Re: [R] a simple reshape

Rui Barradas Fri, 03 Apr 2020 14:31:08 -0700

Hello,

That function is new (tidyr 1.0.0) and exists in


packageVersion('tidyverse')
#[1] ‘1.3.0’
packageVersion('tidyr')
#[1] ‘1.0.2’

From the help page ?pivot_wider:

Details

pivot_wider() is an updated approach to spread(), designed to be bothsimpler to use and to handle more use cases. We recommend you usepivot_wider() for new code; spread() isn't going away but is no longerunder active development.



So you must update your packages.

Hope this helps,

Rui Barradas


Às 20:17 de 03/04/20, Yuan Chun Ding escreveu:

Hi Rui,

Thanks a lot,

i got this error, I have library(tidyverse).

Ding

  Error in pivot_wider(., id_cols = "vntr1", names_from = "group", names_prefix = 
"a",  :
   could not find function "pivot_wider"
________________________________________
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: Friday, April 3, 2020 12:08 PM
To: Yuan Chun Ding; r-help mailing list
Subject: Re: [R] a simple reshape

Hello,

It's a bit more complicated than you have coded it.
I will use pivot_wider, it's now the natural way of doing it.


test1 %>%
    group_by(vntr1) %>%
    mutate(group = row_number()) %>%
    ungroup() %>%
    pivot_wider(
      id_cols ="vntr1",
      names_from = "group",
      names_prefix = "a",
      values_from = "val"
    )


Hope this helps,

Rui Barradas

Às 19:57 de 03/04/20, Yuan Chun Ding escreveu:

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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