3
我有一些這樣的:3206.6186522022
相當於Excel的Round函數的
而且我使用了一個Excel工作表中公式爲:
ROUND(3206.6186522022;-2)
這給了我:3200
所以,我怎麼能在c#中實現相同?
我有一些這樣的:3206.6186522022
相當於Excel的Round函數的
而且我使用了一個Excel工作表中公式爲:
ROUND(3206.6186522022;-2)
這給了我:3200
所以,我怎麼能在c#中實現相同?
下面是一個示例LINQPad程序,演示了一種實現方法。我測試了通過Excel中運行的所有這些數字來驗證它的工作方式:
void Main()
{
Verify(ROUND(3206.618652, 2), 3206.62).Dump();
Verify(ROUND(3206.618652, 1), 3206.6).Dump();
Verify(ROUND(3206.618652, 0), 3207).Dump();
Verify(ROUND(3206.618652, -1),).Dump();
Verify(ROUND(3206.618652, -2), 3200).Dump();
Verify(ROUND(3207.618652, 2), 3207.62).Dump();
Verify(ROUND(3207.618652, 1), 3207.6).Dump();
Verify(ROUND(3207.618652, 0), 3208).Dump();
Verify(ROUND(3207.618652, -1),).Dump();
Verify(ROUND(3207.618652, -2), 3200).Dump();
Verify(ROUND(3205.618652, 2), 3205.62).Dump();
Verify(ROUND(3205.618652, 1), 3205.6).Dump();
Verify(ROUND(3205.618652, 0), 3206).Dump();
Verify(ROUND(3205.618652, -1),).Dump();
Verify(ROUND(3205.618652, -2), 3200).Dump();
Verify(ROUND(-3206.618652, 2), -3206.62).Dump();
Verify(ROUND(-3206.618652, 1), -3206.6).Dump();
Verify(ROUND(-3206.618652, 0), -3207).Dump();
Verify(ROUND(-3206.618652, -1), -3210).Dump();
Verify(ROUND(-3206.618652, -2), -3200).Dump();
Verify(ROUND(-3207.618652, 2), -3206.62).Dump();
Verify(ROUND(-3207.618652, 1), -3206.6).Dump();
Verify(ROUND(-3207.618652, 0), -3207).Dump();
Verify(ROUND(-3207.618652, -1), -3210).Dump();
Verify(ROUND(-3207.618652, -2), -3200).Dump();
Verify(ROUND(-3205.618652, 2), -3205.62).Dump();
Verify(ROUND(-3205.618652, 1), -3205.6).Dump();
Verify(ROUND(-3205.618652, 0), -3206).Dump();
Verify(ROUND(-3205.618652, -1), -3210).Dump();
Verify(ROUND(-3205.618652, -2), -3200).Dump();
Verify(ROUND(3205.4, 0), 3204).Dump();
Verify(ROUND(3205.6, 0), 3205).Dump();
Verify(ROUND(-4.4, 0), -4).Dump();
Verify(ROUND(-4.5, 0), -5).Dump();
Verify(ROUND(-4.6, 0), -5).Dump();
Verify(ROUND(4.4, 0), 4).Dump();
Verify(ROUND(4.5, 0), 5).Dump();
Verify(ROUND(4.6, 0), 5).Dump();
}
public static string Verify(double value, double expected)
{
if (Math.Abs(value - expected) < 1e-8)
return string.Empty;
return value + " is not equal (enough) to " + expected;
}
public static double ROUND(double value, int decimals)
{
if (decimals < 0)
{
var factor = Math.Pow(10, -decimals);
return ROUND(value/factor, 0) * factor;
}
return Math.Round(value, decimals, MidpointRounding.AwayFromZero);
}
僅供參考,這段代碼也可以(針對相同的測試數據檢查):
public static double Round(double value, int digits)
{
double pow = Math.Pow(10, digits);
return Math.Truncate(value * pow + Math.Sign(value)*0.5)/pow;
}
它的工作..非常感謝..!! –
@VishalSuthar對於某些值而言這並不完全正確。爲了解決這個問題,把'Math.Round(value,decimals);'改爲'返回Math.Round(value,decimals,MidpointRounding.AwayFromZero);'然後把ROUND(4.5,0)返回5。這將是正確的。 –
我應該指出,差異的出現是因爲默認'Math.Round()'使用「銀行家舍入」(即,圓到最近的偶數),但Excel不會。通過指定'MidpointRounding.AwayFromZero',您可以像Excel一樣使用它。 –