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Getting the index of the row where the new value starts

I have a simple data.frame

one as shown below. I want to get all the indices of the rows where the new one starts origin

. In this case it would be 1, 5, and 8. Is there a way to do this without a loop?

df <- data.frame(origin=c(rep('2016-01-01', 4), rep('2016-02-01',3), rep('2016-03-01',2)), 
  date=c('2016-01-01','2016-02-01','2016-03-01','2016-04-01','2016-02-01','2016-03-01','2016-04-01','2016-03-01','2016-04-01'),
  val=rnorm(9))

df$date <- as.Date(df$date)
df$origin <- as.Date(df$origin)

df
      origin       date        val
1 2016-01-01 2016-01-01 -2.0856573
2 2016-01-01 2016-02-01 -0.5930160
3 2016-01-01 2016-03-01  0.5370460
4 2016-01-01 2016-04-01  1.5539720
5 2016-02-01 2016-02-01  0.4866211
6 2016-02-01 2016-03-01 -0.1443780
7 2016-02-01 2016-04-01 -0.9286197
8 2016-03-01 2016-03-01 -0.6311255
9 2016-03-01 2016-04-01  1.1667005
+3


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3 answers 


which(!duplicated(df$origin))
#[1] 1 5 8

If the values ​​may be repeated (or not sorted), use the following information to find where the series of new values ​​begins.



which(c(TRUE, as.character(df$origin)[-NROW(df)] != as.character(df$origin)[-1]))
#[1] 1 5 8
+3


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Another option using rle

and cumsum

. We c()

a 1

at the beginning because this is the beginning, and then we remove the last element from the loop (since there are no new elements after it). A bit esoteric, but:



date_runs <- rle(as.character(df$origin))
cumsum(c(1,date_runs[[1]][-length(date_runs[[1]])]))
##[1] 1 5 8
+1


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You can use functions in a package dplyr

:

library(dplyr)
df %>%
  group_by(origin) %>%
  slice(1)
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