[R] Replace NAs in split lists
Jeff Newmiller
jdnewmil at dcn.davis.ca.us
Mon Jan 8 15:19:40 CET 2018
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.pikal at 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-bounces at r-project.org] On Behalf Of Jeff
>> Newmiller
>> Sent: Monday, January 8, 2018 9:13 AM
>> To: r-help at r-project.org; Ek Esawi <esawiek at 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 <esawiek at 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 <esawiek at 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 at 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 at 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...
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>rocks...1k
>>
>> ______________________________________________
>> R-help at 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.
>
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