I have a matrix like so:
m <- matrix(c(1,1,1,1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,
1,1,1,1,1,1,1,2,0,1,1,1,1,1,1,1,1,1), nrow = 12, ncol = 12)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 1 1 1 1 1 1 1 1 1 1 1 1
[2,] 1 1 2 1 1 2 1 1 2 1 1 2
[3,] 1 1 0 1 1 0 1 1 0 1 1 0
[4,] 1 1 1 1 1 1 1 1 1 1 1 1
[5,] 1 1 1 1 1 1 1 1 1 1 1 1
[6,] 1 1 1 1 1 1 1 1 1 1 1 1
[7,] 1 1 1 1 1 1 1 1 1 1 1 1
[8,] 1 1 1 1 1 1 1 1 1 1 1 1
[9,] 0 1 1 0 1 1 0 1 1 0 1 1
[10,] 1 1 1 1 1 1 1 1 1 1 1 1
[11,] 1 1 1 1 1 1 1 1 1 1 1 1
[12,] 1 1 1 1 1 1 1 1 1 1 1 1
and I want to find the index locations where 0 is completely surrounded by 1s in a 3x3 window. I can find all the zeros with:
which(m == 0) but this will also return places where a 2 surrounds a 0 such as at index location m[3,3]
w <- which(m == 0, arr.ind = TRUE)
w
# row col
# [1,] 9 1
# [2,] 3 3
# [3,] 9 4
# [4,] 3 6
# [5,] 9 7
# [6,] 3 9
# [7,] 9 10
# [8,] 3 12
We don't need to know which zeroes are on a boundary, so filter out those:
w <- w[ w[,1] > 1 & w[,1] < (nrow(m)-1) & w[,2] > 2 & w[,2] < (ncol(m)-1), ]
w
# row col
# [1,] 3 3
# [2,] 9 4
# [3,] 3 6
# [4,] 9 7
# [5,] 3 9
# [6,] 9 10
Now we can take those inner indices and build 3x3 submatrices into a list. Here are the first couple (of six):
Map(function(rn,cn) m[rn+(-1:1),cn+(-1:1)], w[,1], w[,2])[1:2]
# [[1]]
# [,1] [,2] [,3]
# [1,] 1 2 1
# [2,] 1 0 1
# [3,] 1 1 1
# [[2]]
# [,1] [,2] [,3]
# [1,] 1 1 1
# [2,] 1 0 1
# [3,] 1 1 1
Now we can just filter out the ones where there is only one non-1 entry in the matrix.
Filter(function(m3) sum(m3 != 1) == 1, Map(function(rn,cn) m[rn+(-1:1),cn+(-1:1)], w[,1], w[,2]))
# [[1]]
# [,1] [,2] [,3]
# [1,] 1 1 1
# [2,] 1 0 1
# [3,] 1 1 1
# [[2]]
# [,1] [,2] [,3]
# [1,] 1 1 1
# [2,] 1 0 1
# [3,] 1 1 1
# [[3]]
# [,1] [,2] [,3]
# [1,] 1 1 1
# [2,] 1 0 1
# [3,] 1 1 1
Since you need to just count the occurrences, add length(...) around that, and you have your answer.
(If you're curious, the reason I went with sum(m3!=1)==1 is because I wasn't certain if you wanted the border submatrices as well. If you wanted those, then the number of 1s would be reduced, not "8" as a typical 3x3 would be. But we know that there should always be exactly one non-1 in the submatrix: the center 0.)
To get just the indices that match,
w[mapply(function(rn,cn) sum(m[rn+(-1:1),cn+(-1:1)] != 1) == 1,
w[,1], w[,2]),]
# row col
# [1,] 9 4
# [2,] 9 7
# [3,] 9 10
Related
i have to put a new vector (in this example zero-vector) into an exisiting matrix. The problem is that I have an iterative process and the positions and number of vectors to insert change. I have not been able to come up with a function that a) works and b) is efficient enough for huge amounts of data.
A non-dynamic approach using simply cbind() is
old <- matrix(1,10,10) #original matrix
vec <- matrix(5,10,1) #vector 1 to insert
vec2 <- matrix(8,10,1) #vector 2 to insert
old
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 1 1 1 1 1 1 1 1 1 1
[2,] 1 1 1 1 1 1 1 1 1 1
[3,] 1 1 1 1 1 1 1 1 1 1
[4,] 1 1 1 1 1 1 1 1 1 1
[5,] 1 1 1 1 1 1 1 1 1 1
[6,] 1 1 1 1 1 1 1 1 1 1
[7,] 1 1 1 1 1 1 1 1 1 1
[8,] 1 1 1 1 1 1 1 1 1 1
[9,] 1 1 1 1 1 1 1 1 1 1
[10,] 1 1 1 1 1 1 1 1 1 1
#assume that the positions to insert are 4 and 8
goal <- cbind(old[,c(1:3)],
vec,
old[,4:6], #attention, now old column 6 is new column 7
vec2,
old[,7:ncol(old)])
goal
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 1 1 1 5 1 1 1 8 1 1 1 1
[2,] 1 1 1 5 1 1 1 8 1 1 1 1
[3,] 1 1 1 5 1 1 1 8 1 1 1 1
[4,] 1 1 1 5 1 1 1 8 1 1 1 1
[5,] 1 1 1 5 1 1 1 8 1 1 1 1
[6,] 1 1 1 5 1 1 1 8 1 1 1 1
[7,] 1 1 1 5 1 1 1 8 1 1 1 1
[8,] 1 1 1 5 1 1 1 8 1 1 1 1
[9,] 1 1 1 5 1 1 1 8 1 1 1 1
[10,] 1 1 1 5 1 1 1 8 1 1 1 1
However, I could not think of something that works with both changing positions and number of vectors to insert.
