Java years to months

How to convert months to years-and-months in java

This will get you the average jobs sum by month (accross years and names). The class tracks a span of time unrelated to the timeline in terms of years, months and days.

How to convert months to years-and-months in java

As a beginner in Java, my goal is to convert months into years. If I take 18 months and divide it by 12, the result would be 1.5 years. However, I would like to represent it in the format of 1 year and 6 months.

Employ the modulo operator represented by % .

int months = 18; int years = months / 12; // 1 int remainingMonths = months % 12; // 6 

java.time

Java 8 and later versions come with the java.time framework, which replaces the previously troublesome date-time classes, including java.util.Date , .Calendar , and java.text.SimpleDateFormat . The Joda-Time team also recommends transitioning to java.time.

For additional information, refer to the Oracle Tutorial and explore numerous examples and explanations on Stack Overflow.

The ThreeTen-Backport library brings a lot of the java.time features to older versions of Java, such as Java 6 and 7. Additionally, this library has been adjusted for Android in the ThreeTenABP repository on GitHub, developed by Jake Wharton.

The java.time library is expanded by the ThreeTen-Extra project, which serves as a testing platform for potential java.time additions. Here, you can discover several helpful classes, including Interval , YearWeek , YearQuarter , and others.

Period

The Period category monitors a time period that is not associated with years, months, or days on the timeline. It is recommended to make it a practice to invoke normalized to ensure that it computes eighteen months as one year and six months.

Period period = Period.ofMonths( 18 ).normalized(); 

Obtain the count of both years and months.

int years = period.getYears(); int months = period.getMonths(); 

To represent the value in standard ISO 8601 format for «durations,» use toString . This format uses P to mark the beginning, while T separates the years-months-days portion from the hours-minutes-seconds portion (which is irrelevant in this case).

String output = period.toString(); 

TimeSpan Years and Months, If I subtract the years, months and days separately I can get the age without too much fuss. My college already worked out the code,

Convert years to months

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Convert months into mixed years and months

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Days/Months/Years between two dates

I possess two dates that require comparison. The first date is obtained from another application’s table, which is in the format of YYYYMMDD. The second date is in local time. My objective is to calculate the time difference in terms of Days/Months/Years.

 my $t = Time::Piece->strptime($expi, "%Y %m %d"); #First date my $today = Time::Piece->new; #Localtime my $diff = $t - $today; print $diff; 

However, the result is a string of nonsensical digits.

Assign the value «Fri Dec 31 00:00:00 9999» to the variable $t.

The current date and time is stored as a variable named $today, which contains the value «Wed Feb 19 14:40:55 2020».

Any ideas? Thanks in advance.

The maximum difference that Time::Piece can provide is in seconds, and it is impossible to convert it into months, days, or any other unit of time.

use DateTime qw( ); use DateTime::Format::Strptime qw( ); my $format = DateTime::Format::Strptime->new( pattern => '%Y%m%d', time_zone => 'floating', on_error => 'croak', ); my $today = DateTime ->now( time_zone => 'local' ) ->set_time_zone('floating') ->truncate( to => 'day' ); my $dt = $format->parse_datetime('20200101'); my $diff = $today - $dt; printf("%d years, %d months and %d days\n", $diff->in_units(qw( years months days ))); 

0 years, 1 months and 18 days 

How to convert number days to years, months, days?, If you’re using Java 8 with Groovy, you could do: import java.time.* int totalDays = 364 Period diff = LocalDate.now().with < now

Get average of months over years

My objective is to determine the mean number of ‘JOBS’ per month. This entails calculating the average number of jobs generated in May (from 2011 to 2020), June, and so on.

NAME CLOSEDATE JOBS month year A 2019-01-01 2 1 2019 B 2019-01-01 23 1 2019 C 2018-05-24 2 5 2018 D 2019-05-23 200 5 2019 E 2020-05-23 40 5 2020 F 2020-05-14 23 5 2020 G 2020-06-12 93 6 2020 

I tried running pd.pivot_table(proj, index=[‘month’],values=[‘JOBS’],aggfunc=[np.sum,np.mean]) but it only provided the average number of jobs for each record in the month, instead of the average of aggregate month.

The desired outcome for May in the given dataset is 66.25 jobs, which is obtained by calculating the average of the four values: 2, 200, 40, and 23.

