如何舍入到最接近的0.5？

Question

我必须显示评级,为此我需要增加如下:

如果数字是1.0则应该等于1
如果数字是1.1应该等于1
如果数字是1.2应该等于1
如果数字是1.3应该等于1.5
如果数字是1.4应该等于1.5 如果数字是1.4应该等于1.5
如果数字是1.5应该等于1.5
如果数字是1.6应该等于1.5
如果数字是1.7应该等于1.5
如果数字是1.8应该等于2.0
如果数字是1.8应该等于2.0 如果数字是1.9应该等于2.0
如果数字是2.0应该等于2.0
如果数字是2.1应该等于2.0
等等......

有没有一种简单的方法来计算所需的值？

Answer 1

将您的评分乘以2,然后使用舍入Math.Round(rating, MidpointRounding.AwayFromZero),然后将该值除以2.

Math.Round(value * 2, MidpointRounding.AwayFromZero) / 2

Answer 2

乘以2,舍入,然后除以2

如果你想要最近的四分之一,乘以4,除以4,等等

Answer 3

以下是我编写的几种方法,它们总是向上或向下舍入到任何值.

public static Double RoundUpToNearest(Double passednumber, Double roundto)
{
    // 105.5 up to nearest 1 = 106
    // 105.5 up to nearest 10 = 110
    // 105.5 up to nearest 7 = 112
    // 105.5 up to nearest 100 = 200
    // 105.5 up to nearest 0.2 = 105.6
    // 105.5 up to nearest 0.3 = 105.6

    //if no rounto then just pass original number back
    if (roundto == 0)
    {
        return passednumber;
    }
    else
    {
        return Math.Ceiling(passednumber / roundto) * roundto;
    }
}

public static Double RoundDownToNearest(Double passednumber, Double roundto)
{
    // 105.5 down to nearest 1 = 105
    // 105.5 down to nearest 10 = 100
    // 105.5 down to nearest 7 = 105
    // 105.5 down to nearest 100 = 100
    // 105.5 down to nearest 0.2 = 105.4
    // 105.5 down to nearest 0.3 = 105.3

    //if no rounto then just pass original number back
    if (roundto == 0)
    {
        return passednumber;
    }
    else
    {
        return Math.Floor(passednumber / roundto) * roundto;
    }
}