I don't know. You seem to be posting in HTML so your code is mangled. Can you post plain text and use the reprex package to make sure it produces the errorin a clean R session? -- Sent from my phone. Please excuse my brevity.
On January 8, 2018 8:03:45 AM PST, Ek Esawi <esaw...@gmail.com> wrote: >Thank you Jeff. Your code works, as usual , perfectly. I am just >wondering why if i put the whole code in one line, i get an error >message. >sdf2 <- lapply( sdf, function(z){z$Value ><-ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z}) >error. unexpected symbol in sdf2 > >Thanks again > >EK > > >On Mon, Jan 8, 2018 at 3:12 AM, Jeff Newmiller ><jdnew...@dcn.davis.ca.us> wrote: >> Upon closer examination I see that you are not using the split >version of >> df1 as I usually would, so here is a reproducible example: >> >> #---- >> df1 <- read.table( text= >> "ID ID_2 Firist Value >> 1 a aa TRUE 2 >> 2 a ab FALSE NA >> 3 a ac FALSE NA >> 4 b aa TRUE 5 >> 5 b ab FALSE NA >> ", header=TRUE, as.is=TRUE ) >> >> sdf <- split( df1, df1$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#.. Dead: OO#.. >Playing >> Research Engineer (Solar/Batteries O.O#. #.O#. with >> /Software/Embedded Controllers) .OO#. .OO#. >rocks...1k >> >--------------------------------------------------------------------------- ______________________________________________ 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.