Question

假设我有一个在已知最小值和最大值之间有10个值的刻度。如何在最小值和最大值之间获得最接近的值。例如：

min = 0, max = 10, value = 2.75 -> expected: value = 3
min = 5, max = 6, value = 5.12 -> expected: value = 5.1
min = 0, max = 1, value = 0.06 -> expected: value = 0.1

Answer 1

你可以使用这样的东西

function nearest(value, min, max, steps){
    var zerone = Math.round((value-min)*steps/(max-min))/steps; // bring to 0-1 range
    return zerone*(max-min) + min;
}

console.log(nearest(2.75, 0, 10, 10)); // 3
console.log(nearest(5.12, 5, 6, 10)); // 5.1
console.log(nearest(0.06, 0, 1, 10)); // 0.1

http://jsfiddle.net/gaby/4RN37/1/

演示

Answer 2

你的情景对我来说没什么意义。为什么.06轮到1而不是.1但是5.12轮到5.1并且具有相同的比例（1整数）？这令人困惑。

无论哪种方式，如果你想要舍入精确的小数位数，请检查出来： http://www.javascriptkit.com/javatutors/round.shtml

var original=28.453

1) //round "original" to two decimals
var result=Math.round(original*100)/100  //returns 28.45

2) // round "original" to 1 decimal
var result=Math.round(original*10)/10  //returns 28.5

3) //round 8.111111 to 3 decimals
var result=Math.round(8.111111*1000)/1000  //returns 8.111

通过本教程，您应该能够完全按照自己的意愿行事。

Answer 3

也许更容易理解：

var numberOfSteps = 10;
var step = (max - min) / numberOfSteps;
var difference = start - min;
var stepsToDifference = Math.round(difference / step);
var answer = min + step * stepsToDifference;

这也允许您更改序列中的步骤数。

Answer 4

我建议这样的事情：

var step = (max - min) / 10;
return Math.round(value / step) * step;

Answer 5

例如，我遇到的问题是我得到5.7999997而不是废弃的5.8。这是我的第一个修复程序（针对Java ...）。

public static float nearest(float val, float min, float max, int steps) {
    float step = (max - min) / steps;
    float diff = val - min;
    float steps_to_diff = round(diff / step);
    float answer = min + step * steps_to_diff;
    answer = ((int) (answer * steps)) / (float) steps;
    return answer;
}

但是在nearest(6.5098, 0, 10, 1000)上使用它，我会得到6.509而不是想要的6.51。这为我解决了问题（当值真的很大时要注意是否有溢出）：

public static float nearest(float val, float min, float max, int steps) {
    val *= steps;
    min *= steps;
    max *= steps;
    float step = (max - min) / steps;
    float diff = val - min;
    float steps_to_diff = round(diff / step);
    float answer = min + step * steps_to_diff;
    return answer / (float) steps;
}

Answer 6

var step = 10;
return Math.ceil(x / step) * step;

JavaScript：在比例上舍入到最接近的值

6 个答案: