64位double可以精确地表示整数+/- 253。

鉴于这一事实,我选择使用双类型作为我所有类型的单一类型,因为我的最大整数是一个无符号的32位数字。

但现在我必须打印这些伪整数,但问题是它们也和实际的双精度数混合在一起。

那么如何在Java中很好地打印这些double呢?

我试过String。format("%f", value),这很接近,除了我得到了很多小值的末尾零。

下面是%f的输出示例

232.00000000
0.18000000000
1237875192.0
4.5800000000
0.00000000
1.23450000

我想要的是:

232
0.18
1237875192
4.58
0
1.2345

当然,我可以写一个函数来修剪这些零,但由于字符串操作,这是大量的性能损失。我能用其他格式的代码做得更好吗?


Tom E.和Jeremy S.的答案是不可接受的,因为他们都任意舍入到小数点后两位。请先理解问题再回答。


请注意字符串。Format (Format, args…)依赖于语言环境(见下面的答案)。


当前回答

这是另一个答案,它有一个选项,只有当小数不为零时才附加小数。

   /**
     * Example: (isDecimalRequired = true)
     * d = 12345
     * returns 12,345.00
     *
     * d = 12345.12345
     * returns 12,345.12
     *
     * ==================================================
     * Example: (isDecimalRequired = false)
     * d = 12345
     * returns 12,345 (notice that there's no decimal since it's zero)
     *
     * d = 12345.12345
     * returns 12,345.12
     *
     * @param d float to format
     * @param zeroCount number decimal places
     * @param isDecimalRequired true if it will put decimal even zero,
     * false will remove the last decimal(s) if zero.
     */
    fun formatDecimal(d: Float? = 0f, zeroCount: Int, isDecimalRequired: Boolean = true): String {
        val zeros = StringBuilder()

        for (i in 0 until zeroCount) {
            zeros.append("0")
        }

        var pattern = "#,##0"

        if (zeros.isNotEmpty()) {
            pattern += ".$zeros"
        }

        val numberFormat = DecimalFormat(pattern)

        var formattedNumber = if (d != null) numberFormat.format(d) else "0"

        if (!isDecimalRequired) {
            for (i in formattedNumber.length downTo formattedNumber.length - zeroCount) {
                val number = formattedNumber[i - 1]

                if (number == '0' || number == '.') {
                    formattedNumber = formattedNumber.substring(0, formattedNumber.length - 1)
                } else {
                    break
                }
            }
        }

        return formattedNumber
    }

其他回答

如果这个想法是将存储为双精度的整数打印出来,就像它们是整数一样,否则以最小的必要精度打印双精度:

public static String fmt(double d)
{
    if(d == (long) d)
        return String.format("%d",(long)d);
    else
        return String.format("%s",d);
}

生产:

232
0.18
1237875192
4.58
0
1.2345

并且不依赖于字符串操作。

0.0 -> 0% 1.0 -> 100% 0.1 -> 10% 0.11 -> 11% 0.01 -> 1% 0.111 -> 11.1% 0.001 -> 0.1% 0.1111 -> 11.11% 0.0001 -> 0.01%

添加".replace()"是因为我总是使用错误的分隔符

import java.text.NumberFormat

fun Double.formating(): String {
    val defaultFormat: NumberFormat = NumberFormat.getPercentInstance()
    defaultFormat.minimumFractionDigits = 0
    defaultFormat.maximumFractionDigits = 2

    return defaultFormat.format(this).replace(",", ".")
}

我在JSF应用程序中使用它来格式化数字,而不带后面的零。最初的内置格式化程序要求您指定小数位数的最大数量,如果您有太多小数位数,这在这里也很有用。

/**
 * Formats the given Number as with as many fractional digits as precision
 * available.<br>
 * This is a convenient method in case all fractional digits shall be
 * rendered and no custom format / pattern needs to be provided.<br>
 * <br>
 * This serves as a workaround for {@link NumberFormat#getNumberInstance()}
 * which by default only renders up to three fractional digits.
 *
 * @param number
 * @param locale
 * @param groupingUsed <code>true</code> if grouping shall be used
 *
 * @return
 */
public static String formatNumberFraction(final Number number, final Locale locale, final boolean groupingUsed)
{
    if (number == null)
        return null;

    final BigDecimal bDNumber = MathUtils.getBigDecimal(number);

    final NumberFormat numberFormat = NumberFormat.getNumberInstance(locale);
    numberFormat.setMaximumFractionDigits(Math.max(0, bDNumber.scale()));
    numberFormat.setGroupingUsed(groupingUsed);

