R: Apply Function to Matrix with Elements of Vector as Argument

R: Apply function to matrix with elements of vector as argument

Mapply is definitely a possibility. This should work:

mapply(MYFUNC, x = as.data.frame(t(df)), Var = var2)

#V1        V2        V3        V4        V5        V6        V7        V8        V9       V10 
#5.0795111 2.8693537 1.8285747 1.3640238 0.8300597 0.6280441 0.7706310 0.6720132 0.5719003 0.4259674 

The issue I think you were running into is that mapply takes either vectors or lists. In R matrices aren't lists, but data.frames are. All you need to do is transpose your matrix and convert to a data.frame and then mapply should work. Each column in a data.frame is an element in the list which is why we have to transpose it (so that each row will be mapped to each element in the vector).

how to apply function with two matrices and one vector as input variables

?mapply may be helpful. Does this work for you?

Below fun is a function that accepts three arguments.

x1 <- read.table(
  text = ' 1.103990 1.345141 1.671156 2.041531 2.435917
  1.031078 1.109853 1.229888 1.380356 1.552349
  1.018405 1.065593 1.139852 1.236245 1.349988
  1.013136 1.046822 1.100638 1.171764 1.257230
  1.010249 1.036439 1.078646 1.135048 1.203625
  1.008425 1.029847 1.064566 1.111308 1.168612
  1.007169 1.025289 1.054776 1.094688 1.143918'
)
x2 <- read.table(
  text = ' 6.27530 15.29211  28.49757  46.41790  69.23123
10.96466 23.60472  39.23650  58.71576  82.53972
14.17965 29.67335  47.61181  68.85091  93.98208
16.78984 34.69621  54.71981  77.67461 104.17505
19.04558 39.07866  61.00511  85.59340 113.45587
21.06106 43.01689  66.70069  92.83912 122.03283
22.89981 46.62361  71.94674  99.55885 130.04526'
)
x1 <- t(x1) # Expensive operation in case of large data.frames
x2 <- t(x2) # Expensive operation in case of large data.frames
x3 <- 1:5
fun <- function(x1, x2, x3) {
  sqrt(x1**2 + x2**2 + x3**2)
}
mapply(fun, x1, x2, x3)
[1]   6.449665  15.480892  28.703732  46.634636  69.454279  11.058340  23.715282  39.370237  58.868038  82.705593  14.251302  29.759758
[13]  47.719846  68.978084  94.124672  16.850079  34.769568  54.813037  77.786362 104.302549  19.098553  39.143529  61.088353  85.694332
[25] 113.572370  21.108888  43.075671  66.776608  92.931896 122.140809  22.943751  46.677749  72.016984  99.645185 130.146372

Using apply with a different function argument for each element's evaluation

You could use mapply where x would be your matrix column, and y the constant. I didn't bother with converting the matrix into a list the smart way, so I have to use unlist inside the function.

mat <- matrix(1:5, nrow = 10, ncol = 3, byrow = TRUE)

mat.list <- apply(mat, MARGIN = 2, FUN = list)

mapply(FUN = function(x, y) {
  sqrt(mean((unlist(x) - y)^2))
}, x = mat.list, y = list(1, 2, 3))

[1] 2.449490 1.732051 1.414214

Using mapply to apply a function with two arguments to every row of two matrices

If we want to apply on each row, then split the matrix by row and pass as a list

mapply(f, asplit(M1, 1), asplit(M2, 1))

Note that mapply on a matrix (or a vector) will loop over each element i.e. here the unit is a single element whereas in data.frame/data.table/tibble, the single unit is a column. By splitting by row (asplit - MARGIN = 1), we get a list of vectors and here the unit is a list element

As @Adam mentioned in the comments, it may needs to be transposed (Not clear without testing with f)

Generating a matrix by applying function to all pairs

Either vectorize F so that you can use it with outer or do something like below

sapply(c('A', 'B', 'C', 'D', 'E'), function(a)
           sapply(c('a', 'b', 'c', 'd', 'e'), function(b)
                     F(a,b)))
#    A   B   C   D   E
#a -32 -31 -30 -29 -28
#b -33 -32 -31 -30 -29
#c -34 -33 -32 -31 -30
#d -35 -34 -33 -32 -31
#e -36 -35 -34 -33 -32

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