I'm sure this is trivial - nonetheless, any help would be appreciated.
The problem is simple: given a matrix, I'd like to get TRUE if the matrix in question has at least one element equal to zero. So, checking
A <- matrix(c(1, 2, 3, 4, 5, 0), nrow = 2, ncol = 3, byrow = TRUE)
> A
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 0
would return TRUE, while
B <- matrix(c(1, 2, 3, 4, 5, 6), nrow = 2, ncol = 3, byrow = TRUE)
> B
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 4 5 6
would return FALSE.
Something like
if ( A == 0 ) { cat("\nZero detected")}
gives a warning. Is there a simple way to do this?
The warning is generated because you're presenting a vector of logical to if, which expects a single value. any is a function to tell if any of the logical values are TRUE:
any(A==0)
## [1] TRUE
any(B==0)
## [1] FALSE
There's also a function all which determines if all of the values in a logical vector are TRUE.
Try
0 %in% A
It should return TRUE or FALSE. It works for NA too:
x = matrix(1:24, ncol = 4)
x[3, 3] = NA
NA %in% x
#TRUE
Related
I'm stumped using indexing in R. I have two matrices, one with logical values and one with data. I want to use the first to index into the second one. However, I've noticed that R reorders my values when doing so.
My data looks roughly like this:
ind <- matrix(c(TRUE, FALSE, FALSE, FALSE, FALSE, FALSE, FALSE, TRUE, FALSE, FALSE, TRUE, FALSE), nrow = 3, ncol = 4, byrow = TRUE)
data <- matrix(c(1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4), nrow = 3, ncol = 4, byrow = TRUE)
Now, when indexing result <- data[ind], I was obtaining the following: c(1, 3, 4) when I was trying to obtain c(1, 4, 3).
How can I prevent R from reordering columwise? I'd appreciate any input. I'm sure it's an easy fix - I just don't know which.
Thank you!
When converting matrices to and from vectors, R makes the transformation columnwise.
as.vector(data)
[1] 1 1 1 2 2 2 3 3 3 4 4 4
As this happens to both your ind and your data this is generally not a problem.
So to retain your order, you have to transpose both matrices:
> t(data)[t(ind)]
[1] 1 4 3
PS: have you tried which with a matrix?
> which(arr.ind = T, ind)
row col
[1,] 1 1
[2,] 3 3
[3,] 2 4
Here is another base R trick (but I recommend the answer by #c0bra)
> rowSums(data * ind)
[1] 1 4 3
I have two matrices, one of them has a NA value and I want to use a function that only runs if there are NAs present in the data, so if I run the function it should only work on df2 and not df1. How would I do this?
df1 <- matrix(1:4, nrow = 2, ncol = 2)
df2 <- matrix(1,2,3,NA, nrow = 2, ncol = 2)
Based on the comment above, here is a complete answer (assuming I understand what you are getting at). The function is set up to do something or not to the matrix depending on whether it has NA values.
df1 <- matrix(1:4, nrow = 2, ncol = 2)
df2 <- matrix(c(1,2,3,NA), nrow = 2, ncol = 2)
myfunc <- function(m) {
ret <- m
if (all(!is.na(m))) {
print("This matrix has no NAs")
} else {
print("This matrix has NAs")
}
return(ret)
}
myfunc(df1)
# [1] "This matrix has no NAs"
# [,1] [,2]
# [1,] 1 3
# [2,] 2 4
myfunc(df2)
# [1] "This matrix has NAs"
# [,1] [,2]
# [1,] 1 3
# [2,] 2 NA
Given a random matrix (any size!), write a function that determines whether or not that matrix is a Toeplitz Matrix. In linear algebra, a Toeplitz matrix is one in which the elements on any given diagonal from top left to bottom right are identical.
Here is an example:
x <- structure(c(1, 5, 4, 7, 2, 1, 5, 4, 3, 2, 1, 5, 4, 3, 2, 1, 8,
4, 3, 2), .Dim = 4:5)
[,1] [,2] [,3] [,4] [,5]
[1,] 1 2 3 4 8
[2,] 5 1 2 3 4
[3,] 4 5 1 2 3
[4,] 7 4 5 1 2
So our function should receive such matrix and return TRUE if it meets the conditions.
