[bug] utils.py 的 make_underlying_symbol有个小错误

[IcyCC]

def make_underlying_symbol(id_or_symbol):
    id_or_symbol = bytes2str(id_or_symbol)
    if six.PY2:
        return filter(lambda x: x not in '0123456789 ', id_or_symbol).upper()
    else:
        return ''.join(list(filter(lambda x: x not in '0123456789 ', 'rb1705'))).upper()

python3的所有的make_underlying_symbol 都会返回RB

[Cuizi7]

pr merged