问题
My data looks like this:
I am trying to make it look like this:
I would like to do this in tidyverse using %>%-chaining.
df <-
structure(list(id = c(2L, 2L, 4L, 5L, 5L, 5L, 5L), start_end = structure(c(2L,
1L, 2L, 2L, 1L, 2L, 1L), .Label = c(\"end\", \"start\"), class = \"factor\"),
date = structure(c(6L, 7L, 3L, 8L, 9L, 10L, 11L), .Label = c(\"1979-01-03\",
\"1979-06-21\", \"1979-07-18\", \"1989-09-12\", \"1991-01-04\", \"1994-05-01\",
\"1996-11-04\", \"2005-02-01\", \"2009-09-17\", \"2010-10-01\", \"2012-10-06\"
), class = \"factor\")), .Names = c(\"id\", \"start_end\", \"date\"
), row.names = c(3L, 4L, 7L, 8L, 9L, 10L, 11L), class = \"data.frame\")
What I have tried:
data.table::dcast( df, formula = id ~ start_end, value.var = \"date\", drop = FALSE ) # does not work because it summarises the data
tidyr::spread( df, start_end, date ) # does not work because of duplicate values
df$id2 <- 1:nrow(df)
tidyr::spread( df, start_end, date ) # does not work because the dataset now has too many rows.
These questions do not answer my question:
Using spread with duplicate identifiers for rows (because they summarise)
R: spread function on data frame with duplicates (because they paste the values together)
Reshaping data in R with "login" "logout" times (because not specifically asking for/answered using tidyverse and chaining)
回答1:
We can use tidyverse. After grouping by 'start_end', 'id', create a sequence column 'ind' , then spread from 'long' to 'wide' format
library(dplyr)
library(tidyr)
df %>%
group_by(start_end, id) %>%
mutate(ind = row_number()) %>%
spread(start_end, date) %>%
select(start, end)
# id start end
#* <int> <fctr> <fctr>
#1 2 1994-05-01 1996-11-04
#2 4 1979-07-18 NA
#3 5 2005-02-01 2009-09-17
#4 5 2010-10-01 2012-10-06
Or using tidyr_1.0.0
chop(df, date) %>%
spread(start_end, date) %>%
unnest(c(start, end))
来源：https://stackoverflow.com/questions/43259380/spread-with-duplicate-identifiers-using-tidyverse-and