R table by matrix row

Question

For a matrix (as.matrix), how can I generate a table where rows are equal to rows of the matrix?

>table(matrix)

and

&g         


        

Answer 1

you can use apply over the rows, and then use mapply with an ifelse statement to get back your matrix.

Assuming X is your matrix:

# this will get you the values, just not in a nice matrix
tables.list <- apply(X, 1, table)

# unique values
vals <- sort(unique(c(X)))

# this will get you the matrix
results <- t(mapply(function(v, t)
   ifelse(v %in% names(t), t[as.character(v)], 0), list(vals), tables.list ))

# give it names
dimnames(results) <- list(rownames(X), vals)

results

#   2 4 5 6 7 8
# a 0 1 2 1 0 0
# b 0 0 4 0 0 0
# c 0 0 0 2 1 1
# d 1 0 0 3 0 0
# e 0 1 1 0 2 0

Answer 2

One alternative is to convert your matrix to a long data.frame (using stack), at which point you can easily use table:

Here's your data:

mymat <- structure(c(5L, 5L, 8L, 2L, 7L, 5L, 5L, 7L, 6L, 7L, 4L, 5L, 6L, 
            6L, 5L, 6L, 5L, 6L, 6L, 4L), .Dim = c(5L, 4L), .Dimnames = list(
              c("a", "b", "c", "d", "e"), c("1", "2", "3", "4")))

This is what it looks like as a long data.frame:

head(stack(data.frame(t(mymat))))
#   values ind
# 1      5   a
# 2      5   a
# 3      4   a
# 4      6   a
# 5      5   b
# 6      5   b

Here's how we can use that to create the table you want:

with(stack(data.frame(t(mymat))), table(ind, values))
#    values
# ind 2 4 5 6 7 8
#   a 0 1 2 1 0 0
#   b 0 0 4 0 0 0
#   c 0 0 0 2 1 1
#   d 1 0 0 3 0 0
#   e 0 1 1 0 2 0

Answer 3

I used apply too:

t(apply(mat, 1, function(x) table(factor(x, levels = unique(sort(c(mat)))))))

R > mat  = matrix(sample(1:8, 20, replace = T), 5, 4)
R > mat
     [,1] [,2] [,3] [,4]
[1,]    5    6    1    4
[2,]    4    3    4    8
[3,]    4    8    4    3
[4,]    3    3    5    1
[5,]    1    1    3    1
R > t(apply(mat, 1, function(x) table(factor(x, levels = unique(sort(c(mat)))))))
     1 3 4 5 6 8
[1,] 1 0 1 1 1 0
[2,] 0 1 2 0 0 1
[3,] 0 1 2 0 0 1
[4,] 1 2 0 1 0 0
[5,] 3 1 0 0 0 0

Answer 4

## source data
x=as.matrix(read.table(text="
   1  2  3  4 
a  5  5  4  6    
b  5  5  5  5     
c  8  7  6  6   
d  2  6  6  6     
e  7  7  5  4
"))

# result

table(rep(rownames(x),ncol(x)),c(x))

#   2 4 5 6 7 8
# a 0 1 2 1 0 0
# b 0 0 4 0 0 0
# c 0 0 0 2 1 1
# d 1 0 0 3 0 0
# e 0 1 1 0 2 0