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kev
Asked: 2020-04-22 04:58:32 +0800 CST 2020-04-22 04:58:32 +0800 CST

Error in spread function in R

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When I try to use spread(instrumento, precio)it, it returns an error Each row of output must be identified by a unique combination of keys. Keys are shared for 851 rowsI need to work with the different instruments, for this I try to take them as variables

         fecha instrumento precio
2   2020-04-20      YPFD   397.50
3   2020-04-20      YPFD   397.50
4   2020-04-20      YPFD   397.50
5   2020-04-17      YPFD   401.85
6   2020-04-17      YPFD   402.00
7   2020-04-17      YPFD   401.00
8   2020-04-16      YPFD   400.60
r

1 Answers

  1. Best Answer

    To begin with, with the data you show, it doesn't make sense to make a spread(), from what I understand, what you are looking for is that each row is a date and each column an ​​instrument, however it has multiple prices for the same instrument/date and the condition to be able to do a spreadis that there is only one value per row/column, in your case date/instrument. It occurs to me that what you need in the first place is some aggregation function, for example:

    library("tidyverse")

read.table(text="fecha instrumento precio
2020-04-20      YPFD   397.50
2020-04-20      YPFD   397.50
2020-04-20      YPFD   397.50
2020-04-17      YPFD   401.85
2020-04-17      YPFD   402.00
2020-04-17      YPFD   401.00
2020-04-16      YPFD   400.60", header = TRUE, stringsAsFactors=FALSE) %>% 
  group_by(fecha, instrumento) %>% 
  summarize(precio=max(precio)) %>% 
  spread(instrumento, precio) 


# A tibble: 3 x 2
# Groups:   fecha [3]
  fecha       YPFD
  <chr>      <dbl>
1 2020-04-16  401.
2 2020-04-17  402 
3 2020-04-20  398.
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