Cut elements from the beginning and end of an R vector












4















For time series analysis I handle data that often contains leading and trailing zero elements. In this example, there are 3 zeros at the beginning an 2 at the end. I want to get rid of these elements, and filter for the contents in the middle (that also may contain zeros)



vec <- c(0, 0, 0, 1, 2, 0, 3, 4, 0, 0)


I did this by looping from the beginning and end, and masking out the unwanted elements.



mask <- rep(TRUE, length(vec))



# from begin

i <- 1

while(vec[i] == 0 && i <= length(vec)) {

  mask[i] <- FALSE

  i <- i+1

}



# from end

i <- length(vec)

while(i >= 1 && vec[i] == 0) {

  mask[i] <- FALSE

  i <- i-1

}



cleanvec <- vec[mask]

cleanvec

[1] 1 2 0 3 4


This works, but I wonder if there is a more efficient way to do this, avoiding the loops.






r







share|improve this question























  • Maybe this: stackoverflow.com/questions/32581950/…

    – Andreas Neumeier
    Dec 30 '18 at 17:15











  • This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

    – Ric S
    Dec 30 '18 at 17:32
















4















For time series analysis I handle data that often contains leading and trailing zero elements. In this example, there are 3 zeros at the beginning an 2 at the end. I want to get rid of these elements, and filter for the contents in the middle (that also may contain zeros)



vec <- c(0, 0, 0, 1, 2, 0, 3, 4, 0, 0)


I did this by looping from the beginning and end, and masking out the unwanted elements.



mask <- rep(TRUE, length(vec))



# from begin

i <- 1

while(vec[i] == 0 && i <= length(vec)) {

  mask[i] <- FALSE

  i <- i+1

}



# from end

i <- length(vec)

while(i >= 1 && vec[i] == 0) {

  mask[i] <- FALSE

  i <- i-1

}



cleanvec <- vec[mask]

cleanvec

[1] 1 2 0 3 4


This works, but I wonder if there is a more efficient way to do this, avoiding the loops.






r







share|improve this question























  • Maybe this: stackoverflow.com/questions/32581950/…

    – Andreas Neumeier
    Dec 30 '18 at 17:15











  • This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

    – Ric S
    Dec 30 '18 at 17:32














4












4








4








For time series analysis I handle data that often contains leading and trailing zero elements. In this example, there are 3 zeros at the beginning an 2 at the end. I want to get rid of these elements, and filter for the contents in the middle (that also may contain zeros)



vec <- c(0, 0, 0, 1, 2, 0, 3, 4, 0, 0)


I did this by looping from the beginning and end, and masking out the unwanted elements.



mask <- rep(TRUE, length(vec))



# from begin

i <- 1

while(vec[i] == 0 && i <= length(vec)) {

  mask[i] <- FALSE

  i <- i+1

}



# from end

i <- length(vec)

while(i >= 1 && vec[i] == 0) {

  mask[i] <- FALSE

  i <- i-1

}



cleanvec <- vec[mask]

cleanvec

[1] 1 2 0 3 4


This works, but I wonder if there is a more efficient way to do this, avoiding the loops.






r







share|improve this question














For time series analysis I handle data that often contains leading and trailing zero elements. In this example, there are 3 zeros at the beginning an 2 at the end. I want to get rid of these elements, and filter for the contents in the middle (that also may contain zeros)



vec <- c(0, 0, 0, 1, 2, 0, 3, 4, 0, 0)


I did this by looping from the beginning and end, and masking out the unwanted elements.



mask <- rep(TRUE, length(vec))



# from begin

i <- 1

while(vec[i] == 0 && i <= length(vec)) {

  mask[i] <- FALSE

  i <- i+1

}



# from end

i <- length(vec)

while(i >= 1 && vec[i] == 0) {

  mask[i] <- FALSE

  i <- i-1

}



cleanvec <- vec[mask]

cleanvec

[1] 1 2 0 3 4


This works, but I wonder if there is a more efficient way to do this, avoiding the loops.






r




r






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 30 '18 at 17:13









wotuzu17wotuzu17

977




977













  • Maybe this: stackoverflow.com/questions/32581950/…

    – Andreas Neumeier
    Dec 30 '18 at 17:15











  • This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

    – Ric S
    Dec 30 '18 at 17:32



















  • Maybe this: stackoverflow.com/questions/32581950/…

    – Andreas Neumeier
    Dec 30 '18 at 17:15











  • This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

    – Ric S
    Dec 30 '18 at 17:32

















Maybe this: stackoverflow.com/questions/32581950/…

– Andreas Neumeier
Dec 30 '18 at 17:15





Maybe this: stackoverflow.com/questions/32581950/…

– Andreas Neumeier
Dec 30 '18 at 17:15













This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

– Ric S
Dec 30 '18 at 17:32





This is referred to lists, not vectors; furthermore, it removes all zeros in the list elements, whereas here the question asks for removing zeros at the beginning and at the end of the vector.

– Ric S
Dec 30 '18 at 17:32












3 Answers
3






active

oldest

votes


















6














vec[ min(which(vec != 0)) : max(which(vec != 0)) ]


Basically the which(vec != 0) part gives the positions of the numbers that are different from 0, and then you take the min and max of them.






share|improve this answer



















  • 1





    good solution, but this will cause an error if vec consists only of zeros.

    – wotuzu17
    Dec 30 '18 at 18:26











  • You are right, maybe you can set an if statement to check whether that conditions occurs.

