how to insert new variable in (*args,**kwargs) section?

Question

i want to make a decoration method to assign the variable which the function would use but wouldn't be deliver by itself.

for example add new variable y in lambda r,i wrote code in this way but didn't work.

r = lambda x:x+y

def foo(func):
    def wrapped(*args,**kwargs):
        y = 3
        return func(y=y,*args,**kwargs)
    return wrapped

r = foo(r)
print(r(444))

this wouldn't work too

r = lambda x:x+y

def foo(func):
    def wrapped(*args,**kwargs):
        y = 3
        return func(*args,**kwargs)
    return wrapped

r = foo(r)
print(r(444))

`r` can accept just one argument, you're passing too many args. — Ashwini Chaudhary, Oct 29 '13 at 11:05

score 6 · Answer 1 · edited May 23 '17 at 12:10

6

kwargs is a casual python dict type, so you can just set the value of the y key to be 3

r = lambda x, y=0:x+y

def foo(func):
    def wrapped(*args,**kwargs):
        print type(kwargs) #  will output <type 'dict'>
        kwargs['y'] = 3
        return func(*args,**kwargs)
    return wrapped

In Understanding kwargs in Python this is explained in details.

edited May 23 '17 at 12:10

Community

1
1

answered Oct 29 '13 at 11:03

Alexander Zhukov

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This is not going to work and will raise : `TypeError: () got an unexpected keyword argument 'y'` – Ashwini Chaudhary Oct 29 '13 at 11:10
r takes only two arguments where as you can pass variable number of positional and keyword arguments to the argument of `func`. – ajay Oct 29 '13 at 11:25

score 3 · Answer 2 · answered Oct 29 '13 at 11:09

Problem is that the function r can accept only one argument, you need to change its definition to accept more args:

r = lambda x, y=0, *args, **kwargs: x + y

def foo(func):
    def wrapped(*args,**kwargs):
        y = 3
        return func(y=y, *args,**kwargs)
    return wrapped

r = foo(r)
print(r(444))
#447

how to insert new variable in (*args,**kwargs) section?

2 Answers2