I have a list of matrices that I like to divide the values in each matrix by a different value.
l1 <- list(1,2,3,4,5,6)
l2 <- list(7,8,9,10,11,12)
mat <- Map(
function(x, y) outer(unlist(x), unlist(y), `+`) / 2,
split(l1, ceiling(seq_along(l1) / 3)),
split(l2, ceiling(seq_along(l2) / 3))
)
For example the output below shows one of the elements in the mat list:
$`1`
[,1] [,2] [,3]
[1,] 4.0 4.5 5.0
[2,] 4.5 5.0 5.5
[3,] 5.0 5.5 6.0
I would like to divide the values in the matrix by another matrix with different values
Maybe a matrix that looks like this (I wasn't sure how to create a matrix in r)
2 1 2
3 2 3
1 2 3
My desired output would then look like this:
[,1] [,2] [,3]
[1,] 4.0/2 4.5/1 5.0/2
[2,] 4.5/3 5.0/2 5.5/3
[3,] 5.0/1 5.5/2 6.0/3
How could I do create this output? How do I create a matrix with my desired values in R?
Thank you.
If your matrices are the same dimensions you can divide them with the / operator.
# create matrix to divide by
mat_div <- matrix(c(2,3,1,1,2,2,2,3,3), nrow = 3)
# divide list of matricies
lapply(mat, `/`, mat_div)
#------
$`1`
[,1] [,2] [,3]
[1,] 2.0 4.50 2.500000
[2,] 1.5 2.50 1.833333
[3,] 5.0 2.75 2.000000
$`2`
[,1] [,2] [,3]
[1,] 3.5 7.50 4.000000
[2,] 2.5 4.00 2.833333
[3,] 8.0 4.25 3.000000
We can use Map
mat <- Map(`/`, mat, list(mat2))
-otuput
mat
$`1`
[,1] [,2] [,3]
[1,] 2.0 4.50 2.500000
[2,] 1.5 2.50 1.833333
[3,] 5.0 2.75 2.000000
$`2`
[,1] [,2] [,3]
[1,] 3.5 7.50 4.000000
[2,] 2.5 4.00 2.833333
[3,] 8.0 4.25 3.000000
data
mat2 <- cbind(c(2, 3, 1), c(1, 2, 2), c(2, 3, 3))
Related
I have two lists A_1 and A_2, each contains two matrices.
A_1 <- list(a=matrix(1:8, 2), b=matrix(2:9, 2))
A_2 <- list(a=matrix(10:17, 2), b=matrix(5:12, 2))
I'd like to calculate element-wise averages of these two lists which results a list of
tibble::lst((A_1$a + A_2$a)/2, (A_1$b + A_2$b)/2)
I used
purrr::pmap(list(A_1 , A_2), mean)
but got
Error in mean.default(.l[[1L]][[i]], .l[[2L]][[i]], ...) :
'trim' must be numeric of length one`
or
purrr::map2(A_1, A_2, mean)
Error in mean.default(.x[[i]], .y[[i]], ...) :
'trim' must be numeric of length one`
In base R, We could use:
A <-list(A_1, A_2)
lapply(Reduce(\(x, y)Map('+', x, y), A), '/', length(A))
$a
[,1] [,2] [,3] [,4]
[1,] 5.5 7.5 9.5 11.5
[2,] 6.5 8.5 10.5 12.5
$b
[,1] [,2] [,3] [,4]
[1,] 3.5 5.5 7.5 9.5
[2,] 4.5 6.5 8.5 10.5
This code is generic in that we can use to find the mean of several lists.
Note that A_1 and A_2 must have the same number of matrices, not necessarily 2. Can be 10 etc. Also note that each corresponding matrix has the same dimensions. Example below:
B_1 <- list(matrix(c(1,2,3,4), 2), matrix(c(1,3,4,2), 2),
matrix(c(1:10), 5), matrix(c(1:20), 5))
B_2 <- lapply(B_1, '*', 2) # In this case, its B_1 * 2
B_3 <- lapply(B_2, '*', 3) #
Now you could use the code provide above:
B <-list(B_1, B_2, B_3)
lapply(Reduce(\(x, y)Map('+', x, y), B), '/', length(B))
Your mistake is in using the second matrix as trim= argument of mean whcih is the second. You need to concatenate the matrices. Example:
mean(1:3, 2:4)
# Error in mean.default(1:3, 2:4) : 'trim' must be numeric of length one
mean(c(1:3, 2:4))
# [1] 2.5
As solution you may use Map
Map(\(x, y) (x + y)/2, A_1, A_2)
# $a
# [,1] [,2] [,3] [,4]
# [1,] 5.5 7.5 9.5 11.5
# [2,] 6.5 8.5 10.5 12.5
#
# $b
# [,1] [,2] [,3] [,4]
# [1,] 3.5 5.5 7.5 9.5
# [2,] 4.5 6.5 8.5 10.5
Or, why not using arrays?
