Hi Yes, that is why I mentioned "the example". Of course if there are missing values in other columns or data layout is different, na.locf could give undesired result.
However it is simple and works with data similar to the example. Cheers Petr > -----Original Message----- > From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us] > Sent: Monday, January 8, 2018 4:45 PM > To: Eric Berger <ericjber...@gmail.com> > Cc: PIKAL Petr <petr.pi...@precheza.cz>; r-help@r-project.org; Ek Esawi > <esaw...@gmail.com> > Subject: Re: [R] Replace NAs in split lists > > "Enforce" is overstating it... results will differ if there are no non-NA > values for > a given ID, and there is a potential further discrepancy if there are multiple > non-NA values. But these issues were not identified by the OP, so may not be > relevant in their case. > -- > Sent from my phone. Please excuse my brevity. > > On January 8, 2018 6:41:33 AM PST, Eric Berger <ericjber...@gmail.com> > wrote: > >You can enforce these assumptions by sorting on multiple columns, which > >leads to > > > >na.locf(df1[ order(df1$ID,df1$Value), ]) > > > > > > > >On Mon, Jan 8, 2018 at 4:19 PM, Jeff Newmiller > ><jdnew...@dcn.davis.ca.us> > >wrote: > > > >> 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 ..... ..... > >Go > >> >Live... > >> >> DCN:<jdnew...@dcn.davis.ca.us> Basics: ##.#. ##.#. > >Live > >> >Go... > >> >> Live: OO#.. 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