I don't get exactly that error message,
> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
Error: unexpected symbol in "ifelse(is.na(z$Value),z$Value[!is.na
(z$Value)][1],z$Value)z"
The 'symbol' in "unexpected symbol" refers to a "name" ('z' in this case).
The problem is usually at the end of the code snippet shown in the error
message as in
> f(a)b # missing newline or semicolon?
Error: unexpected symbol in "f(a)b"
or
> f(a b) # missing comma?
Error: unexpected symbol in "f(a b"
You need a newline or semicolon between the end of 'ifelse(...)' and 'z' or
maybe
you don't want the z there at all.
After you fix that it will complain about the unmatched right brace and
parenthesis.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Jan 8, 2018 at 8:55 AM, Ek Esawi <esaw...@gmail.com> wrote:
> OPS! Sorry i did indeed posted the code in HTML; should have known better.
>
> ifelse(is.na(z$Value),z$Value[!is.na(z$Value)][1],z$Value)z})
> error. unexpected symbol in sdf2
>
> On Mon, Jan 8, 2018 at 11:44 AM, Jeff Newmiller
> <jdnew...@dcn.davis.ca.us> wrote:
> > 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.
>
[[alternative HTML version deleted]]
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