通过将行添加到具有相同模式的R中的矩阵进行扩展

Question

I have matrix, but want to extend it with the same pattern. Note that it may be extended for any given number of rows and columns, and is not normally square

    04/06/2012  11/06/2012  18/06/2012  25/06/2012  02/07/2012
26/03/2012  10  11  12  13  14
02/04/2012  9   10  11  12  13
09/04/2012  8   9   10  11  12
16/04/2012  7   8   9   10  11
23/04/2012  6   7   8   9   10
30/04/2012  5   6   7   8   9
07/05/2012  4   5   6   7   8
14/05/2012  3   4   5   6   7
21/05/2012  2   3   4   5   6
28/05/2012  1   2   3   4   5

Ie I want to extend it to something like this:

    04/06/2012  11/06/2012  18/06/2012  25/06/2012  02/07/2012
26/03/2012  10  11  12  13  14
02/04/2012  9   10  11  12  13
09/04/2012  8   9   10  11  12
16/04/2012  7   8   9   10  11
23/04/2012  6   7   8   9   10
30/04/2012  5   6   7   8   9
07/05/2012  4   5   6   7   8
14/05/2012  3   4   5   6   7
21/05/2012  2   3   4   5   6
28/05/2012  1   2   3   4   5
04/06/2012  0   1   2   3   4
11/06/2012  NA  0   1   2   3
18/06/2012  NA  NA  0   1   2
25/06/2012  NA  NA  NA  0   1
02/07/2012  NA  NA  NA  NA  0

Answer 1

I'm sure there's a clever way to do this with Reduce or something, but this is what came to mind:

lengthOut <- 6   ## Set to one less than the number of columns you want to create
startAt <- 10    ## Set the maximum value of the FIRST column
vapply(c(0, sequence(lengthOut)), function(x) { 
  x <- (startAt + x):0                   # Create a sequence in the normal manner
  length(x) <- startAt + lengthOut + 1   # Extend the length of that sequence
  x
}, numeric(startAt + lengthOut + 1))     # Specify what to return
#       [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#  [1,]   10   11   12   13   14   15   16
#  [2,]    9   10   11   12   13   14   15
#  [3,]    8    9   10   11   12   13   14
#  [4,]    7    8    9   10   11   12   13 
#  [5,]    6    7    8    9   10   11   12
#  [6,]    5    6    7    8    9   10   11
#  [7,]    4    5    6    7    8    9   10
#  [8,]    3    4    5    6    7    8    9
#  [9,]    2    3    4    5    6    7    8
# [10,]    1    2    3    4    5    6    7
# [11,]    0    1    2    3    4    5    6
# [12,]   NA    0    1    2    3    4    5
# [13,]   NA   NA    0    1    2    3    4
# [14,]   NA   NA   NA    0    1    2    3
# [15,]   NA   NA   NA   NA    0    1    2
# [16,]   NA   NA   NA   NA   NA    0    1
# [17,]   NA   NA   NA   NA   NA   NA    0

Answer 2

Here's another approach

x <- 16:0
matrix(c(sapply(6:1, function(z) rep(lead(x, z))), x), ncol=7)

#      [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#[1,]   10   11   12   13   14   15   16
#[2,]    9   10   11   12   13   14   15
#[3,]    8    9   10   11   12   13   14
#[4,]    7    8    9   10   11   12   13
#[5,]    6    7    8    9   10   11   12
#[6,]    5    6    7    8    9   10   11
#[7,]    4    5    6    7    8    9   10
#[8,]    3    4    5    6    7    8    9
#[9,]    2    3    4    5    6    7    8
#[10,]    1    2    3    4    5    6    7
#[11,]    0    1    2    3    4    5    6
#[12,]   NA    0    1    2    3    4    5
#[13,]   NA   NA    0    1    2    3    4
#[14,]   NA   NA   NA    0    1    2    3
#[15,]   NA   NA   NA   NA    0    1    2
#[16,]   NA   NA   NA   NA   NA    0    1
#[17,]   NA   NA   NA   NA   NA   NA    0

Edit: forgot to mention that I used dplyr::lead

Answer 3

Not sure if this helps:

m1 <- matrix(rep(10:1,each=7)+0:6,ncol=7,byrow=T)
m2 <- matrix(NA,ncol=7,nrow=7)
indx <- 0:6+rep(c(0:-6),each=7)

m2[lower.tri(m2, diag=TRUE)] <- indx[indx>=0]
rbind(m1,t(m2))
#       [,1] [,2] [,3] [,4] [,5] [,6] [,7]
# [1,]   10   11   12   13   14   15   16
# [2,]    9   10   11   12   13   14   15
# [3,]    8    9   10   11   12   13   14
# [4,]    7    8    9   10   11   12   13
# [5,]    6    7    8    9   10   11   12
# [6,]    5    6    7    8    9   10   11
# [7,]    4    5    6    7    8    9   10
# [8,]    3    4    5    6    7    8    9
# [9,]    2    3    4    5    6    7    8
# [10,]   1    2    3    4    5    6    7
# [11,]   0    1    2    3    4    5    6
# [12,]  NA    0    1    2    3    4    5
# [13,]  NA   NA    0    1    2    3    4
# [14,]  NA   NA   NA    0    1    2    3
# [15,]  NA   NA   NA   NA    0    1    2
# [16,]  NA   NA   NA   NA   NA    0    1
# [17,]  NA   NA   NA   NA   NA   NA    0