[R] Sequential Filtering of a Data Set

Lorenzo Isella |orenzo@|@e||@ @end|ng |rom gm@||@com
Sat Apr 27 21:56:14 CEST 2019


Perfect!
Thanks a lot.

Lorenzo

On Fri, Apr 26, 2019 at 04:06:49PM +0100, Rui Barradas wrote:
>Hello,
>
>Something like this?
>
>Map(function(y) {subset(tt2, year >= y)}, 2001:2018)
>
>Or this?
>
>mapply(function(y) {subset(tt2, year >= y)}, 2001:2018, SIMPLIFY = FALSE)
>
>
>Hope this helps,
>
>Rui Barradas
>
>Às 15:51 de 26/04/19, Lorenzo Isella escreveu:
>>Dear All,
>>I must be drowning in a glass of water.
>>Consider the following data set
>>
>>tt2<-structure(list(year = c(2000, 2001, 2002, 2003, 2004, 2005, 
>>2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013, 2014, 2015, 2016, 
>>2017, 2018), country = c("DE", "DE", "DE", "DE", "DE", "DE", "DE", 
>>"DE", "DE", "DE", "DE", "DE", "DE", "DE", "DE", "DE", "DE", "DE", 
>>"DE"), berd = c(35600, 36331.9, 36950, 38029, 38363, 38651.038, 
>>41148, 43034, 46073, 45275, 46929, 51077.2, 53790.1, 53566.2, 
>>56996.5, 60952, 62826, 68644, NA)), row.names = c(NA, -19L), class = 
>>c("tbl_df", "tbl", "data.frame"))
>>
>>
>>I would like to obtain a list of it, where every element of the list
>>contains the subset of tt2 for which year>=2000,
>>year>=2001....year>=2018 etc...
>>It seems something I can tackle with map or map2 from purrr, but so
>>far I am banging my head against the wall.
>>Anyone can help me?
>>Regards
>>
>>Lorenzo
>>
>>______________________________________________
>>R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>https://stat.ethz.ch/mailman/listinfo/r-help
>>PLEASE do read the posting guide 
>>http://www.R-project.org/posting-guide.html
>>and provide commented, minimal, self-contained, reproducible code.



More information about the R-help mailing list