def keep_most(terms_d, terms_amout):
    """ keep_most(): is getting a SHORTED dictionary of Terms-Frequencies and
        the amount of Terms to return as arguments. It is returning the number
        of Terms equal to the argument 'terms_amount' with the Highest Frequency. """
    terms_l = [(v, k) for (k, v) in terms_d.iteritems()]
    terms_l.sort()
    terms_l.reverse()

    most_terms_l = terms_l[0: terms_amout]
    terms_d = dict([(k, v) for (v, k) in most_terms_l])

    return terms_d