AA_1 <- array(unlist(A_1), dim=c(dim(A_1$a), length(A_1)))
AA_2 <- array(unlist(A_2), dim=c(dim(A_2$a), length(A_2)))
(AA_1 + AA_2)/2
# , , 1
#
# [,1] [,2] [,3] [,4]
# [1,] 5.5 7.5 9.5 11.5
# [2,] 6.5 8.5 10.5 12.5
#
# , , 2
#
# [,1] [,2] [,3] [,4]
# [1,] 3.5 5.5 7.5 9.5
# [2,] 4.5 6.5 8.5 10.5
in base R:
item_names <- names(A_1)
structure(
lapply(item_names, function(name){
0.5 * (A_1[[name]] + A_2[[name]])
## or, if you want the scalar mean:
## mean(A_1[[name]] + A_2[[name]])
}),
names = item_names
)
#> $a
#> [,1] [,2] [,3] [,4]
#> [1,] 5.5 7.5 9.5 11.5
#> [2,] 6.5 8.5 10.5 12.5
#>
#> $b
#> [,1] [,2] [,3] [,4]
#> [1,] 3.5 5.5 7.5 9.5
#> [2,] 4.5 6.5 8.5 10.5
I have a function return list of list, I would like to find the standard deviation of the matrices of my output. The output of my function is a list of two list. I tried this code but it return me NAN. Since my function is complex, then I use this example from another question please see here since it is quite close to what I am trying to do.
> A <- matrix(c(1:9), 3, 3)
> A
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
> B <- matrix(c(2:10), 3, 3)
> B
[,1] [,2] [,3]
[1,] 2 5 8
[2,] 3 6 9
[3,] 4 7 10
> my.list1 <- list(A, B)
so the mean of the first list is:
[,1] [,2] [,3]
[1,] 1.5 4.5 7.5
[2,] 2.5 5.5 8.5
[3,] 3.5 6.5 9.5
Then the standard deviation will be:
[,1] [,2] [,3]
[1,] 0.7071068 0.7071068 0.7071068
[2,] 0.7071068 0.7071068 0.7071068
[3,] 0.7071068 0.7071068 0.7071068
> c <- matrix(c(1:9), 3, 3)
> c
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
> d <- matrix(c(2:10), 3, 3)
> d
[,1] [,2] [,3]
[1,] 2 5 8
[2,] 3 6 9
[3,] 4 7 10
> my.list2 <- list(c, d)
my.list <-list(my.list1,my.list2)
How can I get the standard deviation of my matrices on an element by element for the list?
Try ?rapply
> rapply(my.list, sd)
[1] 2.738613 2.738613 2.738613 2.738613
You could bind your lists into an array, or perhaps make your function return an array(?), then you could use apply() to apply your chosen functions...
A <- matrix(1:9, 3, 3)
B <- matrix(2:10, 3, 3)
my.list1 <- list(A, B)
c <- matrix(1:9, 3, 3)
d <- matrix(2:10, 3, 3)
my.list2 <- list(c, d)
Create array from all 4 lists
my.array1 <- abind::abind(c(my.list1, my.list2), along = 3)
Find the mean() of the required dimension
apply(my.array1, c(1, 2), mean)
apply(my.array1, c(1,2), sd)
Output
[,1] [,2] [,3]
[1,] 1.5 4.5 7.5
[2,] 2.5 5.5 8.5
[3,] 3.5 6.5 9.5
There is a way to apply a function f to every column of a matrix:
M <- matrix(seq(1,16), 4, 4)
apply(M, 2, mean)
#[1] 2.5 6.5 10.5 14.5
But if I want to build a descriptive statistics about matrix I should use more indeces. For example, max, min, mean etc.
But R doesn't allow to do something like this:
apply(M, 2, c(mean, max))
to get this output:
# [,1] [,2] [,3] [,4]
#mean 2.5 6.5 10.5 14.5
#max 4 8 12 16
Would you tell me how to manage with this problem?
apply(M, 2, function(x) c(mean(x), max(x)))
# [,1] [,2] [,3] [,4]
# [1,] 2.5 6.5 10.5 14.5
# [2,] 4.0 8.0 12.0 16.0
Try the following:
f <- c("max", "min", "mean")
sapply(f, function(x) apply(M, 2, x))
max min mean
[1,] 4 1 2.5
[2,] 8 5 6.5
[3,] 12 9 10.5
[4,] 16 13 14.5
This question already has answers here:
Mean of each element of a list of matrices
(3 answers)
How to sum a numeric list elements
(2 answers)
Closed 9 years ago.
