CODEWITHSUNDEEP
rmatrixsubset
samalevy
samalevy

Reputation: 81

Replacing items in a matrix based on column and value in R

I would like to know how to replace observations in a series of matrices that have rules for replacement that differ by matrix column in as automated a way as possible.

I have a large number of matrices to process (~240). Processing requires replacing observations over a certain number with 0. Each matrix has identical dimensions and contains two types of columns, which require a different cut-off at which all observations in that column should be replaced with 0. Matrices differ based on the cut-off at which observations should be replaced with 0. I have figured out a way to process the data but it is very slow and manual, what I would like to know is if there is a way to automate this either through a function or a for loop.

For instance, in the matrix x where observations in columns 3,4 and 9 need to become 0 if greater than 45 and observations in all other columns need to become 0 if greater than 70.

I have thought of two ways to replace the data, both detailed below:

cols <- c(3,4,9)
  
x <- matrix(sample(1:100),10,10)

#attempt 1
x[x>70]  <- 0            # replace all values > 70 with 0 (true for all columns)
x.sub  <- x[,cols]       # subset columns with lower limit
x.sub[x.sub>45]  <- 0    # replace all values > 45 with 0 for the subset (rule 2 columns)
x[,cols]   <- x.sub      # add revised values back to main object

#attempt 2               
x.sub1 <- x[,-cols]      # subset columns that follow rule 1
x.sub1[x.sub1>70] <- 0   # replace all values over 70 with 0
x.sub2  <- x[,cols]      # subset columns that follow rule 2
x.sub2[x.sub3>45] <- 0   # replace all values over 45 with 0
x[,-cols]  <- x.sub1     # add revised values back to main object (rule 1 columns)
x[,cols]   <- x.sub2     # add revised values back to main object (rule 2 columns)

Any help would be greatly appreciated!!

Upvotes: 1

Views: 563

Answers (2)

akrun
akrun

Reputation: 886948

We could also this with negative indexing

x[, -cols] <- (x[, -cols] <=70) * x[, -cols]
x[, cols] <- (x[, cols]  <= 45) * x[, cols]
x
#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,]    0   27    0    0   35   20   24   33    0    38
# [2,]    0   36    0   12   19   17   40   44   18    52
# [3,]    0   10    0   15   60   43    0    0    0     0
# [4,]    2    0   32    0    9   63    0   48   16    69
# [5,]    0    0    5   30    0   47   57    0   45     0
# [6,]   39   61    0    0    0    0   37   56   23    46
# [7,]    0   51    0   42    0    3   28    7   26    11
# [8,]    6    0    0   25   65   13    0   49    0     4
# [9,]    0    0    0    0    0    8   31    0   14    34
#[10,]    0    0    0   29   22    0    1   66   21    41

data

cols <- c(3,4,9)
set.seed(69)
x <- matrix(sample(1:100), 10, 10)

Upvotes: 1

Allan Cameron
Allan Cameron

Reputation: 173793

If we make a reproducible matrix:

cols <- c(3,4,9)

set.seed(69)

x <- matrix(sample(1:100), 10, 10)
x
#>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#>  [1,]   96   27   74   53   35   20   24   33   50    38
#>  [2,]   81   36   54   12   19   17   40   44   18    52
#>  [3,]   91   10   55   15   60   43   88   71   58    82
#>  [4,]    2   92   32   70    9   63   87   48   16    69
#>  [5,]   77   78    5   30   79   47   57   72   45    89
#>  [6,]   39   61   85   75   80   93   37   56   23    46
#>  [7,]   95   51   67   42   99    3   28    7   26    11
#>  [8,]    6   76   62   25   65   13   86   49   59     4
#>  [9,]   97  100   68   83   73    8   31   94   14    34
#> [10,]   90   84   64   29   22   98    1   66   21    41

We can get the indices of the entries that are in the target columns using a little modular arithmetic:

incols <- ((seq_along(x) - 1) %/% nrow(x) + 1) %in% cols

Which allows straightforward subsetting:

x[x > 70 & !incols]  <- 0
x[x > 40 & incols]   <- 0

x
#>       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
#>  [1,]    0   27    0    0   35   20   24   33    0    38
#>  [2,]    0   36    0   12   19   17   40   44   18    52
#>  [3,]    0   10    0   15   60   43    0    0    0     0
#>  [4,]    2    0   32    0    9   63    0   48   16    69
#>  [5,]    0    0    5   30    0   47   57    0    0     0
#>  [6,]   39   61    0    0    0    0   37   56   23    46
#>  [7,]    0   51    0    0    0    3   28    7   26    11
#>  [8,]    6    0    0   25   65   13    0   49    0     4
#>  [9,]    0    0    0    0    0    8   31    0   14    34
#> [10,]    0    0    0   29   22    0    1   66   21    41

Created on 2020-07-03 by the reprex package (v0.3.0)

Upvotes: 1

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