[R] a simple reshape

[Yuan Chun Ding]

Hi Deepayan,

Thank you very much!! Yes, your method also works very well,  I thought about creating an extra time variable, but did not know how to do it.

Ding
________________________________________
From: Deepayan Sarkar [deepayan.sarkar using gmail.com]
Sent: Saturday, April 4, 2020 3:10 AM
To: Yuan Chun Ding
Cc: r-help mailing list
Subject: Re: [R] a simple reshape

Hi,

[For a non-tidyverse solution:]

Your problem is ambiguous without a 'time' variable; e.g., why should
the answer not be

test2  <- data.frame(vntr1=c("v1","v2"),
                     a1 =c(NA, 0.5693),
                     a2 = c(0.02, 0.12),
                     a3 =c(NA, 0.11),
                     a4=c(0.98, 0.04))

? If you do add an artificial time variable, say using

test1 <- transform(test1,
    time = unsplit(lapply(split(vntr1, vntr1), seq_along), vntr1))

to give

> test1
 vntr1  val time
1    v1 0.98    1
2    v1 0.02    2
3    v2 0.59    1
4    v2 0.12    2
5    v2 0.11    3
6    v2 0.04    4

then either reshape() or dcast() easily gives you what you want:

> reshape(test1, v.names = "val", idvar = "vntr1", direction = "wide", timevar = "time")
 vntr1 val.1 val.2 val.3 val.4
1    v1  0.98  0.02    NA    NA
3    v2  0.59  0.12  0.11  0.04

> reshape2::dcast(test1, vntr1 ~ time, value.var="val")
 vntr1    1    2    3    4
1    v1 0.98 0.02   NA   NA
2    v2 0.59 0.12 0.11 0.04

-Deepayan

On Sat, Apr 4, 2020 at 12:28 AM Yuan Chun Ding <ycding using coh.org> wrote:
>
> Hi R users,
>
> I want to do a data reshape from long to wide, I thought it was easy using tidyverse spread function, but it did not work well. Can you help me?
>
> Thank you,
>
> Ding
>
> test1 data frame is long file and test2 is the wide file I want to get
>
> test1 <- data.frame (vntr1=c("v1","v1", "v2","v2","v2","v2"),
>                      val =c(0.98,0.02, 0.59,0.12,0.11,0.04))
>
> test2  <- data.frame(vntr1=c("v1","v2"),
>                      a1 =c(0.98, 0.5693),
>                      a2 = c(0.02, 0.12),
>                      a3 =c(NA, 0.11),
>                      a4=c(NA, 0.04))
>
> the following code does not work
> test2 <-test1 %>%spread(vntr1, val)
>
>  Error: Each row of output must be identified by a unique combination of keys.
> Keys are shared for 6 rows:
> * 1, 2
> * 3, 4, 5, 6
> Do you need to create unique ID with tibble::rowid_to_column()?
> Call `rlang::last_error()` to see a backtrace
>
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