[R] Method Guidance

[Rui Barradas]

Hello,

Here is a base R solution for what I understand of the question.
It involves ave and cumsum. cumsum of the values of Event_A breaks 
Event_B in segments and ave applies a function to each segment. To find 
where are the times B, coerce to logical and have which() take care of 
it. Data in dput format at the end.



ave(as.logical(df1$Event_B), cumsum(df1$Event_A),
     FUN = function(x) {
       y <- integer(length(x))
       y[x] <- which(x) - 1L
       y
     })
#[1] 0 1 0 0 1 0 3 0 0 2


More readable, with an auxiliary function.


aux_fun <- function(x) {
   y <- integer(length(x))
   y[x] <- which(x) - 1L
   y
}

ave(as.logical(df1$Event_B), cumsum(df1$Event_A), FUN = aux_fun)
#[1] 0 1 0 0 1 0 3 0 0 2



Now assign this result to a df1 column. Here I just test for equality.


new <- ave(as.logical(df1$Event_B), cumsum(df1$Event_A), FUN = aux_fun)
identical(new, df1$Time_B)
#[1] TRUE


# Data
df1 <-
structure(list(Time = 1:10, Event_A = c(1L, 0L, 0L, 1L, 0L, 0L,
0L, 1L, 0L, 0L), Event_B = c(1L, 1L, 0L, 0L, 1L, 0L, 1L, 1L,
0L, 1L), Time_B = c(0L, 1L, 0L, 0L, 1L, 0L, 3L, 0L, 0L, 2L)),
class = "data.frame", row.names = c(NA, -10L))



Hope this helps,

Rui Barradas

Às 00:56 de 12/01/22, Jeff Reichman escreveu:
> R-Help Forum
> 
>   
> 
> Looking for a little guidance. Have an issue were I'm trying to determine
> the time between when Event A happened(In days) to when a subsequent Event B
> happens. For Example at Time 1 Evat A happens and subsequently Event B
> happens at the same day (0) and the next day (1) then Event A happens again
> at time 4 and Event B happens the next day and 3 days later so on and so
> forth. I gather there is no function that will do that so I suspect I will
> need to grate so sour of do while loop?  Any suggestions?
> 
>   
> 
>   
> 
> Time      Event_A               Event_B               Time_B
> 
> 1              1                              1
> 0
> 
> 2              0                              1
> 1
> 
> 3              0                              0
> 0
> 
> 4              1                              0
> 0
> 
> 5              0                              1
> 1
> 
> 6              0                              0
> 0
> 
> 7              0                              1
> 3
> 
> 8              1                              1
> 0
> 
> 9              0                              0
> 0
> 
> 10           0                              1                              2
> 
> 
>   
> 
> Jeff Reichman
> 
> 
> 	[[alternative HTML version deleted]]
> 
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