How to obtain a position of last non-zero element





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16















I've got a binary variable representing if event happened or not:



event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)


I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:



last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)


How can I obtain that with base R, tidyverse or any other way?






r tidyverse base







share|improve this question





























    16















    I've got a binary variable representing if event happened or not:



    event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)


    I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:



    last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)


    How can I obtain that with base R, tidyverse or any other way?






    r tidyverse base







    share|improve this question

























      16












      16








      16


      1






      I've got a binary variable representing if event happened or not:



      event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)


      I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:



      last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)


      How can I obtain that with base R, tidyverse or any other way?






      r tidyverse base







      share|improve this question














      I've got a binary variable representing if event happened or not:



      event <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0)


      I need to obtain a variable that would indicate the time when the last event happened. The expected output would be:



      last_event <- c(0, 0, 0, 0, 5, 5, 5, 5, 5, 5, 5, 5, 13, 13, 13, 13)


      How can I obtain that with base R, tidyverse or any other way?






      r tidyverse base




      r tidyverse base






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked yesterday









      jakesjakes

      431315




      431315
























          4 Answers
          4






          active

          oldest

          votes


















          17














          Taking advantage of the fact that you have a binary vector, the following gives your desired output:



          cummax(seq_along(event) * event)





          share|improve this answer



















          • 6





            Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

            – Konrad Rudolph
            yesterday






          • 3





            or without multiplication cummax(ifelse(event, seq_along(event), 0))

            – jogo
            yesterday











          • @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

            – Konrad Rudolph
            yesterday



















          8














          Whenever you need to fill repetitions with a value, think run-length encoding.



          In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:



          lengths = rle(event == 0)$lengths
          
nonzeros = which(event != 0)
          
runs = c(0, rep(nonzeros, each = 2))
          
result = rep(runs, lengths)


          Alternative, substitute the runs in the RLE and then inverse it:



          rle = rle(event == 0)
          
nonzeros = which(event != 0)
          
rle$values = c(0, rep(nonzeros, each = 2))
          
result = inverse.rle(rle)





          share|improve this answer































            1














            You can also do somthing like this-



            > zero.locf <- function(x) {
            
  v <- x!=0
            
  c(0, x[v])[cumsum(v)+1]
            
}
            

            
> zero.locf(1:length(event)*event)
            

            
[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





            share|improve this answer































              1














              Another option is to find the index where event == 1 and repeat it based on length.



              rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
              
#[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





              share|improve this answer
























                Your Answer






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                4 Answers
                4






                active

                oldest

                votes








                4 Answers
                4






                active

                oldest

                votes









                active

                oldest

                votes






                active

                oldest

                votes









                17














                Taking advantage of the fact that you have a binary vector, the following gives your desired output:



                cummax(seq_along(event) * event)





                share|improve this answer



















                • 6





                  Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                  – Konrad Rudolph
                  yesterday






                • 3





                  or without multiplication cummax(ifelse(event, seq_along(event), 0))

                  – jogo
                  yesterday











                • @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                  – Konrad Rudolph
                  yesterday
















                17














                Taking advantage of the fact that you have a binary vector, the following gives your desired output:



                cummax(seq_along(event) * event)





                share|improve this answer



















                • 6





                  Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                  – Konrad Rudolph
                  yesterday






                • 3





                  or without multiplication cummax(ifelse(event, seq_along(event), 0))

                  – jogo
                  yesterday











                • @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                  – Konrad Rudolph
                  yesterday














                17












                17








                17







                Taking advantage of the fact that you have a binary vector, the following gives your desired output:



                cummax(seq_along(event) * event)





                share|improve this answer













                Taking advantage of the fact that you have a binary vector, the following gives your desired output:



                cummax(seq_along(event) * event)






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered yesterday









                mgiormentimgiormenti

                394111




                394111








                • 6





                  Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                  – Konrad Rudolph
                  yesterday






                • 3





                  or without multiplication cummax(ifelse(event, seq_along(event), 0))

                  – jogo
                  yesterday











                • @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                  – Konrad Rudolph
                  yesterday














                • 6





                  Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                  – Konrad Rudolph
                  yesterday






                • 3





                  or without multiplication cummax(ifelse(event, seq_along(event), 0))

                  – jogo
                  yesterday











                • @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                  – Konrad Rudolph
                  yesterday








                6




                6





                Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                – Konrad Rudolph
                yesterday





                Yes! So much more elegant than my solution. I was thinking about cumulative sums but I didn’t think of multiplying the indices by the binary vector.

                – Konrad Rudolph
                yesterday




                3




                3





                or without multiplication cummax(ifelse(event, seq_along(event), 0))

                – jogo
                yesterday





                or without multiplication cummax(ifelse(event, seq_along(event), 0))

                – jogo
                yesterday













                @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                – Konrad Rudolph
                yesterday





                @jogo That solution makes sense if the type of event is logical. It does work even for a numeric vector due to R’s implicit conversions but … eh.

