有没有一种更快的方法将十六进制小数部分转换为十进制?

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英文:

Is there a faster way to convert a hexadecimal fractional part to a decimal one?

问题

我编写了一个能生成圆周率十六进制数字的程序。在某些关键值点,我想将我拥有的十六进制值转换为十进制,并将其保存到文件中。目前我在使用BigDecimal进行数学计算,代码如下:

private static String toDecimal(String hex) {
    String rawHex = hex.replace(".", "");
    BigDecimal base = new BigDecimal(new BigInteger(rawHex, 16));
    BigDecimal factor = new BigDecimal(BigInteger.valueOf(16).pow(rawHex.length() - 1));
    BigDecimal value = base.divide(factor);
    return value.toPlainString().substring(0, hex.length());
}

请注意,这个方法仅适用于十六进制值的整数部分只有一位的情况,包括圆周率在内。请勿将其用于通用情况。

虽然这段代码运行正常,但是对于最新的关键值,即250万位数字,转换过程耗时11.3小时。

有没有更快的方法可以手动完成这个转换?

我尝试过将第一位小数除以16,第二位除以16^2,依此类推,但这很快会变得难以处理。也许有一些方法可以将数字位移到后面,从而保持较低的除数?但是可能需要处理n+1、n+2、n+3等数字,以获得正确的第n位的值。

英文:

I wrote a program that generates digits of pi in hexadecimal. Every so often, at benchmark values, I would like to convert the hexadecimal value I have into a decimal one and save it to a file. Currently I am using BigDecimal to do that math with this code:

private static String toDecimal(String hex) {
    String rawHex = hex.replace(".", "");
    BigDecimal base = new BigDecimal(new BigInteger(rawHex, 16));
    BigDecimal factor = new BigDecimal(BigInteger.valueOf(16).pow(rawHex.length() - 1));
    BigDecimal value = base.divide(factor);
    return value.toPlainString().substring(0, hex.length());
}

Note that this method will only work for hexadecimal values with one digit in the integral part, pi included, do not copy and paste this for general use.

So this code works fine, but for the latest benchmark, 2.5m digits, the conversion took 11.3 hours to complete.

Is there any faster way to do this manually?

I tried dividing the first decimal place by 16, the second by 16^2, etc but this would quickly get out of hand. Maybe some way of bitshifting the digits back to keep the divisor low? But potentially the n+1, n+2, n+3, etc digits need to be processed to get the correct value for n.

答案1

得分: 4

首先,我认为你的函数 toDecimal 是错误的,因为它不能正确地转换输入 ".1a"(它与 16 的倍数相差),例如,对于输入 ".800" 会抛出异常。第三行应该是:

BigDecimal factor = new BigDecimal(BigInteger.valueOf(16).pow(rawHex.length()));

异常出现在:

return value.toPlainString().substring(0, hex.length());

转换后的值可能比输入值短,导致 java.lang.StringIndexOutOfBoundsException 异常。

接下来:

实际上,我没有将这个方法与你当前的方法进行基准测试;我只是提供这个作为“思考的食物”。在这里,我按照孩子在学校学习的方式进行了乘法操作,而在你的情况下,我们需要一个大循环来生成一个数字。但如果你能以某种方式将它改成使用 BigDecimal(目前还不清楚怎么做),那可能比你当前的方法更快(真正需要的是一个 BigHexadecimal 类)。

可以观察到,将一个基数的分数转换为另一个基数可以通过乘法来实现。在这种情况下,我们有以下十六进制分数(可以忽略整数部分,当转换 pi 时,它是 3):

<b>.h<sub>1</sub>h<sub>2</sub>h<sub>3</sub>h<sub>4</sub> ... h<sub>n</sub></b>

其中 <b>h<sub>n</sub></b> 是第 n 个十六进制 "nibble"。

我们希望将上述转换为以下的十进制分数:

<b>.d<sub>1</sub>d<sub>2</sub>d<sub>3</sub>d<sub>4</sub> ... d<sub>n</sub></b>

其中 <b>d<sub>n</sub></b> 是第 n 个十进制数字。

如果我们将这两个数量都乘以 10,我们将得到:

<b>h<sup>'</sup><sub>1</sub>.h<sup>'</sup><sub>2</sub>h<sup>'</sup><sub>3</sub>h<sup>'</sup><sub>4</sub> ... h<sup>'</sup><sub>n</sub></b>

