kwargs

Python 3.2.3. There were some ideas listed here , which work on regular var's, but it seems **kwargs play by different rules... so why doesn't this work and how can I check to see if a key in **kwargs exists? if kwargs['errormessage']: print("It exists") I also think this should work, but it doesn't -- if errormessage in kwargs: print("yeah it's here") I'm guessing because kwargs is iterable? Do I have to iterate through it just to check if a particular key is there? DSM You want if 'errormessage' in kwargs: print("found it") To get the value of errormessage if 'errormessage' in kwargs: print(

I have a python function with this signature: def post_message(self, message, *args, **kwargs): I would like to call the function from c++ and pass to it some kwargs. Calling the function is not the problem. Knowing how to pass the kwargs is. Here is a non-working paraphrased sample: std::string message("aMessage"); boost::python::list arguments; arguments.append("1"); boost::python::dict options; options["source"] = "cpp"; boost::python::object python_func = get_python_func_of_wrapped_object() python_func(message, arguments, options) When I exercise this code, in pdb I get (which is not what

Can I pass a list of kwargs to a method for brevity? This is what i'm attempting to do: def method(**kwargs): #do something keywords = (keyword1 = 'foo', keyword2 = 'bar') method(keywords) Yes. You do it like this: def method(**kwargs): print kwargs keywords = {'keyword1': 'foo', 'keyword2': 'bar'} method(keyword1='foo', keyword2='bar') method(**keywords) Running this in Python confirms these produce identical results: {'keyword2': 'bar', 'keyword1': 'foo'} {'keyword2': 'bar', 'keyword1': 'foo'} As others have pointed out, you can do what you want by passing a dict. There are various ways to

In python, I wrote this function to teach myself how **kwargs works in Python: def fxn(a1, **kwargs): print a1 for k in kwargs: print k, " : ", kwargs[k] I then called this function with fxn(3, a2=2, a3=3, a4=4) Here was the output that my Python interpreter printed: 3 a3 : 3 a2 : 2 a4 : 4 Why did the for loop print the value of a3 before that of a2 even though I fed a2 into my function first? kwargs is a dictionary. Dictionaries are unordered - simply put, the order is unspecified and an implementation detail. Peeking under the hood will show that the order varies wildly depending on the hash

I come from a background in static languages. Can someone explain (ideally through example) the real world advantages of using **kwargs over named arguments ? To me it only seems to make the function call more ambiguous. Thanks. Real-world examples: Decorators - they're usually generic, so you can't specify the arguments upfront: def decorator(old): def new(*args, **kwargs): # ... return old(*args, **kwargs) return new Places where you want to do magic with an unknown number of keyword arguments. Django's ORM does that, e.g.: Model.objects.filter(foo__lt = 4, bar__iexact = 'bar') Alex Martelli

In python, I can define a function as follows: def func(kw1=None,kw2=None,**kwargs): ... In this case, i can call func as: func(kw1=3,kw2=4,who_knows_if_this_will_be_used=7,more_kwargs=Ellipsis) I can also define a function as: def func(arg1,arg2,*args): ... which can be called as func(3,4,additional,arguments,go,here,Ellipsis) Finally, I can combine the two forms def func(arg1,arg2,*args,**kwargs): ... But, what does not work is calling: func(arg1,arg2,*args,kw1=None,kw2=None,**kwargs): #SYNTAX ERROR (in python 2 only, apparently this works in python 3) ... My original thought was that this