[R] how to create a pivot table in r?
Rui Barradas
ru|pb@rr@d@@ @end|ng |rom @@po@pt
Thu Feb 6 06:34:06 CET 2020
Hello,
Function tidyr::pivot_wider is meant for this kind of problem. It is
package tidyr's new way of reshaping from long to wide format. QUoting
from the help page:
Details
pivot_wider() is an updated approach to spread(), designed to be both
simpler to use and to handle more use cases. We recomend you use
pivot_wider() for new code; spread() isn't going away but is no longer
under active development.
In what follows I have use column 'date' to fill the cells, not column
'ObsDate' like in your question. Just change this and you'll get your
expected result.
library(dplyr)
library(tidyr)
library(lubridate)
daT %>%
arrange(ObsSite) %>%
pivot_wider(id_cols = id,
names_from = ObsSite,
values_from = date,
values_fn = list(date = function(x){
paste(x, collapse = ",")
})) %>%
mutate_all(function(x) ifelse(is.na(x), 0, x)) %>%
arrange(id)
Hope this helps,
Rui Barradas
Às 00:37 de 06/02/20, Marna Wagley escreveu:
> Hi R users,
> I was trying to create a pivot table for the following data, in which I
> wanted to put "id" in rows and "ObsSite" in columns and "Obsdate" is in
> the cells.
>
> I used the following code but it took only one date among the two dates.
> For example, the animal (Id2) which was observed in the site7 two time or
> days (07/03/14 & 05/17/2014). see below
> id ObsSite ObsDate
> id1 site7 06/13/13
> id2 site7 07/03/14
> id2 site7 05/17/14
> id4 site4 05/08/14
> id5 site5 06/13/14
> id6 site1 05/30/14
> id6 site1 06/28/13
> id7 site5 06/25/13
>
> I wanted to put both dates in the cell if there is any multiple dates, as
> similar shown below
>
>
>
>
>
> site1 site4 site5 site7
> id1 0 0 0 6/13/13
> id2 0 0 0 7/3/2014, 5/17/2014
> id4 0 5/8/14 0 0
> id5 0 0 6/13/14 0
> id6 5/30/2014, 6/28/2013 0 0 0
> id7 0 0 6/25/13 0
>
> the code I used is given below but it gave me only one date in that cells.
> Is there any way to get both dates in these cells?
> Thanks,
>
> ###
> library(lubridate)
> daT<-structure(list(id = structure(c(1L, 2L, 2L, 3L, 4L, 5L, 5L, 6L
> ), .Label = c("id1", "id2", "id4", "id5", "id6", "id7"), class = "factor"),
> ObsSite = structure(c(4L, 4L, 4L, 2L, 3L, 1L, 1L, 3L), .Label =
> c("site1",
> "site4", "site5", "site7"), class = "factor"), ObsDate =
> structure(c(4L,
> 8L, 2L, 1L, 5L, 3L, 7L, 6L), .Label = c("05/08/14", "05/17/14",
> "05/30/14", "06/13/13", "06/13/14", "06/25/13", "06/28/13",
> "07/03/14"), class = "factor")), .Names = c("id", "ObsSite",
> "ObsDate"), class = "data.frame", row.names = c(NA, -8L))
> daT
> daT$date <- mdy(daT$ObsDate)
> tmp <- split(daT, daT$id)
> head(tmp)
>
> pivotTable <- do.call(rbind, lapply(tmp, function(daT){
> tb <- table(daT$ObsSite)
> idx <- which(tb>0)
> tb1 <- replace(tb, idx, as.character(daT$date))
> }))
>
>
> data.frame(pivotTable)
>
> [[alternative HTML version deleted]]
>
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