Suppose you have list of matrices. What is the most convenient way to calculate the mean matrix on an element by element basic? Suppose we have a list of matrices:
> A <- matrix(c(1:9), 3, 3)
> A
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
> B <- matrix(c(2:10), 3, 3)
> B
[,1] [,2] [,3]
[1,] 2 5 8
[2,] 3 6 9
[3,] 4 7 10
> my.list <- list(A, B)
So the desired output should be:
[,1] [,2] [,3]
[1,] 1.5 4.5 7.5
[2,] 2.5 5.5 8.5
[3,] 3.5 6.5 9.5
You can use:
Reduce("+", my.list) / length(my.list)
According to comments, you want both mean and sd implemented on a list of matrices, and the above ways will not work smoothly for sd. Try this instead :
apply(simplify2array(my.list), 1:2, mean)
apply(simplify2array(my.list), 1:2, sd)
Here is an alternative that should be pretty quick as we are working with base functions designed to work with matrices. We just take your list and use array to turn it into a 3D array then either use apply or just rowMeans...
# Make some data, a list of 3 matrices of 4x4
ll <- replicate( 3 , matrix( sample(5,16,repl=TRUE) , 4 ) , simplify = FALSE )
# Make a 3D array from list of matrices
arr <- array( unlist(ll) , c(4,4,3) )
# Get mean of third dimension
apply( arr , 1:2 , mean )
# [,1] [,2] [,3] [,4]
#[1,] 3.000000 3.666667 3.000000 1.666667
#[2,] 2.666667 3.666667 3.333333 3.666667
#[3,] 4.666667 2.000000 1.666667 3.666667
#[4,] 1.333333 4.333333 3.666667 3.000000
Or you can use rowMeans which is quicker, specifying you want to get the mean over 2 dimensions...
# Get mean of third dimension
rowMeans( arr , dims = 2 )
# [,1] [,2] [,3] [,4]
#[1,] 3.000000 3.666667 3.000000 1.666667
#[2,] 2.666667 3.666667 3.333333 3.666667
#[3,] 4.666667 2.000000 1.666667 3.666667
#[4,] 1.333333 4.333333 3.666667 3.000000
I have a correlation matrix:
a <- matrix(c(1, .8, .8, .8, 1, .8, .8, .8, 1), 3)
## [,1] [,2] [,3]
## [1,] 1.0 0.8 0.8
## [2,] 0.8 1.0 0.8
## [3,] 0.8 0.8 1.0
I would now like to create a covariance matrix from the correlation matrix. How can this be done in R?
I tried:
e1.sd <- 3
e2.sd <- 10
e3.sd <- 3
e.cov <- a * as.matrix(c, e1.sd, e2.sd, e3.sd) %*% t(as.matrix(c(e1.sd, e2.sd, e3.sd)))
But I get the error:
Error in a * as.matrix(c, e1.sd, e2.sd, e3.sd) %*% t(as.matrix(c(e1.sd, :
non-conformable arrays
What am I doing wrong?
If you know the standard deviations of your individual variables, you can:
stdevs <- c(e1.sd, e2.sd, e3.sd)
#stdevs is the vector that contains the standard deviations of your variables
b <- stdevs %*% t(stdevs)
# b is an n*n matrix whose generic term is stdev[i]*stdev[j] (n is your number of variables)
a_covariance <- b * a #your covariance matrix
On the other hand, if you don't know the standard deviations, it's impossible.
require(MBESS)
a <- matrix(c(1,.8,.8,.8,1,.8,.8,.8,1),3)
> cor2cov(a,c(3,10,3))
[,1] [,2] [,3]
[1,] 9.0 24 7.2
[2,] 24.0 100 24.0
[3,] 7.2 24 9.0
Building on S4M's answer, in base R, I would write this function:
cor2cov <- function(V, sd) {
V * tcrossprod(sd)
}
tcrossprod will calculate the product of each combination of elements of the sd vector (equivalent to x %*% t(x)), which we then (scalar) multiply by the variance-covariance matrix
Here's a quick check that the function is correct using the built in mtcars data set:
all.equal(
cor2cov(cor(mtcars), sapply(mtcars, sd)),
cov(mtcars)
)
The answer marked as correct is wrong.
The correct solution seems to be the one provided by MBESS package, so see the post from dayne.
> a
[,1] [,2] [,3]
[1,] 1.0 0.8 0.8
[2,] 0.8 1.0 0.8
[3,] 0.8 0.8 1.0
> b <- c(3,10,3)
> b %*% t(b)
[,1] [,2] [,3]
[1,] 9 30 9
[2,] 30 100 30
[3,] 9 30 9
> c <- b %*% t(b)
> c %*% a
[,1] [,2] [,3]
[1,] 40.2 44.4 40.2
[2,] 134.0 148.0 134.0
[3,] 40.2 44.4 40.2
> cor2cov(cor.mat=a, b )
[,1] [,2] [,3]
[1,] 9.0 24 7.2
[2,] 24.0 100 24.0
[3,] 7.2 24 9.0
> a %*% c
[,1] [,2] [,3]
[1,] 40.2 134 40.2
[2,] 44.4 148 44.4
[3,] 40.2 134 40.2
>