How to get double precision value in python equals to java Float.valueof()-CodePudding

I am trying to migrate a java calculation to python, where I see the difference in the float value calculation results

for example: in Java

public class Main
{
    public static void main(String[] args) {
        long lg_val =  Long.parseLong("16191753860");
        float fg_val = Float.valueOf(lg_val);
        System.out.println(fg_val);
    }
}

result - 1.61917542E10 (prints the double value)

but in python float() method returns same value 16191753860.0

How to get the double-precision calculation in python to get the same results

CodePudding user response：

The python float is a double precision floating point number. It is in fact your Java example that is using the single precision float, which is why it formats in scientific notation (as it has gone above integer precision).

If you want to print out the scientific notation of a float in python regardless of if it's crossed over the precision boundary, use the e format.

f'{lg_val:e}' -> '1.619175e 10'

Or to use a double precision number in Java instead, use

double fg_val = Double.valueOf(lg_val);

CodePudding user response：

You can use numpy.float64(). In your example,

import numpy as np 

np.float64(lg_val)