转换为分数英寸
本文关键字：转换 | 更新日期: 2023-09-27 18:01:48
我需要用以下格式将双精度值(厘米)转换为分数值:3 1/64(英寸)。在阅读了大量相关内容并找到了转换为分数的算法后，我认为它们并不适合我的需要，因为我的分数应该是这些格式:?/2，?/4，?/8，?/16，?/32，?/64。我见过这样的换算表:我认为我最好的解决方案是创建一个键值列表，其中包含表中的所有值，并为每个数字找到列表中的最佳近似值。
例如:3.21 cm。= 1.26378英寸= 1英寸+ 0.26378因此，根据链接的表，0.26378 = 17/64。最后的结果应该是1 17/64英寸。
我的问题是:
-
是一个好主意，有一个表中的值列表，并找到最接近的值，以便给出分数，或者更好地为此创建一个算法?
-
如果它是好的创建一个列表的值，我怎么能找到一个给定的数字在我的列表中最接近的值?
我建议用简单的数学来代替表格
private static string ToFraction64(double value) {
// denominator is fixed
int denominator = 64;
// integer part, can be signed: 1, 0, -3,...
int integer = (int) value;
// numerator: always unsigned (the sign belongs to the integer part)
// + 0.5 - rounding, nearest one: 37.9 / 64 -> 38 / 64; 38.01 / 64 -> 38 / 64
int numerator = (int) ((Math.Abs(value) - Math.Abs(integer)) * denominator + 0.5);
// some fractions, e.g. 24 / 64 can be simplified:
// both numerator and denominator can be divided by the same number
// since 64 = 2 ** 6 we can try 2 powers only
// 24/64 -> 12/32 -> 6/16 -> 3/8
// In general case (arbitrary denominator) use gcd (Greatest Common Divisor):
// double factor = gcd(denominator, numerator);
// denominator /= factor;
// numerator /= factor;
while ((numerator % 2 == 0) && (denominator % 2 == 0)) {
numerator /= 2;
denominator /= 2;
}
// The longest part is formatting out
// if we have an actual, not degenerated fraction (not, say, 4 0/1)
if (denominator > 1)
if (integer != 0) // all three: integer + numerator + denominator
return string.Format("{0} {1}/{2}", integer, numerator, denominator);
else if (value < 0) // negative numerator/denominator, e.g. -1/4
return string.Format("-{0}/{1}", numerator, denominator);
else // positive numerator/denominator, e.g. 3/8
return string.Format("{0}/{1}", numerator, denominator);
else
return integer.ToString(); // just an integer value, e.g. 0, -3, 12...
}
const double cmInInch = 2.54;
// 1 17/64
Console.Write(ToFraction64(3.21 / cmInInch));
// -1 17/64
Console.Write(ToFraction64(-1.26378));
// 3 1/4
Console.Write(ToFraction64(3.25001));
// 3 1/4
Console.Write(ToFraction64(3.24997));
// 5
Console.Write(ToFraction64(5.000001));
// -1/8
Console.Write(ToFraction64(-0.129));
// 1/8
Console.Write(ToFraction64(0.129));
以Dmitry Bychenko的代码为例，如果分数等于1，我们需要添加"else if(分母==分子)"，如果值为正，则加1，如果值为负(ex分母/分子= 64/64)，则去掉1
private static string ToFraction64(double value)
{
// denominator is fixed
int denominator = 64;
// integer part, can be signed: 1, 0, -3,...
