我編寫了一個(gè)程序，它總是計(jì)算兩個(gè)日期之間的天數(shù)差異。我現(xiàn)在想總結(jié)所有差異天數(shù)，并將其除以計(jì)數(shù)，以便得到平均天數(shù)。有人能幫我嗎？有沒有辦法甚至避免 for 循環(huán)。import java.text.ParseException;import java.text.SimpleDateFormat;import java.util.Date;public class AVGDateCalculation { public static void main (String [] args) throws ParseException{ String formatted = "2014-04-28 ,2014-04-28 ," + "2015-10-26 ,2015-10-30 ," + "2015-07-30 ,2015-07-30 ," + "2015-04-14 ,2015-04-20 ," + "2013-11-14 ,2013-11-18 ," + "2014-04-16 ,2014-04-22 ," + "2014-11-19 ,2014-11-21 ," + "2019-10-01 ,2019-10-01 "; String[] parts = formatted.split(","); SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd"); int count = 0; int a; long difference = 0L; float daysBetween = 0.00f; float averageDays = 0.00f; Date dateBefore = null; Date dateAfter = null; for (a = 0; a<parts.length; a+=1) { dateBefore = sdf.parse(parts[a++]); count++; dateAfter = sdf.parse(parts[a+=0]); difference = dateAfter.getTime() - dateBefore.getTime(); daysBetween = (difference / (1000*60*60*24)); averageDays = (count / daysBetween); System.out.println(String.valueOf(dateAfter) + " - " + `String.valueOf(dateBefore));` System.out.println(String.valueOf(dateAfter.getTime()) + " - " + String.valueOf(dateBefore.getTime())); System.out.println(String.valueOf(daysBetween)); System.out.println(String.valueOf(averageDays) + " days"); System.out.println(String.valueOf(count)); } }}
2 回答
江戶川亂折騰
TA貢獻(xiàn)1851條經(jīng)驗(yàn) 獲得超5個(gè)贊
我寧愿使用LocalDate而不是SimpleDateFormat，因?yàn)樵谀抢锬憧梢愿p松地計(jì)算差異和許多其他事情。要分割成對(duì)出現(xiàn)的日期，您可以使用正則表達(dá)式在每隔一個(gè)逗號(hào)分隔字符串。
import java.time.LocalDate;
import java.time.format.DateTimeFormatter;
import java.time.temporal.ChronoUnit;
import java.util.LongSummaryStatistics;
import java.util.function.Function;
import java.util.regex.Pattern;
public class Test{
public static void main(String[] args){
String formatted
= "2014-04-28 ,2014-04-28 ,"
+ "2015-10-26 ,2015-10-30 ,"
+ "2015-07-30 ,2015-07-30 ,"
+ "2015-04-14 ,2015-04-20 ,"
+ "2013-11-14 ,2013-11-18 ,"
+ "2014-04-16 ,2014-04-22 ,"
+ "2014-11-19 ,2014-11-21 ,"
+ "2019-10-01 ,2019-10-01 ";
//a formater for your date pattern
DateTimeFormatter dtf = DateTimeFormatter.ofPattern("yyyy-MM-dd");
//a function to calculate the date differnce for a given pair of dates of form "2014-04-28 ,2014-04-28"
Function<String,Long> dateDiff = s -> ChronoUnit.DAYS.between(
LocalDate.parse(s.split(",")[0].trim(), dtf),
LocalDate.parse(s.split(",")[1].trim(), dtf));
//split your original string at each second ',' with below regex
LongSummaryStatistics statistics = Pattern.compile("(?<!\\G[\\d -]+),")
.splitAsStream(formatted)
.mapToLong(s -> dateDiff.apply(s))
.summaryStatistics();
System.out.println(statistics);
}
}
統(tǒng)計(jì)數(shù)據(jù)包含總和、計(jì)數(shù)、最小值、最大值和平均值。如果你只對(duì)平均值感興趣，也可以直接調(diào)用
double average = Pattern.compile("(?<!\\G[\\d -]+),")
.splitAsStream(formatted)
.mapToLong(s -> dateDiff.apply(s))
.average().getAsDouble();
哈士奇WWW
TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超6個(gè)贊
博士
ChronoUnit? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?// An enum delineating granularities of time.
.DAYS? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? // `DAYS` is one of the objects pre-defined on that enum.
.between(? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? // Calculates elapsed time.
? ? LocalDate.parse( "2015-10-26" ) ,? ? // `LocalDate` represents a date-only value, without time-of-day, without time zone.
? ? LocalDate.parse( "2015-10-30" )? ? ? // `LocalDate` by default parses strings that comply with standard ISO 8601 formats.
)? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? // Returns a `long`, the number of days elapsed. Uses Half-Open approach, where the beginning is *inclusive* while the ending is *exclusive*.?
4
錯(cuò)誤的班級(jí)
該類java.util.Date代表 UTC 中的某個(gè)時(shí)刻，而不是日期。此外，那個(gè)可怕的類在幾年前就被現(xiàn)代的java.time類取代了。
LocalDate
該類LocalDate表示僅日期值，沒有時(shí)間、時(shí)區(qū)或相對(duì)于 UTC 的偏移量。
您的輸入符合 ISO 8601，因此您可以直接解析。無需定義格式化模式。
LocalDate?start?=?LocalDate.parse(?"2015-10-26"?)?;
使用 計(jì)算經(jīng)過的時(shí)間Period。
Period?p?=?Period.of(?start?,?stop?)?;
或者直接詢問天數(shù)。
long?days?=?ChronoUnit.DAYS.between(?start?,?stop?)?;
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