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Jun 20th, 2019
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  1. dt %>%
  2. arrange(label) %>%
  3. group_by(label) %>%
  4. mutate(total = n(),
  5. percent = sum(match == 1) /total)
  6.  
  7. # A tibble: 10 x 6
  8. # Groups: label [6]
  9. id label predicted match total percent
  10. <dbl> <dbl> <dbl> <dbl> <int> <dbl>
  11. 1 3 1 1 1 3 0.667
  12. 2 8 1 1 1 3 0.667
  13. 3 10 1 4 0 3 0.667
  14. 4 6 2 2 1 1 1
  15. 5 7 3 3 1 1 1
  16. 6 5 4 2 0 1 0
  17. 7 2 5 5 1 2 0.5
  18. 8 4 5 3 0 2 0.5
  19. 9 1 6 6 1 2 0.5
  20. 10 9 6 4 0 2 0.5
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