<b>d<sub>1</sub>.d<sub>2</sub>d<sub>3</sub>d<sub>4</sub> ... d<sub>n</sub></b>

乘以 10 只是将十进制分数向左移动一位。

我们必须注意,小数点左边的数量必须相等,即 <b>d<sub>1</sub></b> == <b>h<sup>'</sup><sub>1</sub></b>。因此,我们反复将我们的十六进制分数乘以 10,每次都将整数部分作为下一个十进制数字进行转换。我们重复这个过程,直到新的十六进制分数变为 0 或产生了一些任意数量的十进制数字:

查看 Java 示例

class Test {

    private static String toDecimal(String hex, int numberDigits) {
        /* converts a string such as "13.1a" in base 16 to "19.1015625" in base 10 */
        int index = hex.indexOf('.');
        assert index != -1;
        StringBuilder decimal = new StringBuilder((index == 0) ? "" : String.valueOf(Integer.parseInt(hex.substring(0, index), 16)));
        decimal.append('.');
        int l = hex.length() - index - 1;
        assert l >= 1;
        int firstIndex = index + 1;
        int hexDigits[] = new int[l];
        for (int i = 0; i < l; i++) {
            hexDigits[i] = Integer.parseInt(hex.substring(i + firstIndex, i + firstIndex + 1), 16);
        }
        while (numberDigits != 0 && l != 0) {
            int carry = 0;
            boolean allZeroes = true;
            for (int i = l - 1; i >= 0; i--) {
                int value = hexDigits[i] * 10 + carry;
                if (value == 0 && allZeroes) {
                    l = i;
                }
                else {
                    allZeroes = false;
                    carry = (int)(value / 16);
                    hexDigits[i] = value % 16;
                }
            }
            numberDigits--;
            if (carry != 0 || (numberDigits != 0 && l != 0))
                decimal.append("0123456789".charAt(carry));
        }
        return decimal.toString();
    }

    public static void main(String[] args) {
        System.out.println(toDecimal("13.1a", 15));
        System.out.println(toDecimal("13.8", 15));
        System.out.println(toDecimal("13.1234", 15));
    }

}

打印结果:

19.1015625
19.5
19.07110595703125
英文:

First, I believe your function toDecimal is wrong as it doesn't correctly convert input &quot;.1a&quot; (it's off by a factor of 16), for example, and throws an exception for input &quot;.800&quot;. The third line should be:

BigDecimal factor = new BigDecimal(BigInteger.valueOf(16).pow(rawHex.length()));

The exception arises from:

return value.toPlainString().substring(0, hex.length());

The converted value could be shorter than the input value and you get a java.lang.StringIndexOutOfBoundsException.

Moving on:

In truth I have not benchmarked this against your current method; I just offer this as "food for thought." Here I am doing the multiplications as a child is taught to do it in school and in your case we have a large loop just to produce one digit. But if you could somehow adapt this to use BigDecimal (it's not clear how you would) it might be quicker than your current approach (what's really needed is a BigHexadecimal class).

It can be observed that converting a fraction from one base to another can be done using multiplications. In this case we have the following hexadecimal fraction (we can ignore the integer portion, which is 3 when converting pi):

<b>.h<sub>1</sub>h<sub>2</sub>h<sub>3</sub>h<sub>4</sub> ... h<sub>n</sub></b>

where <b>h<sub>n</sub></b> is the n<sup>th</sup> hexadecimal "nibble".

We wish to convert the above to the following decimal fraction:

<b>.d<sub>1</sub>d<sub>2</sub>d<sub>3</sub>d<sub>4</sub> ... d<sub>n</sub></b>

where <b>d<sub>n</sub></b> is the n<sup>th</sup> decimal digit.

If we were to multiply both quantities by 10, we would get:

<b>h<sup>'</sup><sub>1</sub>.h<sup>'</sup><sub>2</sub>h<sup>'</sup><sub>3</sub>h<sup>'</sup><sub>4</sub> ... h<sup>'</sup><sub>n</sub></b>
<br>
The primes (<sup>`</sup>) denote that we have completely new hexadecimal nibble values following the multiplication.

and

<b>d<sub>1</sub>.d<sub>2</sub>d<sub>3</sub>d<sub>4</sub> ... d<sub>n</sub></b>
<br>
The multiplication by 10 just shifts the decimal fraction one place to the left.