It seems like there might be an uncomplicated element or a specific technique to arrange the table in the following manner:

Year Jan Feb Mar . Dec 2011 1000 4322 5322 2343 2012 3423 4322 5322 2343 . 1645 4322 5322 2343 2020 7895 3432 9999 2343 AVG. 3491 4099 6491 

Calculate the total for every month/year and then find the mean value across all years.

by_month = ( proj.groupby([proj.CLOSEDATE.dt.year, proj.CLOSEDATE.dt.month]) # create the groupby object .JOBS.sum() # select only the JOBS column and aggregate by sum .unstack(0) # drop the 'year' level form MultiIndex and use as columns .mean(axis=1) # areage across the years we just unstacked to axis1 .rename('avg_jobs') .rename_axis('month') ) print(by_month) month 1 25.000000 5 88.333333 6 93.000000 Name: avg_jobs, dtype: float64 

Get the average monthly job sum across years and names without creating individual columns for year/month unless they are required for other computations.

by_month = ( proj.groupby('month') # create the groupby object .JOBS.mean() # select only the JOBS column and aggregate by mean ) print(by_month) month 1 12.50 5 66.25 6 93.00 Name: JOBS, dtype: float64 

Year Decade Century Millennium Time Measurement Relations, Year Decade Century Millennium Time Measurement Relations How many years are in Duration: 4:50

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Java Convert Days Months Years to Period

In this Java core tutorial we learn how to convert a number of years, months and days to java.time.Period object in Java programming language.

How to convert days months and years to Period in Java

In Java, we can use the Period.of(int years, int months, int days) static method to instantiate a new Period object from number of years, months and days as the example Java code below.

import java.time.Period; public class ConvertDaysMonthsYearsToPeriodExample1  public static void main(String. args)  int numberOfDays = 70; int numberOfMonths = 5; int numberOfYears = 4; Period period = Period.of(numberOfYears, numberOfMonths, numberOfDays); System.out.println("Number of days: " + numberOfDays); System.out.println("Number of months: " + numberOfMonths); System.out.println("Number of years: " + numberOfYears); System.out.println("Period: " + period); > >

Number of days: 70 Number of months: 5 Number of years: 4 Period: P4Y5M70D

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Add or Subtract Days, Months & Years to Date in Java

Java examples to add or subtract days, months or years from a given date using various date-time classes. If your requirement is to add or subtract only the business days then read the linked article.

1. Add or Subtract Days, Months, Years to Date since Java 8

This recommended approach if we are using JDK 1.8 or later.

New java.time classes LocalDate , LocalDateTime and ZonedDateTime have following plus methods to add days to a date.

• plusDays(long n) – adds n days to date.
• plusWeeks(long n) – adds n weeks to date.
• plusMonths(long n) – adds n months to date.
• plusYears(long n) – adds n years to date.

Similarly, use the following minus methods to subtract days from a date.

• minusDays(long n) – subtracts n days from date.
• minusWeeks(long n) – subtracts n weeks from date.
• minusMonths(long n) – subtracts n months from date.
• minusYears(long n) – subtracts n years from date.

Before returning the modified date, these methods modify the other date fields as well to ensure that the result date is a valid date.

These methods throw DateTimeException if the result exceeds the supported date range.

//1. Add and substract 1 day from LocalDate LocalDate today = LocalDate.now(); //Today LocalDate tomorrow = today.plusDays(1); //Plus 1 day LocalDate yesterday = today.minusDays(1); //Minus 1 day //2. Add and substract 1 month from LocalDateTime LocalDateTime now = LocalDateTime.now(); //Current Date and Time LocalDateTime sameDayNextMonth = now.plusMonths(1); //2018-08-14 LocalDateTime sameDayLastMonth = now.minusMonths(1); //2018-06-14 //3. Add and substract 1 year from LocalDateTime LocalDateTime sameDayNextYear = now.plusYears(1); //2019-07-14 LocalDateTime sameDayLastYear = now.minusYears(1); //2017-07-14

2. Add or Subtract Days from java.util.Date

Till Java 7, the only good way to add days to Date was using Calendar class.

The calendar.add(int field, int amount) method takes two arguments, i.e., field type and the field value. We can use this method to add days, months or any time unit in the underlying Date class.

• To add a time unit, pass a positive number in the method.
• To subtract a time unit, pass a negative number in the method.

Date today = new Date(); System.out.println(today); Calendar cal = Calendar.getInstance(); cal.setTime(today); // Adding time cal.add(Calendar.YEAR, 2); cal.add(Calendar.MONTH, 2); cal.add(Calendar.DATE, 2); cal.add(Calendar.DAY_OF_MONTH, 2); // Subtracting time cal.add(Calendar.YEAR, -3); cal.add(Calendar.MONTH, -3); cal.add(Calendar.DATE, -3); cal.add(Calendar.DAY_OF_MONTH, -3); // convert calendar to date Date modifiedDate = cal.getTime(); System.out.println(modifiedDate); 

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