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Remove isolated elements of a vector


How to access the last value in a vector?Concatenating two std::vectorsHow to find out if an item is present in a std::vector?How do I erase an element from std::vector<> by index?Test if a vector contains a given elementWhy is Java Vector (and Stack) class considered obsolete or deprecated?Counting the number of elements with the values of x in a vectorWhat is the easiest way to initialize a std::vector with hardcoded elements?Appending a vector to a vectorReshaping vector with indices in python













9















I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
The components in the vector are ordered increasingly, and there are no repetitions.



For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.



I've tried applying



   for (p in 1:(length(index)-2))
if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}


index<-index[index!=0]


where index is my vector of interest, but there's some problem with the logical condition.
Could you please give me some hints?



Thanks in advance.










share|improve this question





























    9















    I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
    What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
    The components in the vector are ordered increasingly, and there are no repetitions.



    For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.



    I've tried applying



       for (p in 1:(length(index)-2))
    if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}


    index<-index[index!=0]


    where index is my vector of interest, but there's some problem with the logical condition.
    Could you please give me some hints?



    Thanks in advance.










    share|improve this question



























      9












      9








      9








      I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
      What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
      The components in the vector are ordered increasingly, and there are no repetitions.



      For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.



      I've tried applying



         for (p in 1:(length(index)-2))
      if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}


      index<-index[index!=0]


      where index is my vector of interest, but there's some problem with the logical condition.
      Could you please give me some hints?



      Thanks in advance.










      share|improve this question
















      I have a vector of integers and I want to filter it by eliminating the components that are "isolated".
      What do I mean by "isolated"? those components that does not lie in an 4-neighbourhood of other component.
      The components in the vector are ordered increasingly, and there are no repetitions.



      For example if I have c(1,2,3,8,15,16,17) then I need to eliminate 8 because is not in a 4-neighbourhood of other element.



      I've tried applying



         for (p in 1:(length(index)-2))
      if((index[p+1]>3+index[p])&(index[p+2]>3+index[p+1])){index[p+1]<-0}


      index<-index[index!=0]


      where index is my vector of interest, but there's some problem with the logical condition.
      Could you please give me some hints?



      Thanks in advance.







      r if-statement vector






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 hours ago







      Ramiro Scorolli

















      asked 2 hours ago









      Ramiro ScorolliRamiro Scorolli

      1697




      1697
























          2 Answers
          2






          active

          oldest

          votes


















          6














          You can achieve it with a combination of outer and colSums, i.e.



          x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
          #[1] 8


          To eliminate the values, we can do,



          i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
          x[!i1]
          #[1] 1 2 3 15 16 17


          where,



          x <- c(1,2,3,8,15,16,17)





          share|improve this answer

































            3














            We keep values where preceding or next difference is lower or equal to 4 :



            v <- c(1,2,3,8,15,16,17)
            v[c(FALSE, abs(diff(v)) <= 4) | c(abs(diff(v)) <= 4, FALSE)]





            share|improve this answer























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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

              oldest

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              6














              You can achieve it with a combination of outer and colSums, i.e.



              x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
              #[1] 8


              To eliminate the values, we can do,



              i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
              x[!i1]
              #[1] 1 2 3 15 16 17


              where,



              x <- c(1,2,3,8,15,16,17)





              share|improve this answer






























                6














                You can achieve it with a combination of outer and colSums, i.e.



                x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
                #[1] 8


                To eliminate the values, we can do,



                i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
                x[!i1]
                #[1] 1 2 3 15 16 17


                where,



                x <- c(1,2,3,8,15,16,17)





                share|improve this answer




























                  6












                  6








                  6







                  You can achieve it with a combination of outer and colSums, i.e.



                  x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
                  #[1] 8


                  To eliminate the values, we can do,



                  i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
                  x[!i1]
                  #[1] 1 2 3 15 16 17


                  where,



                  x <- c(1,2,3,8,15,16,17)





                  share|improve this answer















                  You can achieve it with a combination of outer and colSums, i.e.



                  x[colSums(abs(outer(x, x, `-`)) >= 4) == length(x)-1]
                  #[1] 8


                  To eliminate the values, we can do,



                  i1 <- colSums(outer(x, x, FUN = function(i, j) abs(i - j) >= 4)) == length(x) - 1
                  x[!i1]
                  #[1] 1 2 3 15 16 17


                  where,



                  x <- c(1,2,3,8,15,16,17)






                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 2 hours ago

























                  answered 2 hours ago









                  SotosSotos

                  30.1k51640




                  30.1k51640

























                      3














                      We keep values where preceding or next difference is lower or equal to 4 :



                      v <- c(1,2,3,8,15,16,17)
                      v[c(FALSE, abs(diff(v)) <= 4) | c(abs(diff(v)) <= 4, FALSE)]





                      share|improve this answer




























                        3














                        We keep values where preceding or next difference is lower or equal to 4 :



                        v <- c(1,2,3,8,15,16,17)
                        v[c(FALSE, abs(diff(v)) <= 4) | c(abs(diff(v)) <= 4, FALSE)]





                        share|improve this answer


























                          3












                          3








                          3







                          We keep values where preceding or next difference is lower or equal to 4 :



                          v <- c(1,2,3,8,15,16,17)
                          v[c(FALSE, abs(diff(v)) <= 4) | c(abs(diff(v)) <= 4, FALSE)]





                          share|improve this answer













                          We keep values where preceding or next difference is lower or equal to 4 :



                          v <- c(1,2,3,8,15,16,17)
                          v[c(FALSE, abs(diff(v)) <= 4) | c(abs(diff(v)) <= 4, FALSE)]






                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 2 hours ago









                          ClemsangClemsang

                          32429




                          32429






























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