Force python to not output a float in standard form / scientific notation / exponential form

print '{0:.10f}'.format(1.0e-9)

String formatting in the documentation.

Everyone suggesting the use of the f string format code is implicitly assuming that it's okay to fix the number of digits after the decimal point. That seems like a very shaky assumption to me. However, if you don't make that assumption, there is no built-in mechanism to do what you want. This is the best hack I came up with when faced with a similar problem (in a PDF generator -- numbers in PDF can't use exponential notation). You probably want to take all the bs off the strings, and there may be other Python3-isms in here.

_ftod_r = re.compile(
    br'^(-?)([0-9]*)(?:\.([0-9]*))?(?:[eE]([+-][0-9]+))?$')
def ftod(f):
    """Print a floating-point number in the format expected by PDF:
    as short as possible, no exponential notation."""
    s = bytes(str(f), 'ascii')
    m = _ftod_r.match(s)
    if not m:
        raise RuntimeError("unexpected floating point number format: {!a}"
                           .format(s))
    sign = m.group(1)
    intpart = m.group(2)
    fractpart = m.group(3)
    exponent = m.group(4)
    if ((intpart is None or intpart == b'') and
        (fractpart is None or fractpart == b'')):
        raise RuntimeError("unexpected floating point number format: {!a}"
                           .format(s))

    # strip leading and trailing zeros
    if intpart is None: intpart = b''
    else: intpart = intpart.lstrip(b'0')
    if fractpart is None: fractpart = b''
    else: fractpart = fractpart.rstrip(b'0')

    if intpart == b'' and fractpart == b'':
        # zero or negative zero; negative zero is not useful in PDF
        # we can ignore the exponent in this case
        return b'0'

    # convert exponent to a decimal point shift
    elif exponent is not None:
        exponent = int(exponent)
        exponent += len(intpart)
        digits = intpart + fractpart
        if exponent <= 0:
            return sign + b'.' + b'0'*(-exponent) + digits
        elif exponent >= len(digits):
            return sign + digits + b'0'*(exponent - len(digits))
        else:
            return sign + digits[:exponent] + b'.' + digits[exponent:]

    # no exponent, just reassemble the number
    elif fractpart == b'':
        return sign + intpart # no need for trailing dot
    else:
        return sign + intpart + b'.' + fractpart

This is pretty standard print formatting, specifically for a float:

print "%.9f" % 1.0e-9