Get the difference between dates in terms of weeks months quarters and years

Question

I have two dates let  s say 14 01 2013 and 26 03 2014   I would like to get the difference between those two dates in terms of weeks     months in the example 14   quarters 4  and years 1    Do you know the best way to get this

User · Answer

For weeks  you can use function difftime   date1  lt - strptime  14 01 2013   format   d  m  Y   date2  lt - strptime  26 03 2014   format   d  m  Y   difftime date2 date1 units  weeks   Time difference of 62 28571 weeks   But difftime doesn t work with duration over weeks  The following is a very suboptimal solution using cut POSIXt for those durations but you can work around it   seq1  lt - seq date1 date2  by  days   nlevels cut seq1  months    15 nlevels cut seq1  quarters    5 nlevels cut seq1  years    2   This is however the number of months  quarters or years spanned by your time interval and not the duration of your time interval expressed in months  quarters  years  since  those do not have a constant duration   Considering the comment you made on  SvenHohenstein answer I would think you can use nlevels cut seq1  months    - 1 for what you re trying to achieve

User · Answer

All the existing answers are imperfect  IMO  and either make assumptions about the desired output or don t provide flexibility for the desired output    Based on the examples from the OP  and the OP s stated expected answers  I think these are the answers you are looking for  plus some additional examples that make it easy to extrapolate      This only requires base R and doesn t require zoo or lubridate   Convert to Datetime Objects  date strings   c  14 01 2013    26 03 2014   datetimes   strptime date strings  format     d  m  Y     convert to datetime objects   Difference in Days  You can use the diff in days to get some of our later answers  diff in days   difftime datetimes 2   datetimes 1   units    days     days diff in days  Time difference of 435 9583 days   Difference in Weeks  Difference in weeks is a special case of units    weeks  in difftime    diff in weeks   difftime datetimes 2   datetimes 1   units    weeks     weeks diff in weeks  Time difference of 62 27976 weeks   Note that this is the same as dividing our diff in days by 7  7 days in a week   as double diff in days  7   1  62 27976   Difference in Years  With similar logic  we can derive years from diff in days  diff in years   as double diff in days  365   absolute years diff in years   1  1 194406   You seem to be expecting the diff in years to be  1   so I assume you just want to count absolute calendar years or something  which you can easily do by using floor      get desired output  given your definition of  years  floor diff in years    1  1   Difference in Quarters    get desired output for quarters  given your definition of  quarters  floor diff in years   4    1  4   Difference in Months  Can calculate this as a conversion from diff years    months  defined as absolute calendar months  this might be what you want  given your question details  months diff   diff in years 12 floor month diff    1  14   I know this question is old  but given that I still had to solve this problem just now  I thought I would add my answers  Hope it helps

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what about this     get difference between dates   01 12 2013   and   31 12 2013      weeks difftime strptime  26 03 2014   format     d  m  Y    strptime  14 01 2013   format     d  m  Y   units  weeks   Time difference of 62 28571 weeks    months  as yearmon strptime  26 03 2014   format     d  m  Y   - as yearmon strptime  14 01 2013   format     d  m  Y     12  1  14    quarters  as yearqtr strptime  26 03 2014   format     d  m  Y   - as yearqtr strptime  14 01 2013   format     d  m  Y     4  1  4    years year strptime  26 03 2014   format     d  m  Y   - year strptime  14 01 2013   format     d  m  Y     1  1   as yearmon   and as yearqtr   are in package zoo  year   is in package lubridate  What do you think

User · Answer

Here the still lacking lubridate answer  although Gregor s function is built on this package  The lubridate timespan documentation is very helpful for understanding the difference between periods and duration  I also like the lubridate cheatsheet and this very useful thread library lubridate   dates  lt - c dmy  14 01 2013    dmy  26 03 2014     span  lt - dates 1   --  dates 2   creating an interval object   creating period objects  as period span  unit    year      gt   1   quot 1y 2m 12d 0H 0M 0S quot  as period span  unit    month     gt   1   quot 14m 12d 0H 0M 0S quot  as period span  unit    day     gt   1   quot 436d 0H 0M 0S quot   Periods do not accept weeks as units  But you can convert durations to weeks  as duration span   dweeks 1   makes duration object  in seconds  and divides by duration of a week  in seconds    gt   1  62 28571  Created on 2019-11-04 by the reprex package  v0 3 0

