CODEWITHSUNDEEP
rmatrix
jonem
jonem

Reputation: 459

Convert list of indices and values to matrix

I have a list of indices and values as follows

1 3 2.1
2 1 1.1
2 2 0.2
2 3 0.4
1 2 0.3
1 1 3.2
3 3 4.0
3 1 0.2
3 2 0.1

where the first two columns represent the (i,j)'th index of a matrix that I wish to populate with the corresponding value. That is, I want the above to generate the following matrix

3.2 0.3 2.1
1.1 0.2 0.4
0.2 0.1 4.0

Is there a way to do this in R without resorting to a for loop?

Upvotes: 1

Views: 545

Answers (2)

akrun
akrun

Reputation: 887951

An option would be to create a matrix of 0s and fill the values of the third column of dataset with the indices from the first two columns

m1  <- matrix(0, 3, 3)
m1[as.matrix(df1[1:2])] <- df1[,3]
m1
#     [,1] [,2] [,3]
#[1,]  3.2  0.3  2.1
#[2,]  1.1  0.2  0.4
#[3,]  0.2  0.1  4.0

Or with sparseMatrix

library(Matrix)
sparseMatrix(i = df1$col1, j = df1$col2, x = df1$col3)
# 3 x 3 sparse Matrix of class "dgCMatrix"
            
#[1,] 3.2 0.3 2.1
#[2,] 1.1 0.2 0.4
#[3,] 0.2 0.1 4.0

Another option is xtabs

xtabs(col3 ~ col1 + col2, df1)
# col2
#col1   1   2   3
#   1 3.2 0.3 2.1
#   2 1.1 0.2 0.4
#   3 0.2 0.1 4.0

If we need an efficient option may be dcast

library(data.table)
dcast(setDT(df1), col1 ~ col2, value.var = 'col3')

Or using tidyverse

library(tidyverse)
df1 %>% 
  spread(col2, col3)

NOTE: All the methods work even if there are unequal number of indices

data

df1 <- structure(list(col1 = c(1L, 2L, 2L, 2L, 1L, 1L, 3L, 3L, 3L), 
    col2 = c(3L, 1L, 2L, 3L, 2L, 1L, 3L, 1L, 2L), col3 = c(2.1, 
    1.1, 0.2, 0.4, 0.3, 3.2, 4, 0.2, 0.1)), 
    class = "data.frame", row.names = c(NA, 
-9L))

Upvotes: 1

jay.sf
jay.sf

Reputation: 73802

You may bring the matrix in right order and apply matrix.

matrix(m[order(m[,2], m[,1]), 3], 3)
#      [,1] [,2] [,3]
# [1,]  3.2  0.3  2.1
# [2,]  1.1  0.2  0.4
# [3,]  0.2  0.1  4.0

Data

m <- structure(c(1, 2, 2, 2, 1, 1, 3, 3, 3, 3, 1, 2, 3, 2, 1, 3, 1, 
2, 2.1, 1.1, 0.2, 0.4, 0.3, 3.2, 4, 0.2, 0.1), .Dim = c(9L, 3L
))

Upvotes: 1

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