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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 results(self, times='all', t_precision=12, **kwargs):
 r""" Fetches the calculated quantity from the algorithm and returns it as an array. Parameters times : scalar or list Time steps to be returned. The default value is 'all' which results in returning all time steps. If a scalar is given, only the corresponding time step is returned. If a range is given (e.g., 'range(0, 1, 1e-3)'), time steps in this range are returned. t_precision : integer The time precision (number of decimal places). Default value is 12. Notes ----- The keyword steps is interpreted in the same way as times. """

        if 'steps' in kwargs.keys():
            times = kwargs['steps']
        t_pre = t_precision
        quantity = self.settings['quantity']
        q = [k for k in list(self.keys()) if quantity in k]
        if times == 'all':
            t = q
        elif type(times) in [float, int]:
            n = int(-dc(str(round(times, t_pre))).as_tuple().exponent *
                    (round(times, t_pre) != int(times)))
            t_str = (str(int(round(times, t_pre)*10**n)) +
                     ('e-'+str(n))*(n != 0))
            t = [k for k in q if t_str == k.split('@')[-1]]
        elif 'range' in times:
            t = times.replace(' ', '')
            t = t[6:-1]
            t = t.split(',')
            out = np.arange(float(t[0]), float(t[1]), float(t[2]))
            out = np.append(out, float(t[1]))
            out = np.unique(out)
            out = np.around(out, decimals=t_pre)
            t = []
            for i in out:
                n = int(-dc(str(round(i, t_pre))).as_tuple().exponent *
                        (round(i, t_pre) != int(i)))
                j = (str(int(round(i, t_pre)*10**n))+('e-'+str(n))*(n != 0))
                t_str = [k for k in q if j == k.split('@')[-1]]
                t += (t_str)
        d = {k: self[k] for k in t}
        return d