    // Convert back for locale percent formatter
    return numberFormat.format(bDNumber);
}

/**
 * Formats the given Number as percent with as many fractional digits as
 * precision available.<br>
 * This is a convenient method in case all fractional digits shall be
 * rendered and no custom format / pattern needs to be provided.<br>
 * <br>
 * This serves as a workaround for {@link NumberFormat#getPercentInstance()}
 * which does not renders fractional digits.
 *
 * @param number Number in range of [0-1]
 * @param locale
 *
 * @return
 */
public static String formatPercentFraction(final Number number, final Locale locale)
{
    if (number == null)
        return null;

    final BigDecimal bDNumber = MathUtils.getBigDecimal(number).multiply(new BigDecimal(100));

    final NumberFormat percentScaleFormat = NumberFormat.getPercentInstance(locale);
    percentScaleFormat.setMaximumFractionDigits(Math.max(0, bDNumber.scale() - 2));

    final BigDecimal bDNumberPercent = bDNumber.multiply(new BigDecimal(0.01));

    // Convert back for locale percent formatter
    final String strPercent = percentScaleFormat.format(bDNumberPercent);

    return strPercent;
}

使用给定的十进制长度…

public static String getLocaleFloatValueDecimalWithLength(Locale loc, float value, int length) {
        //make string from float value
        return String.format(loc, (value % 1 == 0 ? "%.0f" : "%."+length+"f"), value);
    }

简而言之:

如果你想摆脱尾随零和区域问题,那么你应该使用:

double myValue = 0.00000021d;

DecimalFormat df = new DecimalFormat("0", DecimalFormatSymbols.getInstance(Locale.ENGLISH));
df.setMaximumFractionDigits(340); //340 = DecimalFormat.DOUBLE_FRACTION_DIGITS

System.out.println(df.format(myValue)); //output: 0.00000021

解释:

为什么其他答案不适合我:

Double.toString() or System.out.println or FloatingDecimal.toJavaFormatString uses scientific notations if double is less than 10^-3 or greater than or equal to 10^7 double myValue = 0.00000021d; String.format("%s", myvalue); //output: 2.1E-7 by using %f, the default decimal precision is 6, otherwise you can hardcode it, but it results in extra zeros added if you have fewer decimals. Example: double myValue = 0.00000021d; String.format("%.12f", myvalue); // Output: 0.000000210000 by using setMaximumFractionDigits(0); or %.0f you remove any decimal precision, which is fine for integers/longs but not for double double myValue = 0.00000021d; System.out.println(String.format("%.0f", myvalue)); // Output: 0 DecimalFormat df = new DecimalFormat("0"); System.out.println(df.format(myValue)); // Output: 0 by using DecimalFormat, you are local dependent. In the French locale, the decimal separator is a comma, not a point: double myValue = 0.00000021d; DecimalFormat df = new DecimalFormat("0"); df.setMaximumFractionDigits(340); System.out.println(df.format(myvalue)); // Output: 0,00000021 Using the ENGLISH locale makes sure you get a point for decimal separator, wherever your program will run.

为什么使用340然后setMaximumFractionDigits?

两个原因:

setMaximumFractionDigits接受一个整数,但是它的实现有DecimalFormat允许的最大数字。DOUBLE_FRACTION_DIGITS等于340 翻倍。MIN_VALUE = 4.9E-324,因此使用340位数字,您肯定不会四舍五入的双精度和损失精度