TypeError: 'kwarg' is an invalid keyword argument for this function
alex23
wuwei23 at gmail.com
Tue Oct 14 01:55:11 EDT 2014
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.
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