We must note that the quantities to the left of the decimal point must be equal, i.e. <b>d<sub>1</sub></b> == <b>h<sup>'</sup><sub>1</sub></b>. Thus we repeatedly multiply our hexadecimal fraction by 10 and each time we do we take the integer portion as the next decimal digit for our conversion. We repeat this until either our new hexadecimal fraction becomes 0 or some arbitrary number of decimal digits have been produced:

See Java Demo

class Test {
private static String toDecimal(String hex, int numberDigits) {
/* converts a string such as &quot;13.1a&quot; in base 16 to &quot;19.1015625&quot; in base 10 */
int index = hex.indexOf(&#39;.&#39;);
assert index != -1;
StringBuilder decimal = new StringBuilder((index == 0) ? &quot;&quot; : String.valueOf(Integer.parseInt(hex.substring(0, index), 16)));
decimal.append(&#39;.&#39;);
int l = hex.length() - index - 1;
assert l &gt;= 1;
int firstIndex = index + 1;
int hexDigits[] = new int[l];
for (int i = 0; i &lt; l; i++) {
hexDigits[i] = Integer.parseInt(hex.substring(i + firstIndex, i + firstIndex + 1), 16);
}
while (numberDigits != 0 &amp;&amp; l != 0) {
int carry = 0;
boolean allZeroes = true;
for (int i = l - 1; i &gt;= 0; i--) {
int value = hexDigits[i] * 10 + carry;
if (value == 0 &amp;&amp; allZeroes) {
l = i;
}
else {
allZeroes = false;
carry = (int)(value / 16);
hexDigits[i] = value % 16;
}
}
numberDigits--;
if (carry != 0 || (numberDigits != 0 &amp;&amp; l != 0))
decimal.append(&quot;0123456789&quot;.charAt(carry));
}
return decimal.toString();
}
public static void main(String[] args) {
System.out.println(toDecimal(&quot;13.1a&quot;, 15));
System.out.println(toDecimal(&quot;13.8&quot;, 15));
System.out.println(toDecimal(&quot;13.1234&quot;, 15));
}
}

Prints:

19.1015625
19.5
19.07110595703125

答案2

得分: 0

感谢@Booboo提供了解决这个问题的方法。我稍微改进了他的代码,以便在所有情况下都能工作。我想在这里发布它,供将来的访问者参考。

/**
 * 将十六进制数字字符串转换为十进制数字字符串。
 *
 * @param hex      十六进制数字字符串。
 * @param accuracy 要在十进制数字字符串中返回的小数位数。
 * @return 十进制数字字符串。
 */
public static String hexToDecimal(String hex, int accuracy) {
    if (!hex.matches("[0-9A-Fa-f.\\-]+") || (accuracy < 0)) {
        return "";
    }
    
    boolean negative = hex.startsWith("-");
    hex = hex.replaceAll("^-", "");
    String integral = hex.contains(".") ? hex.substring(0, hex.indexOf(".")) : hex;
    String fraction = hex.contains(".") ? hex.substring(hex.indexOf(".") + 1) : "";
    if (integral.contains("-") || fraction.contains(".") || fraction.contains("-")) {
        return "";
    }
    
    StringBuilder decimal = new StringBuilder();
    decimal.append(negative ? "-" : "");
    decimal.append(integral.isEmpty() ? "0" : new BigDecimal(new BigInteger(integral, 16)).toPlainString());
    if (fraction.isEmpty() || (accuracy == 0)) {
        return decimal.toString();
    }
    decimal.append(".");
    
    int numberDigits = accuracy;
    int length = Math.min(fraction.length(), numberDigits);
    int[] hexDigits = new int[numberDigits];
    Arrays.fill(hexDigits, 0);
    IntStream.range(0, length).boxed().parallel().forEach(i -> hexDigits[i] = Integer.parseInt(String.valueOf(fraction.charAt(i)), 16));
    
    while ((numberDigits != 0)) {
        int carry = 0;
        for (int i = length - 1; i >= 0; i--) {
            int value = hexDigits[i] * 10 + carry;
            carry = value / 16;
            hexDigits[i] = value % 16;
        }
        decimal.append(carry);
        numberDigits--;
    }
    return decimal.toString();
}

/**
 * 将十六进制数字字符串转换为十进制数字字符串。
 *
 * @param hex 十六进制数字字符串。
 * @return 十进制数字字符串。
 * @see #hexToDecimal(String, int)
 */
public static String hexToDecimal(String hex) {
    String fraction = hex.contains(".") ? hex.substring(hex.indexOf(".") + 1) : "";
    return hexToDecimal(hex, fraction.length());
}

public static void main(String[] args) {
    // 整数部分
    Assert.assertEquals("0", hexToDecimal("0"));
    // ... 其他测试 ...
}

请注意,由于我只返回翻译好的部分,因此只提供了代码的一部分示例。如果您需要完整的代码,请将缺失的部分添加到您的项目中。

英文:

Thank you to @Booboo for the solution to this problem. I improved on his code a little so it should work in every case. I wanted to post it here for future visitors.

