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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 ic_pos(self, row1, row2=None):
 """Calculate the information content of one position. Returns ------- score : float Information content. """

        if row2 is None:
            row2 = [0.25,0.25,0.25,0.25]

        score = 0
        for a,b in zip(row1, row2):
            if a > 0:
                score += a * log(a / b) / log(2)
        return score