How to extract a vector from a list of vectors via matrix in R










1















Problem:



I have a number of vectors. I would like to sign numbers to these vectors starting from 1 to n. Then, I would like to convert these number to a lower.triangular matrix. After that, I would like to extract the vectors comes at the last row of the matrix.



Steps of the problem:



  1. Sign numbers n vectors from 1 to n.


  2. Convert these numbers to a lower.triangular matrix M.


  3. Then, extract the vectors that match the numbers at the last row of the matrix M.


Example:



Assume that I have a list of 10 vectors (X):



X <- list(x1=c(1:3), x2=(2:5), x3=c(4:2), x4=c(5:7), x5=c(12,34,54), x6=c(3:6), x7=c(3:6), x8=c(3,4,5), x9=c(44,56,7), x10=c(34,5,4))


Then, I would like to order them from 1 to 10, where 1 refers to the first vector, and so on. Then, I will have a vector of these numbers, say x = c(1:10). Then, I would like to convert it to a lower triangular matrix M.



 M <- matrix(0,5,5)

> M[lower.tri(M, diag=FALSE)] <- x
> M
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 0 0
[2,] 1 0 0 0 0
[3,] 2 5 0 0 0
[4,] 3 6 8 0 0
[5,] 4 7 9 10 0


Now, I would like to extract, the last row.



> tail(M, 1)
[,1] [,2] [,3] [,4] [,5]
[5,] 4 7 9 10 0

> newX <- as.vector(tail(M,1))
> newX
[1] 4 7 9 10 0


Now the wanted vectors (to be extracted from the whole vectors) are 4, 7, 9, and 10. In other words, I need to extract, x4, x7, x9, and x10.



Hence, I would like to extract the vectors match these number.



Any idea or help, please?










share|improve this question


























    1















    Problem:



    I have a number of vectors. I would like to sign numbers to these vectors starting from 1 to n. Then, I would like to convert these number to a lower.triangular matrix. After that, I would like to extract the vectors comes at the last row of the matrix.



    Steps of the problem:



    1. Sign numbers n vectors from 1 to n.


    2. Convert these numbers to a lower.triangular matrix M.


    3. Then, extract the vectors that match the numbers at the last row of the matrix M.


    Example:



    Assume that I have a list of 10 vectors (X):



    X <- list(x1=c(1:3), x2=(2:5), x3=c(4:2), x4=c(5:7), x5=c(12,34,54), x6=c(3:6), x7=c(3:6), x8=c(3,4,5), x9=c(44,56,7), x10=c(34,5,4))


    Then, I would like to order them from 1 to 10, where 1 refers to the first vector, and so on. Then, I will have a vector of these numbers, say x = c(1:10). Then, I would like to convert it to a lower triangular matrix M.



     M <- matrix(0,5,5)

    > M[lower.tri(M, diag=FALSE)] <- x
    > M
    [,1] [,2] [,3] [,4] [,5]
    [1,] 0 0 0 0 0
    [2,] 1 0 0 0 0
    [3,] 2 5 0 0 0
    [4,] 3 6 8 0 0
    [5,] 4 7 9 10 0


    Now, I would like to extract, the last row.



    > tail(M, 1)
    [,1] [,2] [,3] [,4] [,5]
    [5,] 4 7 9 10 0

    > newX <- as.vector(tail(M,1))
    > newX
    [1] 4 7 9 10 0


    Now the wanted vectors (to be extracted from the whole vectors) are 4, 7, 9, and 10. In other words, I need to extract, x4, x7, x9, and x10.



    Hence, I would like to extract the vectors match these number.



    Any idea or help, please?










    share|improve this question
























      1












      1








      1








      Problem:



      I have a number of vectors. I would like to sign numbers to these vectors starting from 1 to n. Then, I would like to convert these number to a lower.triangular matrix. After that, I would like to extract the vectors comes at the last row of the matrix.



      Steps of the problem:



      1. Sign numbers n vectors from 1 to n.


      2. Convert these numbers to a lower.triangular matrix M.


      3. Then, extract the vectors that match the numbers at the last row of the matrix M.


      Example:



      Assume that I have a list of 10 vectors (X):



      X <- list(x1=c(1:3), x2=(2:5), x3=c(4:2), x4=c(5:7), x5=c(12,34,54), x6=c(3:6), x7=c(3:6), x8=c(3,4,5), x9=c(44,56,7), x10=c(34,5,4))


      Then, I would like to order them from 1 to 10, where 1 refers to the first vector, and so on. Then, I will have a vector of these numbers, say x = c(1:10). Then, I would like to convert it to a lower triangular matrix M.



       M <- matrix(0,5,5)

      > M[lower.tri(M, diag=FALSE)] <- x
      > M
      [,1] [,2] [,3] [,4] [,5]
      [1,] 0 0 0 0 0
      [2,] 1 0 0 0 0
      [3,] 2 5 0 0 0
      [4,] 3 6 8 0 0
      [5,] 4 7 9 10 0


      Now, I would like to extract, the last row.



      > tail(M, 1)
      [,1] [,2] [,3] [,4] [,5]
      [5,] 4 7 9 10 0

      > newX <- as.vector(tail(M,1))
      > newX
      [1] 4 7 9 10 0


      Now the wanted vectors (to be extracted from the whole vectors) are 4, 7, 9, and 10. In other words, I need to extract, x4, x7, x9, and x10.