                – Konrad Rudolph
                yesterday













                8














                Whenever you need to fill repetitions with a value, think run-length encoding.



                In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:



                lengths = rle(event == 0)$lengths
                
nonzeros = which(event != 0)
                
runs = c(0, rep(nonzeros, each = 2))
                
result = rep(runs, lengths)


                Alternative, substitute the runs in the RLE and then inverse it:



                rle = rle(event == 0)
                
nonzeros = which(event != 0)
                
rle$values = c(0, rep(nonzeros, each = 2))
                
result = inverse.rle(rle)





                share|improve this answer




























                  8














                  Whenever you need to fill repetitions with a value, think run-length encoding.



                  In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:



                  lengths = rle(event == 0)$lengths
                  
nonzeros = which(event != 0)
                  
runs = c(0, rep(nonzeros, each = 2))
                  
result = rep(runs, lengths)


                  Alternative, substitute the runs in the RLE and then inverse it:



                  rle = rle(event == 0)
                  
nonzeros = which(event != 0)
                  
rle$values = c(0, rep(nonzeros, each = 2))
                  
result = inverse.rle(rle)





                  share|improve this answer


























                    8












                    8








                    8







                    Whenever you need to fill repetitions with a value, think run-length encoding.



                    In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:



                    lengths = rle(event == 0)$lengths
                    
nonzeros = which(event != 0)
                    
runs = c(0, rep(nonzeros, each = 2))
                    
result = rep(runs, lengths)


                    Alternative, substitute the runs in the RLE and then inverse it:



                    rle = rle(event == 0)
                    
nonzeros = which(event != 0)
                    
rle$values = c(0, rep(nonzeros, each = 2))
                    
result = inverse.rle(rle)





                    share|improve this answer













                    Whenever you need to fill repetitions with a value, think run-length encoding.



                    In this case, you can determine the run lengths and then repeat the indices of count == 0 an according number of times:



                    lengths = rle(event == 0)$lengths
                    
nonzeros = which(event != 0)
                    
runs = c(0, rep(nonzeros, each = 2))
                    
result = rep(runs, lengths)


                    Alternative, substitute the runs in the RLE and then inverse it:



                    rle = rle(event == 0)
                    
nonzeros = which(event != 0)
                    
rle$values = c(0, rep(nonzeros, each = 2))
                    
result = inverse.rle(rle)






                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered yesterday









                    Konrad RudolphKonrad Rudolph

                    404k1017921041




                    404k1017921041























                        1














                        You can also do somthing like this-



                        > zero.locf <- function(x) {
                        
  v <- x!=0
                        
  c(0, x[v])[cumsum(v)+1]
                        
}
                        

                        
> zero.locf(1:length(event)*event)
                        

                        
[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                        share|improve this answer




























                          1














                          You can also do somthing like this-



                          > zero.locf <- function(x) {
                          
  v <- x!=0
                          
  c(0, x[v])[cumsum(v)+1]
                          
}
                          

                          
> zero.locf(1:length(event)*event)
                          

                          
[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                          share|improve this answer


























                            1












                            1








                            1







                            You can also do somthing like this-



                            > zero.locf <- function(x) {
                            
  v <- x!=0
                            
  c(0, x[v])[cumsum(v)+1]
                            
}
                            

                            
> zero.locf(1:length(event)*event)
                            

                            
[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                            share|improve this answer













                            You can also do somthing like this-



                            > zero.locf <- function(x) {
                            
  v <- x!=0
                            
  c(0, x[v])[cumsum(v)+1]
                            
}
                            

                            
> zero.locf(1:length(event)*event)
                            

                            
[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13






                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered yesterday









                            RushabhRushabh

                            1,299221




                            1,299221























                                1














                                Another option is to find the index where event == 1 and repeat it based on length.



                                rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
                                
#[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                                share|improve this answer




























                                  1














                                  Another option is to find the index where event == 1 and repeat it based on length.



                                  rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
                                  
#[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                                  share|improve this answer


























                                    1












                                    1








                                    1







                                    Another option is to find the index where event == 1 and repeat it based on length.



                                    rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
                                    
#[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13





                                    share|improve this answer













                                    Another option is to find the index where event == 1 and repeat it based on length.



                                    rep(c(0, which(event == 1)), tapply(event, cumsum(event == 1), length))
                                    
#[1]  0  0  0  0  5  5  5  5  5  5  5  5 13 13 13 13






                                    share|improve this answer












                                    share|improve this answer



                                    share|improve this answer










                                    answered yesterday









                                    Ronak ShahRonak Shah

                                    46.6k104269




                                    46.6k104269






























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