How To Check If A Key in **kwargs Exists?

问题:

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?

回答1:

You want

if 'errormessage' in kwargs:
    print("found it")

To get the value of errormessage

if 'errormessage' in kwargs:
    print("errormessage equals " + kwargs.get("errormessage"))

In this way, kwargs is just another dict. Your first example, if kwargs['errormessage'], means "get the value associated with the key "errormessage" in kwargs, and then check its bool value". So if there's no such key, you'll get a KeyError.

Your second example, if errormessage in kwargs:, means "if kwargs contains the element named by "errormessage", and unless "errormessage" is the name of a variable, you'll get a NameError.

I should mention that dictionaries also have a method .get() which accepts a default parameter (itself defaulting to None), so that kwargs.get("errormessage") returns the value if that key exists and None otherwise (similarly kwargs.get("errormessage", 17) does what you might think it does). When you don't care about the difference between the key existing and having None as a value or the key not existing, this can be handy.

回答2:

DSM's and Tadeck's answers answer your question directly.

In my scripts I often use the convenient dict.pop() to deal with optional, and additional arguments. Here's an example of a simple print() wrapper:

def my_print(*args, **kwargs):
    prefix = kwargs.pop('prefix', '')
    print(prefix, *args, **kwargs)

Then:

>>> my_print('eggs')
 eggs
>>> my_print('eggs', prefix='spam')
spam eggs

As you can see, if prefix is not contained in kwargs, then the default '' (empty string) is being stored in the local prefix variable. If it is given, then its value is being used.

This is generally a compact and readable recipe for writing wrappers for any kind of function: Always just pass-through arguments you don't understand, and don't even know if they exist. If you always pass through *args and **kwargs you make your code slower, and requires a bit more typing, but if interfaces of the called function (in this case print) changes, you don't need to change your code. This approach reduces development time while supporting all interface changes.

回答3:

It is just this:

if 'errormessage' in kwargs:
    print("yeah it's here")

You need to check, if the key is in the dictionary. The syntax for that is some_key in some_dict (where some_key is something hashable, not necessarily a string).

The ideas you have linked (these ideas) contained examples for checking if specific key existed in dictionaries returned by locals() and globals(). Your example is similar, because you are checking existence of specific key in kwargs dictionary (the dictionary containing keyword arguments).

回答4:

One way is to add it yourself! How? By merging kwargs with a bunch of defaults. This won't be appropriate on all occasions, for example if the keys are not know to you in advance. However if they are, here is a simple example:

import sys

def myfunc(**kwargs):
    args = {'country':'England','town':'London',
            'currency':'Pound', 'language':'English'}

    diff = set(kwargs.keys()) - set(args.keys())
    if diff:
        print("Invalid args:",tuple(diff),file=sys.stderr)
        return

    args.update(kwargs)            
    print(args)

The defaults are set in the dictionary args, which includes all the keys we are expecting. We first check to see if there are any unexpected keys in kwargs. Then we update args with kwargs which will overwrite any new values that the user has set. We don't need to test if a key exists, we now use args as our argument dictionary and have no further need of kwargs.

回答5:

You can discover those things easily by yourself:

def hello(*args, **kwargs):
    print kwargs
    print type(kwargs)
    print dir(kwargs)

hello(what="world")