    – Ric S
    Dec 30 '18 at 19:15



















0














We could use the range and Reduce to get the sequence



vec[Reduce(`:`, range(which(vec != 0)))]

#[1] 1 2 0 3 4





share|improve this answer































    0














    Take the cumsum forward and backward of abs(vec) and keep only elements > 0. if it were known that all elements of vec were non-negative, as in the question, then we could optionally omit abs.



    vec[cumsum(abs(vec)) > 0 & rev(cumsum(rev(abs(vec)))) > 0]
    
## [1] 1 2 0 3 4





    share|improve this answer

























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      6














      vec[ min(which(vec != 0)) : max(which(vec != 0)) ]


      Basically the which(vec != 0) part gives the positions of the numbers that are different from 0, and then you take the min and max of them.






      share|improve this answer



















      • 1





        good solution, but this will cause an error if vec consists only of zeros.

        – wotuzu17
        Dec 30 '18 at 18:26











      • You are right, maybe you can set an if statement to check whether that conditions occurs.

        – Ric S
        Dec 30 '18 at 19:15
















      6














      vec[ min(which(vec != 0)) : max(which(vec != 0)) ]


      Basically the which(vec != 0) part gives the positions of the numbers that are different from 0, and then you take the min and max of them.






      share|improve this answer



















      • 1





        good solution, but this will cause an error if vec consists only of zeros.

        – wotuzu17
        Dec 30 '18 at 18:26











      • You are right, maybe you can set an if statement to check whether that conditions occurs.

        – Ric S
        Dec 30 '18 at 19:15














      6












      6








      6







      vec[ min(which(vec != 0)) : max(which(vec != 0)) ]


      Basically the which(vec != 0) part gives the positions of the numbers that are different from 0, and then you take the min and max of them.






      share|improve this answer













      vec[ min(which(vec != 0)) : max(which(vec != 0)) ]


      Basically the which(vec != 0) part gives the positions of the numbers that are different from 0, and then you take the min and max of them.







      share|improve this answer












      share|improve this answer



      share|improve this answer










      answered Dec 30 '18 at 17:23









      Ric SRic S

      664318




      664318








      • 1





        good solution, but this will cause an error if vec consists only of zeros.

        – wotuzu17
        Dec 30 '18 at 18:26











      • You are right, maybe you can set an if statement to check whether that conditions occurs.

        – Ric S
        Dec 30 '18 at 19:15














      • 1





        good solution, but this will cause an error if vec consists only of zeros.

        – wotuzu17
        Dec 30 '18 at 18:26











      • You are right, maybe you can set an if statement to check whether that conditions occurs.

        – Ric S
        Dec 30 '18 at 19:15








      1




      1





      good solution, but this will cause an error if vec consists only of zeros.

      – wotuzu17
      Dec 30 '18 at 18:26





      good solution, but this will cause an error if vec consists only of zeros.

      – wotuzu17
      Dec 30 '18 at 18:26













      You are right, maybe you can set an if statement to check whether that conditions occurs.

      – Ric S
      Dec 30 '18 at 19:15





      You are right, maybe you can set an if statement to check whether that conditions occurs.

      – Ric S
      Dec 30 '18 at 19:15













      0














      We could use the range and Reduce to get the sequence



      vec[Reduce(`:`, range(which(vec != 0)))]
      
#[1] 1 2 0 3 4





      share|improve this answer




























        0














        We could use the range and Reduce to get the sequence



        vec[Reduce(`:`, range(which(vec != 0)))]
        
#[1] 1 2 0 3 4





        share|improve this answer


























          0












          0








          0







          We could use the range and Reduce to get the sequence



          vec[Reduce(`:`, range(which(vec != 0)))]
          
#[1] 1 2 0 3 4





          share|improve this answer













          We could use the range and Reduce to get the sequence



          vec[Reduce(`:`, range(which(vec != 0)))]
          
#[1] 1 2 0 3 4






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 30 '18 at 18:46









          akrunakrun

          405k13196269




          405k13196269























              0














              Take the cumsum forward and backward of abs(vec) and keep only elements > 0. if it were known that all elements of vec were non-negative, as in the question, then we could optionally omit abs.



              vec[cumsum(abs(vec)) > 0 & rev(cumsum(rev(abs(vec)))) > 0]
              
## [1] 1 2 0 3 4





              share|improve this answer






























                0














                Take the cumsum forward and backward of abs(vec) and keep only elements > 0. if it were known that all elements of vec were non-negative, as in the question, then we could optionally omit abs.



                vec[cumsum(abs(vec)) > 0 & rev(cumsum(rev(abs(vec)))) > 0]
                
## [1] 1 2 0 3 4





                share|improve this answer




























                  0












                  0








                  0







                  Take the cumsum forward and backward of abs(vec) and keep only elements > 0. if it were known that all elements of vec were non-negative, as in the question, then we could optionally omit abs.



                  vec[cumsum(abs(vec)) > 0 & rev(cumsum(rev(abs(vec)))) > 0]
                  
## [1] 1 2 0 3 4





                  share|improve this answer















                  Take the cumsum forward and backward of abs(vec) and keep only elements > 0. if it were known that all elements of vec were non-negative, as in the question, then we could optionally omit abs.



                  vec[cumsum(abs(vec)) > 0 & rev(cumsum(rev(abs(vec)))) > 0]
                  
## [1] 1 2 0 3 4






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 30 '18 at 23:52

























                  answered Dec 30 '18 at 22:42









                  G. GrothendieckG. Grothendieck

                  147k9130235




                  147k9130235






























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