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)

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

来源：https://stackoverflow.com/questions/43259380/spread-with-duplicate-identifiers-using-tidyverse-and

标签