How To Check If A Key in **kwargs Exists?
Question:
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?
Answers:
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.
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).
You can discover those things easily by yourself:
def hello(*args, **kwargs):
print kwargs
print type(kwargs)
print dir(kwargs)
hello(what="world")
One way is to add it by 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 known 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.
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.
if kwarg.__len__() != 0:
print(kwarg)
solution on checking if any value Exists in kwargs :
def abc(*args ,**kwargs):
print(int(kwargs.__len__())>0) #comment this if no need
return True if kwargs.__len__() != 0 else False
abc() # return False
abc(k = "ww") #return True
abc('a argument')
[if no *args is passed abc('a argument') will raise :- TypeError: abc() takes exactly 0 arguments]
#return False
<adapting answer @YeongHwa Jin> #it will end in Error invalid type conversion.and no reputation to add as comment
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?
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.
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).
You can discover those things easily by yourself:
def hello(*args, **kwargs):
print kwargs
print type(kwargs)
print dir(kwargs)
hello(what="world")
One way is to add it by 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 known 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.
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.
if kwarg.__len__() != 0:
print(kwarg)
solution on checking if any value Exists in kwargs :
def abc(*args ,**kwargs):
print(int(kwargs.__len__())>0) #comment this if no need
return True if kwargs.__len__() != 0 else False
abc() # return False
abc(k = "ww") #return True
abc('a argument')
[if no *args is passed abc('a argument') will raise :- TypeError: abc() takes exactly 0 arguments]
#return False
<adapting answer @YeongHwa Jin> #it will end in Error invalid type conversion.and no reputation to add as comment