In java I have:
Random random = new Random();
double randomNum = random.nextDouble();
which creates a random number between 0 and 1. However I want a number between -1000 and 1000, how would I scale this?
Thanks
2 possibilities:
Note that possibility 2 ensures better randomness and better 'density' of your numbers, at the cost of 2 random calls [which might be expansive, if it is an issue].
嗯,数学?
double randomNum = (random.nextDouble()-0.5d) * 2000;
Random random = new Random();
double randomNum = (random.nextDouble() * 2000.0d) - 1000.0d;
public static double randomInterval(double minValue,double maxValue){
Random random = new Random();
double r;
do {
r = random.nextDouble();
} while (r < minValue || r >= maxValue);
return r;
}
double a = randomInterval(-1000,1000) ;
Try this algorithm:
这会为您提供该范围内的数字
double randomNum = (random.nextDouble() * 2000) -1000;
Random random = new Random();
int randomNum = random.nextInt(2000) - 1000;
Here is a general function you could use to linearly rescale a number between zero and one ( val01
) to a different range ( min
.. max
):
public static double rescale(double val01, double min, double max) {
return val01 * (max - min) + min;
}
public static double doubleBetween(double start, double end) {
Random random = new Random();
// We need 64 bits because double have 53 bits precision, so int is too short
// We have now a value between 0 and Long.MAX_VALUE.
long value = -1L;
while (value < 0)
value = Math.abs(random.nextLong()); // Caution, Long.MIN_VALUE returns negative !
// Cast to double
double valueAsDouble = (double) value;
// Scale so that Long.MAX_VALUE is exactly 1 !
double diff = (end-start)/(double) Long.MAX_VALUE;
return start + valueAsDouble*diff;
}
This will give the correct interval including both ends with full double precision. doubles have a special -0.0 value (the negative zero) which will not be given by this routine.
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