I have a class that has an output() method which returns a matplotlib Figure instance. I have a decorator I wrote that takes that fig instance and turns it into a Django response object.

My decorator looks like this:

class plot_svg(object):
    def __init__(self, view):
        self.view = view

    def __call__(self, *args, **kwargs):
        print args, kwargs
        fig = self.view(*args, **kwargs)
        canvas=FigureCanvas(fig)
        response=HttpResponse(content_type='image/svg+xml')
        canvas.print_svg(response)
        return response

and this is how it was being used:

def as_avg(self):
    return plot_svg(self.output)()

The only reason I has it that way instead of using the "@" syntax is because when I do it with the "@":

@plot_svg
def as_svg(self):
    return self.output()

I get this error:

as_svg() takes exactly 1 argument (0 given)

I'm trying to 'fix' this by putting it in the "@" syntax but I can't figure out how to get it working. I'm thinking it has something to do with self not getting passed where it's supposed to...