To test the function, one can use stats::toeplitz() to generate a toeplitz matrix. So for example, the expected output of our function should be:
> toeplitz_detector(stats::toeplitz(sample(5, 5)))
> [1] TRUE
I've solved the problem by defining the following function:
toeplitz_solver <- function(a) {
# re-order a backwards, because we need to check diagonals from top-left
# to bottom right. if we don't reorder, we'll end up with top-right to
# bottom-left.
a <- a[, ncol(a):1]
# get all i and j (coordinates for every element)
i <- 1:nrow(a)
j <- 1:ncol(a)
# get all combinations of i and j
diags <- expand.grid(i, j)
# the coordinates for the diagonals are the ones where
# the sum is the same, e.g.: (3,2), (4,1), (2,3), (1,4)
sums <- apply(diags, 1, sum)
indexes <- lapply(unique(sums), function(x) {
diags[which(sums == x), ]
})
# indexes is now a list where every element is a list of coordinates
# the first element is a list for every coordinates for the first diag
# so on and so forth
results <- sapply(indexes, function(x) {
y <- a[as.matrix(x)]
return(all(y == y[1]))
})
# if every diagonal meets the condition, it is safe to assume that the
# input matrix is in fact toeplitz.
return(all(results))
}
How can I simply pad (append/prepend) a slice of NA's to a (say) 3D array along (say) dimension 2?
Suppose the initial array is given as
A <- array(1:8,c(2,2,2))
I initially thought this would work:
cbind(A,NA)
but it results in an 8x2 matrix instead of a 2x3x2 array. I then tried
abind(A,NA,along=2)
but that results in an error.
I'm hoping there is a much simpler solution than
dimSlice <- dim(A)
dimSlice[2] <- 1
abind(A,array(NA,dimSlice),along=2)
Background
This padding happens as part of a "remove slice and pad opposite side" operation that shifts an array by one position along some dimension, filling in with NA elements at the vacated positions. The one-dimensional equivalent would be, for example, c(A[-1],NA) for vector A, If there is a simple way to accomplish such an operation without an explicit padding sub-operation, that would be even better.
Subsetting with NAs results in NAs (?Extract):
v = 1:3; m = matrix(1:4, 2, 2); a = array(1:6, c(2, 2, 2))
v[c(NA, 1)]
#[1] NA 1
m[, c(2, NA)]
# [,1] [,2]
#[1,] 3 NA
#[2,] 4 NA
a[, c(1, 2, NA), ]
#, , 1
#
# [,1] [,2] [,3]
#[1,] 1 3 NA
#[2,] 2 4 NA
#
#, , 2
#
# [,1] [,2] [,3]
#[1,] 5 1 NA
#[2,] 6 2 NA
So, to pad with NAs, we could subset using the appropriate indices. Putting the above in a more general function to append/prepend "n" indices with NA in dimension "k" of an array:
pad = function(x, k, n = 1L, append = TRUE)
{
dims = replicate(length(dim(x)), substitute(), simplify = FALSE)
if(append) dims[[k]] = c((n + 1):dim(x)[[k]], rep_len(NA, n))
else dims[[k]] = c(rep_len(NA, n), 1:(dim(x)[[k]] - n))
do.call("[", c(list(x), dims))
}
arr = array(1:24, c(3, 2, 2, 2))
pad(arr, 1, 2, FALSE)
pad(arr, 2)
I would like to make a new matrix from another matrix but only with rows which do not contain 0, how can I do that?
Here is a more vectorized way.
x <- matrix(c(0,0,0,1,1,0,1,1,1,1), ncol = 2, byrow = TRUE)
x[rowSums(x==0)==0,]
I found that it could by done very simply
x <- matrix(c(0,0,0,1,1,0,1,1,1,1), ncol = 2, byrow = TRUE)
y <- cbind (x[which(x[,1]*x[,2] >0), 1:2])
I am only piecing together the great suggestions others have already given. I like the ability to store this as a function and generalize to values besides 1 including categorcal values (also selects positively or negatively using the select argument):
v.omit <- function(dataframe, v = 0, select = "neg") {
switch(select,
neg = dataframe[apply(dataframe, 1, function(y) !any(y %in% (v))), ],
pos = dataframe[apply(dataframe, 1, function(y) any(y %in% (v))), ])
}
Let's try it.
x <- matrix(c(0,0,0,1,1,0,1,1,1,1,NA,1), ncol = 2, byrow = TRUE)
v.omit(x)
v.omit(mtcars, 0)
v.omit(mtcars, 1)
v.omit(CO2, "chilled")
v.omit(mtcars, c(4,3))
v.omit(CO2, c('Quebec', 'chilled'))
v.omit(x, select="pos")
v.omit(CO2, c('Quebec', 'chilled'), select="pos")
v.omit(x, NA)
v.omit(x, c(0, NA))
Please do not mark my answer as the correct one as others have answered before me, this is just to extend the conversation. Thanks for the code and the question.
I'm sure there are better ways, but here's one approach. We'll use apply() and the all() function to create a boolean vector to index into the matrix of interest.
x <- matrix(c(0,0,0,1,1,0,1,1,1,1), ncol = 2, byrow = TRUE)
x
> x
[,1] [,2]
[1,] 0 0
[2,] 0 1
[3,] 1 0
[4,] 1 1
[5,] 1 1
> x[apply(x, 1, function(y) all(y > 0)) ,]
[,1] [,2]
[1,] 1 1
[2,] 1 1