CODEWITHSUNDEEP
rmatrix
user3173922
user3173922

Reputation: 11

Fill matrix with column values in R using colnames and rownames

I have a very large dataset, so I want to avoid loops.

I have three columns of data:

col1 = time presented as 10000, 10001, 10002, 10100, 10101, 10102, 10200, 10201, 10202, 10300, ... (total 18000 times)

col2 = id number 1 2 3 4 ... (total 500 ids)

col3 = reading associated with particular id at particular time. 0.1 0.5 0.6 0.7... Say this is called Data3

10000 1 0.1

10001 1 0.5

10002 1 0.6

10100 1 0.7

10200 1 0.6 (NOTE - some random entries missing)

I want to present this as a matrix (called DataMatrix), but there is missing data, so a simple reshape will not do. I want to have the missing data as NA entries.

DataMatrix is currently an NA matrix of 500 columns and 18000 rows, where the row names and column names are the times and ids respectively.

1 2 3 4 ....

10000 NA NA NA NA ....

10001 NA NA NA NA ....

Is there a way I can get R to go through each row of Data3, completing DataMatrix with the reading Data3[,3] by placing it in the row and column of the matrix whose names relate to the Data3[,1] and Data3[,2]. But without loops.

Thanks to all you smart people out there.

Upvotes: 1

Views: 7520

Answers (2)

BrodieG
BrodieG

Reputation: 52637

Here is a solution with possible id values in 1:10 and times values in 1:20. First, create data:

mx <- matrix(c(sample(1:20, 5), sample(1:10, 5), sample(1:50, 5)), ncol=3, dimnames=list(NULL, c("time", "id", "reading")))
times <- 1:20
ids <- 1:10
mx
#      time id reading
# [1,]    4  3      25
# [2,]    5  4       9
# [3,]    9  7      45
# [4,]   18  1      40
# [5,]   11  8      28

Now, use outer to pass every possible combination of time/id to a look up function that returns the corresponding reading value:

outer(times, ids, 
  function(x, y) {
    mapply(function(x.sub, y.sub) {
      val <- mx[mx[, 1] == x.sub & mx[, 2] == y.sub, 3]
      if(length(val) == 0L) NA_integer_ else val
    },
    x, y)
} )

This produces the (hopefully) desired answer:

#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#  [1,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [2,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [3,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [4,] NA   NA   25   NA   NA   NA   NA   NA   NA   NA   
#  [5,] NA   NA   NA   9    NA   NA   NA   NA   NA   NA   
#  [6,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [7,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [8,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
#  [9,] NA   NA   NA   NA   NA   NA   45   NA   NA   NA   
# [10,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [11,] NA   NA   NA   NA   NA   NA   NA   28   NA   NA   
# [12,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [13,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [14,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [15,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [16,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [17,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [18,] 40   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [19,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA   
# [20,] NA   NA   NA   NA   NA   NA   NA   NA   NA   NA

Upvotes: 1

lukeA
lukeA

Reputation: 54237

If I understood you correctly: 

Data3 <- data.frame(col1=10000:10499,  
                    col2=1:500,
                    col3=round(runif(500),1))

library(reshape2) 
DataMatrix <- dcast(Data3, col1~col2, value.var="col3")
DataMatrix[1:5, 1:5]
#    col1   1   2   3   4
# 1 10000 0.4  NA  NA  NA
# 2 10001  NA 0.6  NA  NA
# 3 10002  NA  NA 0.9  NA
# 4 10003  NA  NA  NA 0.5
# 5 10004  NA  NA  NA  NA

Upvotes: 1

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