/**
* Converts a hex number string to a decimal number string.
*
* @param hex      The hex number string.
* @param accuracy The number of decimal places to return in the decimal number string.
* @return The decimal number string.
*/
public static String hexToDecimal(String hex, int accuracy) {
if (!hex.matches(&quot;[0-9A-Fa-f.\\-]+&quot;) || (accuracy &lt; 0)) {
return &quot;&quot;;
}
boolean negative = hex.startsWith(&quot;-&quot;);
hex = hex.replaceAll(&quot;^-&quot;, &quot;&quot;);
String integral = hex.contains(&quot;.&quot;) ? hex.substring(0, hex.indexOf(&quot;.&quot;)) : hex;
String fraction = hex.contains(&quot;.&quot;) ? hex.substring(hex.indexOf(&quot;.&quot;) + 1) : &quot;&quot;;
if (integral.contains(&quot;-&quot;) || fraction.contains(&quot;.&quot;) || fraction.contains(&quot;-&quot;)) {
return &quot;&quot;;
}
StringBuilder decimal = new StringBuilder();
decimal.append(negative ? &quot;-&quot; : &quot;&quot;);
decimal.append(integral.isEmpty() ? &quot;0&quot; : new BigDecimal(new BigInteger(integral, 16)).toPlainString());
if (fraction.isEmpty() || (accuracy == 0)) {
return decimal.toString();
}
decimal.append(&quot;.&quot;);
int numberDigits = accuracy;
int length = Math.min(fraction.length(), numberDigits);
int[] hexDigits = new int[numberDigits];
Arrays.fill(hexDigits, 0);
IntStream.range(0, length).boxed().parallel().forEach(i -&gt; hexDigits[i] = Integer.parseInt(String.valueOf(fraction.charAt(i)), 16));
while ((numberDigits != 0)) {
int carry = 0;
for (int i = length - 1; i &gt;= 0; i--) {
int value = hexDigits[i] * 10 + carry;
carry = value / 16;
hexDigits[i] = value % 16;
}
decimal.append(carry);
numberDigits--;
}
return decimal.toString();
}
/**
* Converts a hex number string to a decimal number string.
*
* @param hex The hex number string.
* @return The decimal number string.
* @see #hexToDecimal(String, int)
*/
public static String hexToDecimal(String hex) {
String fraction = hex.contains(&quot;.&quot;) ? hex.substring(hex.indexOf(&quot;.&quot;) + 1) : &quot;&quot;;
return hexToDecimal(hex, fraction.length());
}
public static void main(String[] args) {
//integer
Assert.assertEquals(&quot;0&quot;, hexToDecimal(&quot;0&quot;));
Assert.assertEquals(&quot;1&quot;, hexToDecimal(&quot;1&quot;));
Assert.assertEquals(&quot;9&quot;, hexToDecimal(&quot;9&quot;));
Assert.assertEquals(&quot;15&quot;, hexToDecimal(&quot;F&quot;));
Assert.assertEquals(&quot;242&quot;, hexToDecimal(&quot;F2&quot;));
Assert.assertEquals(&quot;33190&quot;, hexToDecimal(&quot;81A6&quot;));
Assert.assertEquals(&quot;256&quot;, hexToDecimal(&quot;100&quot;));
Assert.assertEquals(&quot;1048576&quot;, hexToDecimal(&quot;100000&quot;));
Assert.assertEquals(&quot;5191557193152165532727847676938654&quot;, hexToDecimal(&quot;FFF6AA0322BC458D5D11A632099E&quot;));
Assert.assertEquals(&quot;282886881332428154466487121231991859970997056152877088222&quot;, hexToDecimal(&quot;B897A12C89896321C454A7DD9E150233CBB87A9F0233DDE&quot;));
Assert.assertEquals(&quot;-256&quot;, hexToDecimal(&quot;-100&quot;));
Assert.assertEquals(&quot;-144147542&quot;, hexToDecimal(&quot;-8978456&quot;));
Assert.assertEquals(&quot;-332651442596728389665499138728075237402&quot;, hexToDecimal(&quot;-FA42566214321CC67445D58EE874981A&quot;));
Assert.assertEquals(&quot;33190&quot;, hexToDecimal(&quot;81a6&quot;));
//decimal
Assert.assertEquals(&quot;0.10&quot;, hexToDecimal(&quot;0.1a&quot;));
Assert.assertEquals(&quot;0.5&quot;, hexToDecimal(&quot;0.8&quot;));
