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group consecutive dates in a row

5 messages · Ben Bolker, Bert Gunter, Gabor Grothendieck +1 more

Stefano Sofia
# 
Dear R users,

I have a data frame with a single column of POSIXct elements, like


mydf <- data.frame(data_POSIX=as.POSIXct(c("2012-02-05", "2012-02-06", "2012-02-07", "2012-02-13", "2012-02-21"), format = "%Y-%m-%d", tz="Etc/GMT-1"))


I need to transform it in a two-columns data frame where I can get rid of consecutive dates. It should appear like


data_POSIX_init data_POSIX_fin

2012-02-05 2012-02-07

2012-02-13 NA

2012-02-21 NA


I started with two "while cycles" and so on, but this is not an efficient way to do it.

Could you please give me an hint on how to proceed?


Thank you for your precious attention and help

Stefano


         (oo)
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Stefano Sofia PhD
Civil Protection - Marche Region - Italy
Meteo Section
Snow Section
Via del Colle Ameno 5
60126 Torrette di Ancona, Ancona (AN)
Uff: +39 071 806 7743
E-mail: stefano.sofia at regione.marche.it
---Oo---------oO----------------------------------------

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Ben Bolker
# 
rle(as.numeric(diff(mydf$data_POSIX)))  should get you started, I think?
On 2023-08-07 12:41 p.m., Stefano Sofia wrote:
Bert Gunter
# 
Here is another way to obtain the day differences that is the argument
of rle() . It is perhaps more reliable in that it uses methods for
class POSIXct rather than depending on the underlying class structure
and conversion via as.numeric. In theory, the methods won't change or
any changes will be documented, whereas class implementations are
allowed to be fluid and undocumented at the R user level (e.g. the
Help system) afaik. In this case, I don't think that would happen, so
I am merely being pedantic, but I hope I do not offend by making the
point.

But I also wanted to say that the OP could have (imo) looked up the
methodology as I describe below, rather than post here. Or perhaps he
did, but got stymied because he did not know about rle(), a somewhat
esoteric base R function. In which case, the search query "how to find
runs of identical values in an R vector" immediately yielded a hit on
rle(). So my message is: **DO** first try to use R's internal Help
before posting, especially for base R related tasks: it is really a
superb resource (again imo). *

OK, enough sermonizing. Here's what I did. Since the data are POSIXct
class, I went to ?POSIXct and browsed through it until I found
"difftime for time intervals" in the *See Also* section. Following
that link to ?difftime showed me this was what was needed, which is:

difftime(mydf[-1,1], mydf[-nrow(mydf), 1], units = "days")
Time differences in days
[1] 1 1 6 8

Cheers,
Bert

*... and I think it is likely that there are time series related
packages that also could be used, perhaps more immediately, to do what
was requested. But that would require a more diligent search. Though
with new LLM's and generative AI for Help systems becoming available,
the degree of diligence required is rapidly decreasing.
On Mon, Aug 7, 2023 at 9:52?AM Ben Bolker <bbolker at gmail.com> wrote:
Gabor Grothendieck
# 
It is best to use Date, rather than POSIXct, class if there are no times.

Use the cumsum expression shown to group the dates and then summarize
each group.

We assume that the dates are already sorted in ascending order.

  library(dplyr)

  mydf <- data.frame(date = as.Date(c("2012-02-05", "2012-02-06",
"2012-02-07", "2012-02-13", "2012-02-21")))

  mydf %>%
    group_by(grp = cumsum(c(0, diff(date)) > 1)) %>%
    summarize(start = first(date), end = last(date)) %>%
    ungroup %>%
    select(-grp)
  ## # A tibble: 3 ? 2
  ##   start      end
  ##   <date>     <date>
  ## 1 2012-02-05 2012-02-07
  ## 2 2012-02-13 2012-02-13
  ## 3 2012-02-21 2012-02-21

or with only base R:

  smrz <- function(x) with(x, data.frame(start = min(date), end = max(date)))
  do.call("rbind", by(mydf, cumsum(c(0, diff(mydf$date)) > 1), smrz))
  ##        start        end
  ## 0 2012-02-05 2012-02-07
  ## 1 2012-02-13 2012-02-13
  ## 2 2012-02-21 2012-02-21


On Mon, Aug 7, 2023 at 12:42?PM Stefano Sofia
<stefano.sofia at regione.marche.it> wrote:

  
    

      
      
    

  


  


      

    

    
3 days later

    

      
      

        


  

        

      Stefano Sofia
    

    

      
      #
    

  


  

      
Thank you for your hints.

All of them have been useful, and you solved my problem.

I understood the role of rle, but I think that for my task its use is not fundamental.


I will put more attention on looking for the existing documentation.

Thank you again

Stefano


         (oo)
--oOO--( )--OOo--------------------------------------
Stefano Sofia PhD
Civil Protection - Marche Region - Italy
Meteo Section
Snow Section
Via del Colle Ameno 5
60126 Torrette di Ancona, Ancona (AN)
Uff: +39 071 806 7743
E-mail: stefano.sofia at regione.marche.it
---Oo---------oO----------------------------------------