In the following code, how do I pass the dictionary to func2. How should func2 be called?
def func2(a,**c):
if len(c) > 0:
print len(c)
print c
u={'a':1,'b':2}
func2(1,u)
3 Answers
Just as it accepts them:
func2(1,**u)
3 Comments
Adrien Plisson
additionally:
def func2(a,c): ... and func2(1,u) would also work. using variable arguments or standard arguments depends on what you are trying to achieve.
Rajeev
func2(1,**u) i get an error saying Typeerror:func2() got multiple values for keyword argument 'a'
Rajeev
I got the above error because dictionary had the key 'a' and one of the parameters was also 'a'. when i changed it the error got resolved..
This won't run, because there are multiple parameters for the name a.
But if you change it to:
def func2(x,**c):
if len(c) > 0:
print len(c)
print c
Then you call it as:
func2(1, a=1, b=2)
or
u={'a':1,'b':2}
func2(1, **u)
1 Comment
Rajeev
ya true i realized and have updated in of the comment.so parameters names and the dict key are not suppose to be the same.Thanks..
This might help you:
def fun(*a, **kw):
print a, kw
a=[1,2,3]
b=dict(a=1, b=2, c=3)
fun(*a)
fun(**kw)
fun(*a, **kw)
forloop still won't work. Get rid of theand 'a' in cand it would, not sure what that clause is trying to do?