int integer = (int)value;
// numerator: always unsigned (the sign belongs to the integer part)
// + 0.5 - rounding, nearest one: 37.9 / 64 -> 38 / 64; 38.01 / 64 -> 38 / 64
int numerator = (int)((Math.Abs(value) - Math.Abs(integer)) * denominator + 0.5);
// some fractions, e.g. 24 / 64 can be simplified:
// both numerator and denominator can be divided by the same number
// since 64 = 2 ** 6 we can try 2 powers only
// 24/64 -> 12/32 -> 6/16 -> 3/8
// In general case (arbitrary denominator) use gcd (Greatest Common Divisor):
// double factor = gcd(denominator, numerator);
// denominator /= factor;
// numerator /= factor;
while ((numerator % 2 == 0) && (denominator % 2 == 0))
{
numerator /= 2;
denominator /= 2;
}
// The longest part is formatting out
// if we have an actual, not degenerated fraction (not, say, 4 0/1)
if (denominator > 1)
if (integer != 0) // all three: integer + numerator + denominator
return string.Format("{0} {1}/{2}", integer, numerator, denominator);
else if (value < 0) // negative numerator/denominator, e.g. -1/4
return string.Format("-{0}/{1}", numerator, denominator);
else // positive numerator/denominator, e.g. 3/8
return string.Format("{0}/{1}", numerator, denominator);
//if fraction equals to 1 we add 1 if the value is positive or remove 1 if the value is negative (ex denominator/numerator = 64/64)
else if (denominator == numerator)
{
if (value < 0) // negative numerator/denominator, e.g. -1/4
integer--;
else // positive numerator/denominator, e.g. 3/8
integer++;
return integer.ToString();
}
else
return integer.ToString(); // just an integer value, e.g. 0, -3, 12...
}
增加了脚的显示
public static string ToFraction(this double source, int denominator)
{
var divider = denominator;
var inches = (int) Math.Abs(source);
var numerator = (int) ((Math.Abs(source) - Math.Abs(inches)) * divider + 0.5);
while (numerator % 2 == 0 && divider % 2 == 0)
{
numerator /= 2;
divider /= 2;
}
if (divider == numerator)
{
if (source < 0) inches--;
else inches++;
numerator = 0;
}
var feet = Math.DivRem(inches, 12, out inches);
var valueBuilder = new StringBuilder();
if (source + 1d / denominator < 0) valueBuilder.Insert(0, "-");
if (feet > 0)
{
valueBuilder.Append(feet);
valueBuilder.Append("'");
valueBuilder.Append("-");
}
valueBuilder.Append(inches);
if (numerator != 0)
{
valueBuilder.Append(" ");
valueBuilder.Append(numerator);
valueBuilder.Append("/");
valueBuilder.Append(divider);
}
valueBuilder.Append('"');
return valueBuilder.ToString();
}
所有测试通过
[TestCase]
public void FractionTest()
{
Assert.AreEqual("0'"", 0d.ToFraction());
Assert.AreEqual("0'"", (-0d).ToFraction());
Assert.AreEqual("0'"", (-0.00001d).ToFraction());
Assert.AreEqual("1'"", 1d.ToFraction());
Assert.AreEqual("-1'"", (-1d).ToFraction());
Assert.AreEqual("0 1/8'"", 0.129.ToFraction());
Assert.AreEqual("-0 1/8'"", (-0.129).ToFraction());
Assert.AreEqual("-1 1/4'"", (-1.26378).ToFraction());
Assert.AreEqual("5'"", 5.000001.ToFraction());
Assert.AreEqual("3 1/4'"", 3.24997.ToFraction());
Assert.AreEqual("3 1/4'"", 3.25001.ToFraction());
Assert.AreEqual("1'-0'"", 12d.ToFraction());
Assert.AreEqual("1'-0 3/32'"", 12.1d.ToFraction());
Assert.AreEqual("1'-1'"", 13d.ToFraction());
Assert.AreEqual("1'-3 1/8'"", 15.125d.ToFraction());
Assert.AreEqual("1'-0'"", 12.00001d.ToFraction());
Assert.AreEqual("-1'-0'"", (-12.00001d).ToFraction());
Assert.AreEqual("-2'-1 7/32'"", (-25.231d).ToFraction());
}
我的函数更容易，更简单。它从十进制值返回一个包含三个整数{英寸，分子，分母}的数组，fracBase参数表示精度为16,32,64,128 ....
public static int[] GetImpFractions(decimal value, int fracBase = 32)
{
int[] result = { 0, 0, 0 };
result[0] = (int)Math.Truncate(value);
decimal num = (value - (decimal)result[0]);
num *= fracBase;
decimal denom = fracBase;
if (num > 0)
{
while (num % 2 == 0)
{
num /= 2;
denom /= 2;
}
if (num == 1 && denom == 1)
{
result[0] += 1;
num = 0;
denom = 0;
}
result[1] = (int)Math.Truncate(num);
result[2] = (int)Math.Truncate(denom);
}
return result;
}