Yes, you are right if the IDs are always sequentially-adjacent and the first non-NA value appears in the first record for each ID. -- Sent from my phone. Please excuse my brevity.
On January 8, 2018 2:29:40 AM PST, PIKAL Petr <petr.pi...@precheza.cz> wrote: >Hi > >With the example, na.locf seems to be the easiest way. >> library(zoo) > >> na.locf(df1) > ID ID_2 Firist Value >1 a aa TRUE 2 >2 a ab FALSE 2 >3 a ac FALSE 2 >4 b aa TRUE 5 >5 b ab FALSE 5 > >Cheers >Petr > >> -----Original Message----- >> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Jeff >> Newmiller >> Sent: Monday, January 8, 2018 9:13 AM >> To: r-help@r-project.org; Ek Esawi <esaw...@gmail.com> >> Subject: Re: [R] Replace NAs in split lists >> >> Upon closer examination I see that you are not using the split >version of >> df1 as I usually would, so here is a reproducible example: >> >> #---- >> df1 <- read.table( text= >> "ID ID_2 Firist Value >> 1 a aa TRUE 2 >> 2 a ab FALSE NA >> 3 a ac FALSE NA >> 4 b aa TRUE 5 >> 5 b ab FALSE NA >> ", header=TRUE, as.is=TRUE ) >> >> sdf <- split( df1, df1$ID ) >> # note the extra [ 1 ] in case you have more than one non-NA value # >per ID >> sdf2 <- lapply( sdf >> , function( z ) { >> z$Value <- ifelse( is.na( z$Value ) >> , z$Value[ !is.na( z$Value ) ][ 1 >] >> , z$Value >> ) >> z >> } >> ) >> df2 <- do.call( rbind, sdf2 ) >> df2 >> #> ID ID_2 Firist Value >> #> a.1 a aa TRUE 2 >> #> a.2 a ab FALSE 2 >> #> a.3 a ac FALSE 2 >> #> b.4 b aa TRUE 5 >> #> b.5 b ab FALSE 5 >> >> # or using tidyverse methods >> >> library(dplyr) >> #> >> #> Attaching package: 'dplyr' >> #> The following objects are masked from 'package:stats': >> #> >> #> filter, lag >> #> The following objects are masked from 'package:base': >> #> >> #> intersect, setdiff, setequal, union >> df3 <- ( df1 >> %>% group_by( ID ) >> %>% do({ >> mutate( . >> , Value = ifelse( is.na( Value ) >> , Value[ !is.na( Value ) ][ 1 ] >> , Value >> ) >> ) >> }) >> %>% ungroup >> ) >> df3 >> #> # A tibble: 5 x 4 >> #> ID ID_2 Firist Value >> #> <chr> <chr> <lgl> <int> >> #> 1 a aa T 2 >> #> 2 a ab F 2 >> #> 3 a ac F 2 >> #> 4 b aa T 5 >> #> 5 b ab F 5 >> #---- >> >> On Sun, 7 Jan 2018, Jeff Newmiller wrote: >> >> > Why do you want to modify df1? >> > >> > Why not just reassemble the parts as a new data frame and use that >> > going forward in your calculations? That is generally the preferred >> > approach in R so you can re-do your calculations easily if you find >a >> > mistake later. >> > -- >> > Sent from my phone. Please excuse my brevity. >> > >> > On January 7, 2018 7:35:59 PM PST, Ek Esawi <esaw...@gmail.com> >wrote: >> >> I just came up with a solution right after i posted the question, >but >> >> i figured there must be a better and shorter one.than my solution >> >> sdf1[[1]][1,4]<-lapplyresults[[1]] >> >> sdf1[[2]][1,4]<-lapplyresults[[2]] >> >> >> >> EK >> >> >> >> On Sun, Jan 7, 2018 at 10:13 PM, Ek Esawi <esaw...@gmail.com> >wrote: >> >>> Hi all-- >> >>> >> >>> I stumbled on this problem online. I did not like the solution >given >> >>> there which was a long UDF. I thought why cannot split and l/s >apply >> >>> work here. My aim is to split the data frame, use l/sapply, make >> >>> changes on the split lists and combine the split lists to new >data >> >>> frame with the desired changes/output. >> >>> >> >>> The data frame shown below has a column named ID which has 2 >> >> variables >> >>> a and b; i want to replace the NAs on the Value column by 2, >which >> >>> is the only numeric entry, for ID=a and by 5 for ID=b. >> >>> >> >>> I worked out the solution but could not replace the results in >the >> >> split lists. >> >>> >> >>> Original dataframe , df1 >> >>> ID ID_2 Firist Value >> >>> 1 a aa TRUE 2 >> >>> 2 a ab FALSE NA >> >>> 3 a ac FALSE NA >> >>> 4 b aa TRUE 5 >> >>> 5 b ab FALSE NA >> >>> Sdf1 >> >>> $a >> >>> ID ID_2 Firist Value >> >>> 1 a aa TRUE 2 >> >>> 2 a ab FALSE NA >> >>> 3 a ac FALSE NA >> >>> $b >> >>> ID ID_2 Firist Value >> >>> 4 b aa TRUE 5 >> >>> 5 b ab FALSE NA >> >>> Desired results >> >>> ID ID_2 Firist Value >> >>> 1 a aa TRUE 2 >> >>> 2 a ab FALSE 2 >> >>> 3 a ac FALSE 2 >> >>> >> >>> $b >> >>> ID ID_2 Firist Value >> >>> 4 b aa TRUE 5 >> >>> 5 b ab FALSE 5 >> >>> >> >>> My code >> >>> >> >>> sdf <- split(df1,df$ID) >> >>> lapply(sdf, function(z) >> >> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)],z$Value)) >> >>> result: >> >>> $ a: num [1:3] 2 2 2 >> >>> $ b: num [1:2] 5 5 >> >>> >> >>> How could I put these two lists back in the split data frame, >sdf1? >> >>> Then I could use do.call to reassemble a data frame from the >split >> >>> lists, >> >>> >> >>> Thanks, >> >>> EK >> >> >> >> ______________________________________________ >> >> R-help@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. >> > >> > ______________________________________________ >> > R-help@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. >> > >> >> >--------------------------------------------------------------------------- >> Jeff Newmiller The ..... ..... 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