def roundf(x, precision=<NUM_LIT:0>):
if precision > <NUM_LIT:0>:<EOL><INDENT>return float("<STR_LIT>".format(x, precision - <NUM_LIT:1>))<EOL><DEDENT>return x<EOL>
Like round but works with large exponents in floats and high precision
    Based on http://stackoverflow.com/a/6539677/623735
    >>> 234042163./(2**24)
    13.94999998807...
    >>> roundf(234042163./(2**24), 5)
    13.95
    >>> roundf(1234.1234e-123, 5)
    1.2341e-120
    >>> roundf(1234.1234e-123, 3)
    1.23e-120
    >>> roundf(1234123.1234, 5)
    1234100.0
    >>> roundf(1234123.1234, 3)
    1230000.0