Error in calling a function

Question

The definition of a function is the following:

 def testf(t,t1 = 0,*x):
... print t
... print t1
... print x[0]
... return sin(x[0]*t*t1)
... 

I call function as testf(pi,t1=1,1). The error I receive is

SyntaxError: non-keyword arg after keyword arg

Is there anything wrong with function call? I received the same error if I call the function testf(t1=1,t=pi,1) I will not receive error message if I call the function testf(pi,1,1). I have another question about calling this function. It is possible to call the function with default value of t1 and x, which is not None, as a tuple? Thanks a lot for your help.

Answer 1

In your function,

• t is a positional argument

• t1 is a keyword argument, with a default value (also a positional argument at position 1)

• x is a collection of all the other positional arguments

As the error message says, any Python function should have all the positional variables before any keyword arguments. You have to invoke the function like this

testf(math.pi, t1=1)

When you pass a keyword argument, that should be the last argument to the function (or only keyword arguments can follow that).

When you say

testf(math.pi, 1, t1 = 2)

Both 1 and 2 are to be assigned to t1 (t1 is also a positional variable at index 1), which is not possible. So, error will be thrown in this case also.

def testf(t, t1 = 0, *x):
 print t, x, t1
testf(5)
# 5 () 0 -> `t` is 5, `t1` takes the default value, `x` is empty
testf(5, 10, 15, 20)
# 5 (15, 20) 10 -> `t` is 5, `t1` is 10, `x` is 15 and 20
testf(5, t1 = 20)
# 5 () 20 -> `t` is 5, `t1` is 20, `x` is empty
testf(5, 10, t1 = 20)
# TypeError: testf() got multiple values for keyword argument 't1'
testf(5, t1 = 20, 10)
# SyntaxError: non-keyword arg after keyword arg

• stardust: I tried to define the function as you suggested. I got SyntaxError: invalid syntax. Thanks
• thefourtheye: @stardust The actual problem is with the way you called the function, please check the update.
• user2357112: Keyword-only arguments only work in Python 3. The new version of the answer also doesn't work, in Python 2 or 3. Both the second positional argument and the keyword argument are treated as supplying values for t1.
• stardust: I tried all the orders of arguments on my computer. None of them works.
• stardust: Could you please explain to me why the last two calls failed? Thank you.

Answer 2

Positional arguments must correspond to the first n parameters of a function, where n is the number of positional arguments provided. You can't mix a keyword argument in with them; keyword arguments must go after all positional arguments.

In theory, the language could have been designed otherwise, but it would be confusing what the positional arguments would mean under such circumstances, and there is no benefit to motivate the added complexity. For example, it's unclear whether in

def f(a=1, b=1, c=1): pass
f(c=1, 2)

the 2 is supposed to provide the value of a or b.

If you want to call your function, provide t1 positionally:

testf(pi, 1, 1)

Note that the presence or absence of a default value for a parameter, such as t1, has no effect on whether you can provide it as a keyword argument or positional argument.

Answer 3

You should provide non-keyword arguments before a keyword argument

testf(pi,1,t1=1)

some example with args and keyword args:

def testf(t,*x,**keywords):
 t1=keywords['t1'] if 't1' in keywords else 0
 print t,x,keywords,t1
testf(1,5,t1=1)
testf(1,2,3,4)
testf(1,5)
testf(1)

will output

1 (5,) {'t1': 1} 1 <---- t1 = 1
1 (2, 3, 4) {} 0 <---- t1 = 0
1 (5,) {} 0
1 () {} 0