R dataframe - transpose month column to 12 separate columns

Asked May 29 '19 at 08:17

Active May 29 '19 at 08:17

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Probably this is a stupid question, but I can't figure out how to transfer this dataframe in R:

employee <- c('John Doe','John Doe','Jolie Hope','Jolie Hope')
salary <- c(2100,2200,1900,1800)
month <- c("January","Febraury","January","Febraury")
employee.data <- data.frame(employee,salary,month)

To this one in R:

Of course I can change the construction of the dataframe in this example. However my real data has another source ;). Therefore I would like to do this with some simple commands.

asked May 29 '19 at 08:17

Afke

You are right jogo, this is a duplicate. thank you for your reply. – Afke May 29 '19 at 08:23
Try `reshape(employee.data, idvar = "employee", timevar = "month", direction = "wide")` – markus May 29 '19 at 08:23
Or you can do it the tidyverse way: `employee.data %>% spread(month, salary)` – Claudiu Papasteri May 29 '19 at 08:27
The tidyverse way gives me an error: Error: Each row of output must be identified by a unique combination of keys. Keys are shared for 4 rows: – Afke May 29 '19 at 08:30

R dataframe - transpose month column to 12 separate columns

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