Increasing floating point precision in Python

问题

I was working on a project to compute the Leibniz approximation for pi with the below code:

def pi(precision):
    sign = True
    ret = 0
    for i in range(1,precision+1):
        odd = 2 * i - 1
        if sign:
            ret += 1.0 / odd
        else:
            ret -= 1.0 / odd
        sign = not sign
    return ret

However, the output value was always was 12 digits long. How can I increase the precision (e.g. more digits) of the calculation? Does Python support more precise floating points, or will I have to use some external library?

回答1:

Try using Decimal.

Read Arbitrary-precision elementary mathematical functions (Python)^original for more information

回答2:

Python's float type maps to whatever your platform's C compiler calls a double (see http://en.wikipedia.org/wiki/IEEE_floating_point_number).

The Python standard library also comes with an arbitrary-precision decimal module, called decimal: http://docs.python.org/2/library/decimal.html

回答3:

With Python's float, you get 15–17 digits of precision (if you are seeing fewer, you may need to use a different format specifier when printing).

If you need more, you'll need to use a different method (one that only uses integer arithmetic), or a different way to represent floating-point numbers.

See Python floating point arbitrary precision available?

回答4:

The Leibniz formula converges extremely slowly - honestly, you won't live long enough for it get 12 digits of accuracy. Click here for one way to accelerate it enormously.

来源：https://stackoverflow.com/questions/21079282/increasing-floating-point-precision-in-python