Select rows of a matrix that meet a condition

Question

This is easier to do if you convert your matrix to a data frame using as.data.frame(). In that case the previous answers (using subset or m$three) will work, otherwise they will not.

To perform the operation on a matrix, you can define a column by name:

m[m[, "three"] == 11,]

Or by number:

m[m[,3] == 11,]

Note that if only one row matches, the result is an integer vector, not a matrix.

I will choose a simple approach using the dplyr package.

If the dataframe is data.

library(dplyr)
result <- filter(data, three == 11)

m <- matrix(1:20, ncol = 4) 
colnames(m) <- letters[1:4]

The following command will select the first row of the matrix above.

subset(m, m[,4] == 16)

And this will select the last three.

subset(m, m[,4] > 17)

The result will be a matrix in both cases. If you want to use column names to select columns then you would be best off converting it to a dataframe with

mf <- data.frame(m)

Then you can select with

mf[ mf$a == 16, ]

Or, you could use the subset command.

Subset is a very slow function , and I personally find it useless.

I assume you have a data.frame, array, matrix called Mat with A, B, C as column names; then all you need to do is:

• In the case of one condition on one column, lets say column A

  Mat[which(Mat[,'A'] == 10), ]
  

In the case of multiple conditions on different column, you can create a dummy variable. Suppose the conditions are A = 10, B = 5, and C > 2, then we have:

    aux = which(Mat[,'A'] == 10)
    aux = aux[which(Mat[aux,'B'] == 5)]
    aux = aux[which(Mat[aux,'C'] > 2)]
    Mat[aux, ]

By testing the speed advantage with system.time, the which method is 10x faster than the subset method.

If your matrix is called m, just use :

R> m[m$three == 11, ]