User · Answer

Here s a solution    dates  lt - c  14 01 2013    26 03 2014      Date format  dates2  lt - strptime dates  format     d  m  Y    dif  lt - diff as numeric dates2     difference in seconds  dif  60   60   24   7    weeks  1  62 28571 dif  60   60   24   30    months  1  14 53333 dif  60   60   24   30   3    quartes  1  4 844444 dif  60   60   24   365    years  1  1 194521

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I just wrote this for another question  then stumbled here   library lubridate      Calculate age        By default  calculates the typical  age in years   with a     code floor  applied so that you are  e g   5 years old from    5th birthday through the day before your 6th birthday  Set     code floor   FALSE  to return decimal ages  and change  code units     for units other than years      param dob date-of-birth  the day to start calculating age      param age day the date on which age is to be calculated      param units unit to measure age in  Defaults to  code  years    Passed to  link  code duration        param floor boolean for whether or not to floor the result  Defaults to  code TRUE       return Age in  code units   Will be an integer if  code floor   TRUE       examples    my dob  lt - as Date  1983-10-20      age my dob     age my dob  units    minutes      age my dob  floor   FALSE  age  lt - function dob  age day   today    units    years   floor   TRUE        calc age   interval dob  age day    duration num   1  units   units      if  floor  return as integer floor calc age        return calc age      Usage examples   my dob  lt - as Date  1983-10-20    age my dob     1  31  age my dob  floor   FALSE     1  31 15616  age my dob  units    minutes      1  16375680  age seq my dob  length out   6  by    years       1  31 30 29 28 27 26

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try this for a months solution  StartDate  lt - strptime  14 January 2013     d  B  Y    EventDates  lt - strptime c  26 March 2014      d  B  Y    difftime EventDates  StartDate

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A more  precise  calculation  That is  the number of week month quarter year for a non-complete week month quarter year is the fraction of calendar days in that week month quarter year  For example  the number of months between 2016-02-22 and 2016-03-31 is 8 29   31 31   1 27586  explanation inline with code     Calculate precise number of periods between 2 dates         details The number of week month quarter year for a non-complete week month quarter year         is the fraction of calendar days in that week month quarter year          For example  the number of months between 2016-02-22 and 2016-03-31         is 8 29   31 31   1 27586         param startdate start Date of the interval     param enddate end Date of the interval     param period character  It must be one of  day    week    month    quarter  and  year          examples     identical numPeriods as Date  2016-02-15    as Date  2016-03-31     month    15 29   1     identical numPeriods as Date  2016-02-15    as Date  2016-03-31     quarter     15   31   31   29   31      identical numPeriods as Date  2016-02-15    as Date  2016-03-31     year     15   31  366          return exact number of periods between     numPeriods  lt - function startdate  enddate  period         numdays  lt - as numeric enddate - startdate    1     if  grepl  day   period  ignore case TRUE             return numdays         else if  grepl  week   period  ignore case TRUE             return numdays   7              create a sequence of dates between start and end dates     effDaysinBins  lt - cut seq startdate  enddate  by  1 day    period        use the earliest start date of the previous bins and create a breaks of periodic dates with      user s period interval     intervals  lt - seq from as Date min levels effDaysinBins      Y- m- d             by paste  1  period            length out length levels effDaysinBins   1        create a sequence of dates between the earliest interval date and last date of the interval      that contains the enddate     allDays  lt - seq from intervals 1           to intervals intervals  gt  enddate  1  - 1          by  1 day         bin all days in the whole period using previous breaks     allDaysInBins  lt - cut allDays  intervals        calculate ratio of effective days to all days in whole period     sum  tabulate effDaysinBins    tabulate allDaysInBins       numPeriods   Please let me know if you find more boundary cases where the above solution does not work

[r] Get the difference between dates in terms of weeks, months, quarters, and years

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