Java 你如何生成一个带小数位的随机数
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How do you generate a random number with decimal places
提问by user339108
I would like to generate random numbers between 1 & 10, with a decimal place of 1
(e.g. 1.5, 9.4, 6.3, 2.9)
I would like to generate random numbers between 1 & 10, with 2 decimal places
(e.g. 1.51, 9.46, 6.32, 2.93)
I would like to generate random numbers between 1 & 10, with 3 decimal places
(e.g. 1.512, 9.346, 6.329, 2.935)
One solution would be to have a random generator just for the decimal places and then concatenate with the number. Is there a simpler way to achieve this?
一种解决方案是为小数位设置一个随机生成器，然后与数字连接。有没有更简单的方法来实现这一目标？
采纳答案by Sean Patrick Floyd
Although BigDecimal solutions are more flexible, here's a simpler solution using int and double. This method takes a Randomobject, an upper and lower bound and a number of decimal places and returns an accordingly formatted string:
虽然 BigDecimal 解决方案更灵活，但这里有一个使用 int 和 double 的更简单的解决方案。此方法接受一个Random对象、一个上限和下限以及一些小数位，并返回一个相应格式的字符串：
/**
* Generate a decimal string representation of a random number within the
* supplied bounds.
*
* @param random
* the random object (if null, a new one will be created)
* @param lowerBound
* the lower bound, inclusive
* @param upperBound
* the upper bound, inclusive
* @param decimalPlaces
* the decimal places of the result
* @return the formatted string
*/
public static String getRandomValue(final Random random,
final int lowerBound,
final int upperBound,
final int decimalPlaces){
if(lowerBound < 0 || upperBound <= lowerBound || decimalPlaces < 0){
throw new IllegalArgumentException("Put error message here");
}
final double dbl =
((random == null ? new Random() : random).nextDouble() //
* (upperBound - lowerBound))
+ lowerBound;
return String.format("%." + decimalPlaces + "f", dbl);
}
Test Code:
测试代码：
final Random rnd = new Random();
for(int decpl = 0; decpl < 3; decpl++){
for(int low = 0; low < 2; low++){
for(int high = low + 1; high < low + 3; high++){
System.out.println("Random Value between " + low + " and "
+ high + " with " + decpl + " decimal places:");
System.out.println(getRandomValue(rnd, low, high, decpl));
}
}
}
Output:
输出：
Random Value between 0 and 1 with 0 decimal places:
0
Random Value between 0 and 2 with 0 decimal places:
1
Random Value between 1 and 2 with 0 decimal places:
1
Random Value between 1 and 3 with 0 decimal places:
3
Random Value between 0 and 1 with 1 decimal places:
0.6
Random Value between 0 and 2 with 1 decimal places:
1.5
Random Value between 1 and 2 with 1 decimal places:
1.1
Random Value between 1 and 3 with 1 decimal places:
1.9
Random Value between 0 and 1 with 2 decimal places:
0.52
Random Value between 0 and 2 with 2 decimal places:
0.82
Random Value between 1 and 2 with 2 decimal places:
1.75
Random Value between 1 and 3 with 2 decimal places:
1.10
回答by Jigar Joshi
Generate a double and than format according to your decimal need
根据您的十进制需要生成双精度和比格式
回答by Jon Skeet
Well one obvious suggestion is to generate a single integer and divide it appropriately. So for 3DPs, you'd generate a number between 1000 and 9999 (or 10000, depending on the exact bound you wanted) and then divide it by 10000.
一个明显的建议是生成一个整数并适当地划分它。因此，对于 3DP，您将生成 1000 到 9999（或 10000，取决于您想要的确切界限）之间的数字，然后将其除以 10000。
I suggest you use BigDecimalin the actual division, to maintain the decimals exactly.
我建议您BigDecimal在实际除法中使用，以精确保持小数。
回答by christian
1)(int) ((Math.random() * 90) + 10) / 10.0
2)(int) ((Math.random() * 900) + 100) / 100.0
3)(int) ((Math.random() * 9000) + 1000) / 1000.0
1) (int) ((Math.random() * 90) + 10) / 10.0
2) (int) ((Math.random() * 900) + 100) / 100.0
3)(int) ((Math.random() * 9000) + 1000) / 1000.0
