TypeError: 'kwarg' is an invalid keyword argument for this function

[alex23]

On 13/10/2014 8:04 PM, Dave Angel wrote:
> It would also help to spell it the same.  In the OP's
>   implementation,  he defined kwargs, and tried to use it as
>   kwarg.

That's perfectly okay, though: if `kwargs` is the name used to reference 
the dictionary of keyword arguments, `kwarg` would be an instance of a 
keyword argument.