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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def kernel(self):
 """ The 'kernel' is an attempt to get at just the most pithy words in the name """

        stop_words = [ y.lower() for y in self.abbreviations.values() + self.filler_words ]
        kernel = ' '.join([ x for x in self.expand().split() if x.lower() not in stop_words ])

        # this is a hack to get around the fact that this is the only two-word phrase we want to block
        # amongst our stop words. if we end up with more, we may need a better way to do this
        kernel = re.sub(r'\s*United States', '', kernel)

        return kernel