判断字符串是否为有效日期的最快方法 [英] Fastest way to tell if a string is a valid date
问题描述
我在工作中支持一个公共库，它对给定字符串执行许多检查以查看它是否是有效日期。 Java API，commons-lang库和JodaTime都有方法可以解析字符串并将其转换为日期，让您知道它是否实际上是一个有效的日期，但我希望有一种方法在没有实际创建日期对象（或JodaTime库的情况下的DateTime）的情况下进行验证。例如，这是一段简单的示例代码：
I am supporting a common library at work that performs many checks of a given string to see if it is a valid date. The Java API, commons-lang library, and JodaTime all have methods which can parse a string and turn it in to a date to let you know if it is actually a valid date or not, but I was hoping that there would be a way of doing the validation without actually creating a date object (or DateTime as is the case with the JodaTime library). For example here is a simple piece of example code:
public boolean isValidDate(String dateString) {
SimpleDateFormat df = new SimpleDateFormat("yyyyMMdd");
try {
df.parse(dateString);
return true;
} catch (ParseException e) {
return false;
}
}
这对我来说似乎很浪费，我们正在扔掉结果对象。从我的基准测试中，我们在这个公共图书馆中有大约5％的时间用于验证日期。我希望我只是错过了一个明显的API。任何建议都会很棒！
This just seems wasteful to me, we are throwing away the resulting object. From my benchmarks about 5% of our time in this common library is spent validating dates. I'm hoping I'm just missing an obvious API. Any suggestions would be great!
UPDATE
假设我们可以随时使用始终使用相同的日期格式（可能是yyyyMMdd）。我确实考虑过使用正则表达式，但是它需要知道每个月的天数，闰年等...
Assume that we can always use the same date format at all times (likely yyyyMMdd). I did think about using a regex as well, but then it would need to be aware of the number of days in each month, leap years, etc...
结果
解析日期1000万次
Using Java's SimpleDateFormat: ~32 seconds
Using commons-lang DateUtils.parseDate: ~32 seconds
Using JodaTime's DateTimeFormatter: ~3.5 seconds
Using the pure code/math solution by Slanec: ~0.8 seconds
Using precomputed results by Slanec and dfb (minus filling cache): ~0.2 seconds
有一些非常有创意的答案，我很感激！我想现在我只需要决定我需要多少灵活性，我希望代码看起来像。我要说dfb的答案是正确的，因为它纯粹是最快的，这是我原来的问题。谢谢！
There were some very creative answers, I appreciate it! I guess now I just need to decide how much flexibility I need what I want the code to look like. I'm going to say that dfb's answer is correct because it was purely the fastest which was my original questions. Thanks!
推荐答案
如果您真的关心性能而且日期格式非常简单，那么只需预先计算所有有效的字符串并在内存中散列它们。您上面的格式只有大约800万有效组合，最多2050
If you're really concerned about performance and your date format is really that simple, just pre-compute all the valid strings and hash them in memory. The format you have above only has ~ 8 million valid combinations up to 2050
Slanec编辑 - 参考实现
此实现取决于您的特定日期格式。它可以适应任何特定的日期格式（就像我的第一个答案，但更好一点）。
This implementation depends on your specific dateformat. It could be adapted to any specific dateformat out there (just like my first answer, but a bit better).
它使一组所有日期从1900年到2050年（存储为字符串 - 其中有54787个），然后将给定日期与存储日期进行比较。
It makes a set of all dates from 1900 to 2050 (stored as Strings - there are 54787 of them) and then compares the given dates with those stored.
日期设置已创建，它很快就像地狱一样。与我的第一个解决方案相比，快速微基准测试显示了10倍的改进。
Once the dates set is created, it's fast as hell. A quick microbenchmark showed an improvement by a factor of 10 over my first solution.
private static Set<String> dates = new HashSet<String>();
static {
for (int year = 1900; year < 2050; year++) {
for (int month = 1; month <= 12; month++) {
for (int day = 1; day <= daysInMonth(year, month); day++) {
StringBuilder date = new StringBuilder();
date.append(String.format("%04d", year));
date.append(String.format("%02d", month));
date.append(String.format("%02d", day));
dates.add(date.toString());
}
}
}
}
public static boolean isValidDate2(String dateString) {
return dates.contains(dateString);
}
P.S。它可以修改为使用 Set< Integer> 甚至 Trove 的 TIntHashSet 可以大大减少内存使用量（因此允许使用更大的时间跨度），然后性能下降到低于
P.S. It can be modified to use Set<Integer> or even Trove's TIntHashSet which reduces memory usage a lot (and therefore allows to use a much larger timespan), the performance then drops to a level just below my original solution.
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