将一个矩阵的几列与R中另一矩阵的列进行比较

Question

Here I have created a duplicate data set.

set.seed(1234)
m1 = matrix(runif(2000), nrow = 10, ncol = 200)
dim(m1)
[1]  10 200
m2 = matrix(runif(100), nrow = 10, ncol = 10)
dim(m2)
[1] 10 10

I want to compare the first 1:20 columns of m1 against the 1st column of m2 matrix. Similarly, for the next 21:40 columns of m1 against the 2nd column of m2 matrix and so on. Finally, 181:200 columns of matrix m1 against the 10th column of matrix m2 .

I wrote the following code to compare 1st 20 columns of m1 matrix against the 1st column of m2 matrix.

cc = matrix(NA, nrow(m2), ncol(m2))
for (j in 1:ncol(m2)) {
  for (i in 1:nrow(m2)) {
    cc[i, j] = ifelse(m1[i, j] < m2[i,1], 1, 0)
  }
}
ccvalue = data.frame(cc)

How can I improve the above r code do the above comparison. Are there any r function to do?

Thank you in advance.

Answer 1

You can take advantage of the implicit vectorization in R to run the entire matrix of m1 against the columns of m2. You just need to get m2 to repeat columns by subsetting for the same column over and over again. For example, v <-c("A","B","C") you can do v[c(1,1,2,2,3,3)] which equals "A","A","B","B","C","C" .

Test out the following code and let me know if you have any questions:

# we want to compare m1[,c(1,2,3,...)], with m2[,c(1,1,1,...)]
# summing 1,0,...,1,0,... to get 1,1,...,2,2,...
m2_to_compare <- cumsum(rep(c(1,rep(0,19)),10)) 
# length should match m1 columns
length(m2_to_compare) 
(m1 < m2[,m2_to_compare]) * 1 # turns TRUEs and FALSEs into 1s and 0s

Answering Comment:

cc = ifelse(m1 < m2[,m2_to_compare], 1, 0)
# depending on your seed:
sapply(1:10, function(colm) rowSums(cc[,m2_to_compare == colm]))
#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,]    3    1   19    8    9   11   17    2   12    16
# [2,]    2   19   14   10   10   11    9    1    0    14
# [3,]   16   16   17    7    5   20    1   16    2    17
# [4,]   13    2    0   11   20   11    6    5   12     2
# [5,]    0   10    2    1   10   17    3   14    5     7
# [6,]   11    7   17    9   20   18   18   16    7     4
# [7,]   15    3    5    5    8    5    3    3    9     1
# [8,]    0   18    5    8    9   15    9   17    0    20
# [9,]   15   14    5    1    5    0    6   17   19     6
#[10,]    6    1    4   10   11   12    0    9    7     5

Answer 2

A few things to point out:

(1) It would be a good practice to set the seed for matrix m2 as well. Perhaps you overlooked that.

(2) In your code provided, you seem to be only comparing m2 to the first 10 columns of m1 .

If you only mean to compare the 10 columns, you can do it with this:

cc <- (m2 > m1[, c(1:10)])*1