How to send a dictionary to a function that accepts **kwargs?
I have a function that accepts wildcard keyword p开发者_Python百科arameters:
def func(**kargs):
doA
doB
How do I send it a dictionary?
Just use func(**some_dict)
to call it.
This is documented on section 4.7.4 of python tutorial.
Note that the same dict
is not passed into the function. A new copy is created, so some_dict is not kwargs
.
It's not 100% clear by your question, but if you'd like to pass a dict
in through kwargs
, you just make that dict a part of another dict, like so:
my_dict = {} #the dict you want to pass to func
kwargs = {'my_dict': my_dict } #the keyword argument container
func(**kwargs) #calling the function
Then you can catch my_dict
in the function:
def func(**kwargs):
my_dict = kwargs.get('my_dict')
or...
def func(my_dict, **kwargs):
#reference my_dict directly from here
my_dict['new_key'] = 1234
I use the latter a lot when I have the same set of options passed to different functions, but some functions only use some the options (I hope that makes sense...). But there are of course a million ways to go about this. If you elaborate a bit on your problem we could most likely help you better.
func(**mydict)
this will mean kwargs=mydict inside the function
all the keys of mydict must be strings
for python 3.6 just put ** before the dictionary name
def lol(**kwargs):
for i in kwargs:
print(i)
my_dict = {
"s": 1,
"a": 2,
"l": 3
}
lol(**my_dict)
Easy way to pass parameters with a variable, args, and kwargs with Decorator
def printall(func):
def inner(z,*args, **kwargs):
print ('Arguments for args: {}'.format(args))
print ('Arguments for kwargs: {}'.format(kwargs))
return func(*args, **kwargs)
return inner
@printall #<-- you can mark Decorator,it will become to send some variable data to function
def random_func(z,*y,**x):
print(y)
print(x)
return z
z=1
y=(1,2,3)
x={'a':2,'b':2,'c':2}
a = random_func(z,*y,**x)
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