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
>>>
>>---------------------------------------------------------------------------
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