CODEWITHSUNDEEP
rmatrixsplittime-series
Lorenzo Lorenzi
Lorenzo Lorenzi

Reputation: 43

Splitting Matrix into multiple couples of columns in R

using R I have a 10 rows X 6 columns matrix. I need to split it into submatrices throuhg gruoping couples of columns with no overlapping.

i.e. matrix has columns A,B,C,D,E,F and I need to extract 3 different matrices (or data.frames or whatever object within financial packages like zoo or timeSeries) formed by columns AB, CD and EF.

PS: matrix contains financial data series and any couple of columns has a date columns and a NAV column

Upvotes: 4

Views: 6912

Answers (3)

nisetama
nisetama

Reputation: 8893

Another option is to first transpose the matrix and then use split.data.frame to split the matrix by rows:

> t(matrix(1:60,10))|>split.data.frame(rep(1:3,each=2))
$`1`
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]    1    2    3    4    5    6    7    8    9    10
[2,]   11   12   13   14   15   16   17   18   19    20

$`2`
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]   21   22   23   24   25   26   27   28   29    30
[2,]   31   32   33   34   35   36   37   38   39    40

$`3`
     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]   41   42   43   44   45   46   47   48   49    50
[2,]   51   52   53   54   55   56   57   58   59    60

Upvotes: 0

Gavin Simpson
Gavin Simpson

Reputation: 174803

Using some dummy data (note you must have a dataframe as otherwise R would not allow you to hold Date and numeric values in a matrix [unless they were all converted to characters or raw numeric representations])

set.seed(42)
df <- data.frame(A = Sys.Date() + 0:9, B = rnorm(10),
                 C = Sys.Date() - 0:9, D = rnorm(10),
                 E = Sys.Date() - 20:29, F = rnorm(10))

> head(df)
           A          B          C          D          E          F
1 2013-04-05  1.3709584 2013-04-05  1.3048697 2013-03-16 -0.3066386
2 2013-04-06 -0.5646982 2013-04-04  2.2866454 2013-03-15 -1.7813084
3 2013-04-07  0.3631284 2013-04-03 -1.3888607 2013-03-14 -0.1719174
4 2013-04-08  0.6328626 2013-04-02 -0.2787888 2013-03-13  1.2146747
5 2013-04-09  0.4042683 2013-04-01 -0.1333213 2013-03-12  1.8951935
6 2013-04-10 -0.1061245 2013-03-31  0.6359504 2013-03-11 -0.4304691

one easy way to do this is to form an index for the columns you want - here I chose the first column of each pair, 1, 3, 5, etc.

start <- seq(1, by = 2, length = ncol(df) / 2)

Then, we lapply over the indices in start and select from our data frame the ith and ith + 1 columns where i takes each index from start in turn (df[i:(i+1)])

sdf <- lapply(start, function(i, df) df[i:(i+1)], df = df)

which gives:

> sdf
[[1]]
            A           B
1  2013-04-05  1.37095845
2  2013-04-06 -0.56469817
3  2013-04-07  0.36312841
4  2013-04-08  0.63286260
5  2013-04-09  0.40426832
6  2013-04-10 -0.10612452
7  2013-04-11  1.51152200
8  2013-04-12 -0.09465904
9  2013-04-13  2.01842371
10 2013-04-14 -0.06271410

[[2]]
            C          D
1  2013-04-05  1.3048697
2  2013-04-04  2.2866454
....

> str(sdf)
List of 3
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ A: Date[1:10], format: "2013-04-05" "2013-04-06" ...
  ..$ B: num [1:10] 1.371 -0.565 0.363 0.633 0.404 ...
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ C: Date[1:10], format: "2013-04-05" "2013-04-04" ...
  ..$ D: num [1:10] 1.305 2.287 -1.389 -0.279 -0.133 ...
 $ :'data.frame':   10 obs. of  2 variables:
  ..$ E: Date[1:10], format: "2013-03-16" "2013-03-15" ...
  ..$ F: num [1:10] -0.307 -1.781 -0.172 1.215 1.895 ...

An advantage of keeping the sub-data frames in a list is that you can apply a function or other operation to the sub-data frames using a loop or a tools like lapply or sapply for example.

Upvotes: 5

user1981275
user1981275

Reputation: 13370

Are you looking for something like this?

require("zoo")
##matrix with random entries
mat <- matrix(rnorm(60), nrow=10, ncol=6)
colnames(mat) <- LETTERS[1:6]    
## optional: create zoo object 
#mat <- as.zoo(mat) 


##access columns
mat[,c("A", "B")]
mat[,c("C", "D")]
mat[,c("E", "F")]

Note that creating a zoo object for the mock data above is not necessary, but from your question it is a bit unclear how your data looks like.

Upvotes: 1

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