      Hence, I would like to extract the vectors match these number.



      Any idea or help, please?










      share|improve this question














      Problem:



      I have a number of vectors. I would like to sign numbers to these vectors starting from 1 to n. Then, I would like to convert these number to a lower.triangular matrix. After that, I would like to extract the vectors comes at the last row of the matrix.



      Steps of the problem:



      1. Sign numbers n vectors from 1 to n.


      2. Convert these numbers to a lower.triangular matrix M.


      3. Then, extract the vectors that match the numbers at the last row of the matrix M.


      Example:



      Assume that I have a list of 10 vectors (X):



      X <- list(x1=c(1:3), x2=(2:5), x3=c(4:2), x4=c(5:7), x5=c(12,34,54), x6=c(3:6), x7=c(3:6), x8=c(3,4,5), x9=c(44,56,7), x10=c(34,5,4))


      Then, I would like to order them from 1 to 10, where 1 refers to the first vector, and so on. Then, I will have a vector of these numbers, say x = c(1:10). Then, I would like to convert it to a lower triangular matrix M.



       M <- matrix(0,5,5)

      > M[lower.tri(M, diag=FALSE)] <- x
      > M
      [,1] [,2] [,3] [,4] [,5]
      [1,] 0 0 0 0 0
      [2,] 1 0 0 0 0
      [3,] 2 5 0 0 0
      [4,] 3 6 8 0 0
      [5,] 4 7 9 10 0


      Now, I would like to extract, the last row.



      > tail(M, 1)
      [,1] [,2] [,3] [,4] [,5]
      [5,] 4 7 9 10 0

      > newX <- as.vector(tail(M,1))
      > newX
      [1] 4 7 9 10 0


      Now the wanted vectors (to be extracted from the whole vectors) are 4, 7, 9, and 10. In other words, I need to extract, x4, x7, x9, and x10.



      Hence, I would like to extract the vectors match these number.



      Any idea or help, please?







      r






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      asked Nov 15 '18 at 6:37









      MaryamMaryam

      28911




      28911






















          1 Answer
          1






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          oldest

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          2














          You could use paste:



          X[paste0("x", newX[newX != 0])]

          $`x4`
          [1] 5 6 7

          $x7
          [1] 3 4 5 6

          $x9
          [1] 44 56 7

          $x10
          [1] 34 5 4


          paste0("x", newX[newX != 0] will create the character vector "x4", "x7", "x9", "x10" which you can use for indexing the list.






          share|improve this answer




















          • 1





            Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

            – Maryam
            Nov 15 '18 at 7:58






          • 1





            You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

            – LAP
            Nov 15 '18 at 7:59











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          You could use paste:



          X[paste0("x", newX[newX != 0])]

          $`x4`
          [1] 5 6 7

          $x7
          [1] 3 4 5 6

          $x9
          [1] 44 56 7

          $x10
          [1] 34 5 4


          paste0("x", newX[newX != 0] will create the character vector "x4", "x7", "x9", "x10" which you can use for indexing the list.






          share|improve this answer




















          • 1





            Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

            – Maryam
            Nov 15 '18 at 7:58






          • 1





            You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

            – LAP
            Nov 15 '18 at 7:59















          2














          You could use paste:



          X[paste0("x", newX[newX != 0])]

          $`x4`
          [1] 5 6 7

          $x7
          [1] 3 4 5 6

          $x9
          [1] 44 56 7

          $x10
          [1] 34 5 4


          paste0("x", newX[newX != 0] will create the character vector "x4", "x7", "x9", "x10" which you can use for indexing the list.






          share|improve this answer




















          • 1





            Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

            – Maryam
            Nov 15 '18 at 7:58






          • 1





            You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

            – LAP
            Nov 15 '18 at 7:59













          2












          2








          2







          You could use paste:



          X[paste0("x", newX[newX != 0])]

          $`x4`
          [1] 5 6 7

          $x7
          [1] 3 4 5 6

          $x9
          [1] 44 56 7

          $x10
          [1] 34 5 4


          paste0("x", newX[newX != 0] will create the character vector "x4", "x7", "x9", "x10" which you can use for indexing the list.






          share|improve this answer















          You could use paste:



          X[paste0("x", newX[newX != 0])]

          $`x4`
          [1] 5 6 7

          $x7
          [1] 3 4 5 6

          $x9
          [1] 44 56 7

          $x10
          [1] 34 5 4


          paste0("x", newX[newX != 0] will create the character vector "x4", "x7", "x9", "x10" which you can use for indexing the list.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 15 '18 at 7:02

























          answered Nov 15 '18 at 6:56









          LAPLAP

          5,7902723




          5,7902723







          • 1





            Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

            – Maryam
            Nov 15 '18 at 7:58






          • 1





            You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

            – LAP
            Nov 15 '18 at 7:59












          • 1





            Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

            – Maryam
            Nov 15 '18 at 7:58






          • 1





            You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

            – LAP
            Nov 15 '18 at 7:59







          1




          1





          Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

          – Maryam
          Nov 15 '18 at 7:58





          Thank you so much for your very clever help. I thought I will need some lines of codes to done this.

          – Maryam
          Nov 15 '18 at 7:58




          1




          1





          You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

          – LAP
          Nov 15 '18 at 7:59





          You are welcome! Often enough R surprises you with a really simple solution for a seemingly big problem :)

          – LAP
          Nov 15 '18 at 7:59



















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