rules.
d = {'a':i'm a, 'b':(1,2,3),'c':{'a':i'm a,'x':something,'y':
('a','b','c')}}
I'm just designing the algorithm, but I think Python dictionaries can
hold any kind of sequence?
Regards,
Guillermo
--
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Re: Checking if a variable is a dictionary
On Mar 6, 7:10 am, Guillermo [EMAIL PROTECTED] wrote:
Hello,
This is my first post here. I'm getting my feet wet with Python and I
need to know how can I check whether a variable is of type dictionary.
Something like this:
if isdict(a) then print a is a dictionary
Regards,
Guillermo
Or
if type(a)==type({}):
print 'a is a dictionary'
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Re: Checking if a variable is a dictionary
Guillermo wrote:
I'm just designing the algorithm, but I think Python dictionaries
can hold any kind of sequence?
(Watch out, dicts are no sequence types.)
I recommend relying duck typing as long as it's feasible. I. e. if
it can be subscripted like a dict, it is a dict. If this makes
problems, use the type({}) approach.
Regards,
Björn
--
BOFH excuse #56:
Electricians made popcorn in the power supply
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Re: Checking if a variable is a dictionary
On Thu, Mar 6, 2008 at 2:06 PM, Guillermo [EMAIL PROTECTED] wrote: What makes you say you need to know this ? Except for a couple corner cases, you usually don't need to care about this. If you told us more about the actual problem (instead of asking about what you think is the solution), we might be of more help... I want to iterate recursively a dictionary whose elements might be strings or nested tuples or dictionaries and then convert values to a tagged format according to some rules. just do so, if it's not a dict you can always catch the exception, then handle tuples (or the exception), then strings which should work always (of course also error handling here according to your needs) I'm just designing the algorithm, but I think Python dictionaries can hold any kind of sequence? python dicts can hold any kind of object (as a value). What you can't do is have a list as the key, but that is easily circumvented by using a tuple (or any other immutable type). hth martin -- http://tumblr.marcher.name https://twitter.com/MartinMarcher http://www.xing.com/profile/Martin_Marcher http://www.linkedin.com/in/martinmarcher You are not free to read this message, by doing so, you have violated my licence and are required to urinate publicly. Thank you. -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
Guillermo a écrit :
Wow, I think I'm gonna like this forum. Thank you all for the prompt
answers!
Welcome onboard !-)
What makes you say you need to know this ? Except for a couple corner
cases, you usually don't need to care about this. If you told us more
about the actual problem (instead of asking about what you think is the
solution), we might be of more help...
Good point.
I want to iterate recursively a dictionary whose elements might be
strings or nested tuples or dictionaries and then convert values to a
tagged format according to some rules.
If you're absolutely definitively 101% sure that you'll never have
anything else in your dict - IOW : this dict is an internal part of your
module, produced by the module and consumed by the module -, then it
*might* be one of the corner cases where testing type (either directly
or - preferably - via isinstance) is the right thing to do.
d = {'a':i'm a, 'b':(1,2,3),'c':{'a':i'm a,'x':something,'y':
('a','b','c')}}
I'm just designing the algorithm, but I think Python dictionaries can
hold any kind of sequence?
Any Python object (which include classes, modules, functions etc) can be
used as value.
--
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Re: Checking if a variable is a dictionary
Jeffrey Seifried a écrit :
(snip)
if type(a)==type({}):
print 'a is a dictionary'
This instanciates a dict, call type() on it, and discard the dict -
which is useless since the dict type is a builtin. Also, when you want
to test identity, use an identity test.
if type(a) is dict:
print blah blah blah
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Re: Checking if a variable is a dictionary
On Thu, Mar 6, 2008 at 8:06 AM, Guillermo
[EMAIL PROTECTED] wrote:
I want to iterate recursively a dictionary whose elements might be
strings or nested tuples or dictionaries and then convert values to a
tagged format according to some rules.
d = {'a':i'm a, 'b':(1,2,3),'c':{'a':i'm a,'x':something,'y':
('a','b','c')}}
This could be solved with dynamic polymorphism instead of
introspection, which might simplify things depending on how your
dictionary is constructed.
class Value(object):
def to_tagged_format(self):
raise NotImplementedError
class StringValue(Value):
def to_tagged_format(self):
...
class Tuplevalue(Value):
def to_tagged_format(self):
...
class DictValue(Value):
def to_tagged_format(self):
...
for k, v in d.iteritems():
d[k] = v.to_tagged_format()
You can also get the dynamic polymorphism without invoking inheritance
by specifying a protocol that the values in your dict must implement,
instead. Protocols are plentiful in Python, perhaps more popular than
type hierarchies.
--
Neil Cerutti [EMAIL PROTECTED]
--
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Re: Checking if a variable is a dictionary
You can also get the dynamic polymorphism without invoking inheritance by specifying a protocol that the values in your dict must implement, instead. Protocols are plentiful in Python, perhaps more popular than type hierarchies. I'm used to languages with stricter rules than Python. I've read a bit about Python protocols, but where could I get some more info on implementation details? Sounds very interesting. Regards, Guillermo -- http://mail.python.org/mailman/listinfo/python-list
Re: Checking if a variable is a dictionary
On Thu, Mar 6, 2008 at 12:17 PM, Guillermo [EMAIL PROTECTED] wrote: You can also get the dynamic polymorphism without invoking inheritance by specifying a protocol that the values in your dict must implement, instead. Protocols are plentiful in Python, perhaps more popular than type hierarchies. I'm used to languages with stricter rules than Python. I've read a bit about Python protocols, but where could I get some more info on implementation details? Sounds very interesting. A protocol is just an interface that an object agrees to implement. In your case, you would state that every object stored in your special dict must implement the to_tagged_value method with certain agreeable semantics. Python implements sequence types, mapping types, iterators, file streams, and other things with protocols. For example, iterators must support the next and and __iter__ methods. Using a protocol instead instead of a type hierarchy requires less boring boilerplate, and I suppose in Python will usuallly be preferable except when you want to inherit implementation as well as interface. In that case, inheritance often saves boilerplate cide rather than increases it. It's also misnomered as duck-typing (clearly it should be nomed quack-typing). -- Neil Cerutti [EMAIL PROTECTED] -- http://mail.python.org/mailman/listinfo/python-list
