[R] Calculating date difference in days

[William Dunlap]

You can use units<- to change the time units of the difference.  E.g.,

> d <- as.POSIXlt("2018-03-10") -  as.POSIXlt("2018-03-09 02:00:00")
> d
Time difference of 22 hours
> units(d) <- "days"
> d
Time difference of 0.9166667 days
>
> units(d) <- "mins"
> d
Time difference of 1320 mins
> units(d) <- "secs"
> d
Time difference of 79200 secs
> units(d) <- "weeks"
> d
Time difference of 0.1309524 weeks

Bill Dunlap
TIBCO Software
wdunlap tibco.com


On Wed, May 22, 2019 at 2:44 PM <reichmanj using sbcglobal.net> wrote:

> R Help
>
> I have a function to calculate a date difference in days but my results
> come
> back in hours.  I suspect I am using the as.POSIXlt function incorrectly .
>
> Suggestions?
>
> # Start time of data to be considered
> start_day <- "2016-04-30"
>
> # Make event and sequence IDs into factors
> elapsed_days <- function(end_date, start_date){
>   ed <- as.POSIXlt(end_date)
>   sd <- as.POSIXlt(start_date)
>   ed-sd
> }
>
> trans_sequence$eventID <- elapsed_days(trans_sequence$Date, start_day)
>
>
> > trans_sequence
> # A tibble: 39 x 5
> # Groups:   Emitter [15]
>    Emitter Date        SIZE Geohash
> eventID
>      <int> <date>     <int> <chr>
> <time>
>  1       1 2016-05-01    12 A;B;C;D;E;F;G;H;I;J;K;L
> 19 hours
>  2       1 2016-05-02     5 A;B;C;D;E
> 43 hours
>  3       1 2016-05-05    11 A;B;C;D;E;F;G;H;I;J;K
> 115 hours
>  4       2 2016-05-01     9 C;D;E;F;G;H;I;J;K
> 19 hours
>  5       2 2016-05-02     3 F;G;H
> 43 hours
>  6       2 2016-05-05     3 L;M;N
> 115 hours
>  7       3 2016-05-01     3 L;M;N
> 19 hours
>  8       3 2016-05-02     3 I;J;K
> 43 hours
>  9       3 2016-05-04    25
> A;B;C;D;E;F;G;H;I;J;K;L;M;N;O;P;Q;R;S;T;U;V;W;X;Y  91 hours
> 10       3 2016-05-05     7 O;P;Q;R;S;T;U
> 115 hours
>
> Jeff Reichman
>
>
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>
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