CODEWITHSUNDEEP
r
user2806363
user2806363

Reputation: 2593

How to reorder the rows of one matrix with respect to the other matrix?

I have two big matrices A and B with different dimensions. I want to order the rows of matrix B with respect to rows of the matrix A. and add the rows with values 0 to matrix B, if that row does not exist in B but in A

Here is the reproduceable example and expected output:

A<-matrix(c(1:40), ncol=8)
rownames(A)<-c("B", "A", "C", "D", "E")

> A
  [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
B    1    6   11   16   21   26   31   36
A    2    7   12   17   22   27   32   37
C    3    8   13   18   23   28   33   38
D    4    9   14   19   24   29   34   39
E    5   10   15   20   25   30   35   40

> B<-matrix(c(100:108),ncol=3)
 rownames(B)<-c("A", "E", "C")

> B
  [,1] [,2] [,3]
A  100  103  106
E  101  104  107
C  102  105  108

Here is the Expected output :

>B
      [,1] [,2] [,3]
    B    0    0    0
    A  100  103  106
    C  102  105  108
    D    0    0    0
    E  101  104  107
    >

Would someone help me to implement this in R ?

Upvotes: 2

Views: 87

Answers (3)

Ted Petrou
Ted Petrou

Reputation: 61947

Coming from a sql background this is what first came into my head. The other answers are much better.

A1 = cbind(id = rownames(A),as.data.frame(A), stringsAsFactors = FALSE)
B1 = cbind(id = rownames(B),as.data.frame(B), stringsAsFactors = FALSE)
AB = merge(A1, B1, by = "id", all = T)
AB1 = as.matrix(AB[,(dim(A1)[2] + 1) : dim(AB)[2]])
dimnames(AB1) = NULL
AB1[is.na(AB1)] = 0
rownames(AB1) = AB[,1]
(B = AB1[match(AB[,1],A1[,1]),])

Upvotes: 2

Jilber Urbina
Jilber Urbina

Reputation: 61154

Another way to do it

> temp <- A[,seq(ncol(B))]*0            
> temp[rownames(B), ] <- B            
> (B <- temp)
#      [,1] [,2] [,3]
#B    0    0    0
#A  100  103  106
#C  102  105  108
#D    0    0    0
#E  101  104  107

Upvotes: 3

alexis_laz
alexis_laz

Reputation: 13122

A way could be:

res = as.table(array(0, c(nrow(A), ncol(B)), list(rownames(A), NULL)))
B2 = as.data.frame(as.table(B))
res[as.matrix(B2[1:2])] = B2[[3]]
res  ##which can be converted back to a `colnames`less matrix
#    A   B   C
#B   0   0   0
#A 100 103 106
#C 102 105 108
#D   0   0   0
#E 101 104 107

Upvotes: 2

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