如何在java中格式化1200到1.2k
如何在java中格式化1200到1.2k
提问于 2011-01-21 06:12:56
我想用java将以下数字格式化为它们旁边的数字：
1000 to 1k
5821 to 5.8k
10500 to 10k
101800 to 101k
2000000 to 2m
7800000 to 7.8m
92150000 to 92m
123200000 to 123m右边的数字是长整型的，左边的数字是字符串。我该如何处理这件事。我已经为此做了很少的算法，但我想可能已经有一些发明出来的东西在这方面做得更好，而且如果我开始处理数十亿和数万亿的数据，不需要额外的测试:)
其他要求：
- 格式应最多包含4个字符，上面的意思是1.1k is
- ，11.2k不是。同样，7.8m是可以的，19.1m不是。只允许小数点前一位有小数点。小数点前两位表示小数点后没有数字。
- 不需要舍入。(显示的带有k和m的数字更多的是模拟仪表，表示近似，而不是精确的逻辑文章。因此，舍入是不相关的，主要是因为变量的性质，即使在查看缓存的结果时，变量也可以增加或减少几位数。)
回答 24
Stack Overflow用户
回答已采纳
发布于 2015-06-05 16:07:48
下面是，这是一种适用于任何长值的解决方案，我觉得它非常易读(核心逻辑在format方法的最下面三行中完成)。
它利用TreeMap来查找适当的后缀。令人惊讶的是，它比我之前写的使用数组的解决方案效率更高，也更难读。
private static final NavigableMap<Long, String> suffixes = new TreeMap<> ();
static {
suffixes.put(1_000L, "k");
suffixes.put(1_000_000L, "M");
suffixes.put(1_000_000_000L, "G");
suffixes.put(1_000_000_000_000L, "T");
suffixes.put(1_000_000_000_000_000L, "P");
suffixes.put(1_000_000_000_000_000_000L, "E");
}
public static String format(long value) {
//Long.MIN_VALUE == -Long.MIN_VALUE so we need an adjustment here
if (value == Long.MIN_VALUE) return format(Long.MIN_VALUE + 1);
if (value < 0) return "-" + format(-value);
if (value < 1000) return Long.toString(value); //deal with easy case
Entry<Long, String> e = suffixes.floorEntry(value);
Long divideBy = e.getKey();
String suffix = e.getValue();
long truncated = value / (divideBy / 10); //the number part of the output times 10
boolean hasDecimal = truncated < 100 && (truncated / 10d) != (truncated / 10);
return hasDecimal ? (truncated / 10d) + suffix : (truncated / 10) + suffix;
}测试代码
public static void main(String args[]) {
long[] numbers = {0, 5, 999, 1_000, -5_821, 10_500, -101_800, 2_000_000, -7_800_000, 92_150_000, 123_200_000, 9_999_999, 999_999_999_999_999_999L, 1_230_000_000_000_000L, Long.MIN_VALUE, Long.MAX_VALUE};
String[] expected = {"0", "5", "999", "1k", "-5.8k", "10k", "-101k", "2M", "-7.8M", "92M", "123M", "9.9M", "999P", "1.2P", "-9.2E", "9.2E"};
for (int i = 0; i < numbers.length; i++) {
long n = numbers[i];
String formatted = format(n);
System.out.println(n + " => " + formatted);
if (!formatted.equals(expected[i])) throw new AssertionError("Expected: " + expected[i] + " but found: " + formatted);
}
}Stack Overflow用户
发布于 2011-01-21 07:28:25
我知道，这看起来更像是一个C程序，但它是超级轻量级的！
public static void main(String args[]) {
long[] numbers = new long[]{1000, 5821, 10500, 101800, 2000000, 7800000, 92150000, 123200000, 9999999};
for(long n : numbers) {
System.out.println(n + " => " + coolFormat(n, 0));
}
}
private static char[] c = new char[]{'k', 'm', 'b', 't'};
/**
* Recursive implementation, invokes itself for each factor of a thousand, increasing the class on each invokation.
* @param n the number to format
* @param iteration in fact this is the class from the array c
* @return a String representing the number n formatted in a cool looking way.
*/
private static String coolFormat(double n, int iteration) {
double d = ((long) n / 100) / 10.0;
boolean isRound = (d * 10) %10 == 0;//true if the decimal part is equal to 0 (then it's trimmed anyway)
return (d < 1000? //this determines the class, i.e. 'k', 'm' etc
((d > 99.9 || isRound || (!isRound && d > 9.99)? //this decides whether to trim the decimals
(int) d * 10 / 10 : d + "" // (int) d * 10 / 10 drops the decimal
) + "" + c[iteration])
: coolFormat(d, iteration+1));
}它输出：
1000 => 1k
5821 => 5.8k
10500 => 10k
101800 => 101k
2000000 => 2m
7800000 => 7.8m
92150000 => 92m
123200000 => 123m
9999999 => 9.9mStack Overflow用户
发布于 2011-01-21 09:53:37
下面是一个使用DecimalFormat的工程符号的解决方案：
public static void main(String args[]) {
long[] numbers = new long[]{7, 12, 856, 1000, 5821, 10500, 101800, 2000000, 7800000, 92150000, 123200000, 9999999};
for(long number : numbers) {
System.out.println(number + " = " + format(number));
}
}
private static String[] suffix = new String[]{"","k", "m", "b", "t"};
private static int MAX_LENGTH = 4;
private static String format(double number) {
String r = new DecimalFormat("##0E0").format(number);
r = r.replaceAll("E[0-9]", suffix[Character.getNumericValue(r.charAt(r.length() - 1)) / 3]);
while(r.length() > MAX_LENGTH || r.matches("[0-9]+\\.[a-z]")){
r = r.substring(0, r.length()-2) + r.substring(r.length() - 1);
}
return r;
}输出：
7 = 7
12 = 12
856 = 856
1000 = 1k
5821 = 5.8k
10500 = 10k
101800 = 102k
2000000 = 2m
7800000 = 7.8m
92150000 = 92m
123200000 = 123m
9999999 = 10m页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/4753251
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