Assert.assertEquals(&quot;0.0711&quot;, hexToDecimal(&quot;0.1234&quot;));
Assert.assertEquals(&quot;0.528966901&quot;, hexToDecimal(&quot;0.876A5FF4A&quot;));
Assert.assertEquals(&quot;-0.528966901&quot;, hexToDecimal(&quot;-0.876A5FF4A&quot;));
Assert.assertEquals(&quot;-0.00000000&quot;, hexToDecimal(&quot;-0.00000001&quot;));
Assert.assertEquals(&quot;-0.62067648792835838863907521427468&quot;, hexToDecimal(&quot;-0.9EE4A7810C666FF7453D06A44621030E&quot;));
Assert.assertEquals(&quot;0.528966901&quot;, hexToDecimal(&quot;0.876a5ff4a&quot;));
Assert.assertEquals(&quot;0.528966901&quot;, hexToDecimal(&quot;.876a5ff4a&quot;));
Assert.assertEquals(&quot;-0.528966901&quot;, hexToDecimal(&quot;-.876a5ff4a&quot;));
//combined
Assert.assertEquals(&quot;15.33693&quot;, hexToDecimal(&quot;F.56412&quot;));
Assert.assertEquals(&quot;17220744.33934412&quot;, hexToDecimal(&quot;106C488.56DF41A2&quot;));
Assert.assertEquals(&quot;282886881332428154466487121231991859970997056152877088222.62067648792835838863907521427468&quot;, hexToDecimal(&quot;B897A12C89896321C454A7DD9E150233CBB87A9F0233DDE.9EE4A7810C666FF7453D06A44621030E&quot;));
Assert.assertEquals(&quot;-17220744.33934412&quot;, hexToDecimal(&quot;-106C488.56DF41A2&quot;));
Assert.assertEquals(&quot;-282886881332428154466487121231991859970997056152877088222.62067648792835838863907521427468&quot;, hexToDecimal(&quot;-B897A12C89896321C454A7DD9E150233CBB87A9F0233DDE.9EE4A7810C666FF7453D06A44621030E&quot;));
Assert.assertEquals(&quot;-17220744.33934412&quot;, hexToDecimal(&quot;-106c488.56df41a2&quot;));
Assert.assertEquals(&quot;3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808&quot;, hexToDecimal(&quot;3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01377BE5466CF34E90C6CC0AC29B7C97&quot;));
//accuracy
Assert.assertEquals(&quot;-0.00000&quot;, hexToDecimal(&quot;-0.00000001&quot;, 5));
Assert.assertEquals(&quot;-0.000000000232830&quot;, hexToDecimal(&quot;-0.00000001&quot;, 15));
Assert.assertEquals(&quot;-0&quot;, hexToDecimal(&quot;-0.00000001&quot;, 0));
Assert.assertEquals(&quot;282886881332428154466487121231991859970997056152877088222.5&quot;, hexToDecimal(&quot;B897A12C89896321C454A7DD9E150233CBB87A9F0233DDE.9EE4A7810C666FF7453D06A44621030E&quot;, 1));
Assert.assertEquals(&quot;3.14&quot;, hexToDecimal(&quot;3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01377BE5466CF34E90C6CC0AC29B7C97&quot;, 2));
Assert.assertEquals(&quot;3.1415926535&quot;, hexToDecimal(&quot;3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01377BE5466CF34E90C6CC0AC29B7C97&quot;, 10));
Assert.assertEquals(&quot;3.1415926535897932384626433&quot;, hexToDecimal(&quot;3.243F6A8885A308D313198A2E03707344A4093822299F31D0082EFA98EC4E6C89452821E638D01377BE5466CF34E90C6CC0AC29B7C97&quot;, 25));
//invalid
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;0.00000.001&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;0.00000-001&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;156-081.00000001&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;hello&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;9g&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;9G&quot;));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;546.FDA&quot;, -1));
Assert.assertEquals(&quot;&quot;, hexToDecimal(&quot;546.FDA&quot;, -999));
}

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  • 本文由 发表于 2020年8月18日 00:15:12
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