Any help is greatly appreciated, thank you very much.
cbind the vectors onto old and then reorder. If we knew that no were already sorted then we could replace sort(no) with no.
no <- c(4, 8)
vecs <- cbind(vec, vec2)
cbind(old, vecs)[, order(c(1:ncol(old), sort(no) - seq_along(no))) ]
Extending G. Grothendiecks approach and solving the ordering problem:
pos<-c(4,8)
pos<-pos-c(1:length(pos))
cbind(old, vec, vec2)[, order(c(1:ncol(old), c(pos)))]
Edit: Sorry, didn't see the edit of the answer above :)
How to construct a two-circulant matrix?
For example, the following matrix A is two-circulant, i.e every column (expect from the first one) is obtained from the previous one by putting the last two elements as first. Note that the first column is the generator of the matrix.
N=12
k=6
x=c(0,0,0,0,1,1,1,1,2,2,2,2)
A=matrix(0,N,k)
A[,1]=x
for( j in 2:ncol(A) )
{
A[,j]=c(A[11:12,j-1],A[1:10,j-1])
}
> A
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 0 2 2 1 1 0
[2,] 0 2 2 1 1 0
[3,] 0 0 2 2 1 1
[4,] 0 0 2 2 1 1
[5,] 1 0 0 2 2 1
[6,] 1 0 0 2 2 1
[7,] 1 1 0 0 2 2
[8,] 1 1 0 0 2 2
[9,] 2 1 1 0 0 2
[10,] 2 1 1 0 0 2
[11,] 2 2 1 1 0 0
[12,] 2 2 1 1 0 0
Is there any other way to constuct the matrix A? For example by using a function.
You could use the following:
circular_matrix <- function(x, ncol) {
coll <- list(x)
for (i in 1:(ncol-1)) {
current <- coll[[length(coll)]]
coll[[length(coll) + 1]] <- c(tail(current, 2), current[1:(length(current) - 2)])
}
do.call(cbind, coll)
}
circular_matrix(1:10, 5)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 9 7 5 3
# [2,] 2 10 8 6 4
# [3,] 3 1 9 7 5
# [4,] 4 2 10 8 6
# [5,] 5 3 1 9 7
# [6,] 6 4 2 10 8
# [7,] 7 5 3 1 9
# [8,] 8 6 4 2 10
# [9,] 9 7 5 3 1
#[10,] 10 8 6 4 2
a is a matrix.
a<-matrix(c(2,2,1,1,1,2,1,2,1,1,2,1,1,1,1,1,1,2,2,2,1,2,1,1),6)
I want to duplicate matrix "a" 3 times to form a list.
I have tried:
as.list(rep(a,3))
but it doesn't work.
My expected result is as follows:
[[1]]
[,1] [,2] [,3] [,4]
[1,] 2 1 1 2
[2,] 2 2 1 2
[3,] 1 1 1 1
[4,] 1 1 1 2
[5,] 1 2 1 1
[6,] 2 1 2 1
[[2]]
[,1] [,2] [,3] [,4]
[1,] 2 1 1 2
[2,] 2 2 1 2
[3,] 1 1 1 1
[4,] 1 1 1 2
[5,] 1 2 1 1
[6,] 2 1 2 1
[[3]]
[,1] [,2] [,3] [,4]
[1,] 2 1 1 2
[2,] 2 2 1 2
[3,] 1 1 1 1
[4,] 1 1 1 2
[5,] 1 2 1 1
[6,] 2 1 2 1
rep returns the same type it receives, so if you pass it a numeric matrix, it's going to try to return a numeric vector of some sort–not a list. The solution, though, is simple: if you want a list, pass it a list:
rep(list(a), 3)
# [[1]]
# [,1] [,2] [,3] [,4]
# [1,] 2 1 1 2
# [2,] 2 2 1 2
# [3,] 1 1 1 1
# [4,] 1 1 1 2
# [5,] 1 2 1 1
# [6,] 2 1 2 1
#
# [[2]]
# [,1] [,2] [,3] [,4]
# [1,] 2 1 1 2
# [2,] 2 2 1 2
# [3,] 1 1 1 1
# [4,] 1 1 1 2
# [5,] 1 2 1 1
# [6,] 2 1 2 1
#
# [[3]]
# [,1] [,2] [,3] [,4]
# [1,] 2 1 1 2
# [2,] 2 2 1 2
# [3,] 1 1 1 1
# [4,] 1 1 1 2
# [5,] 1 2 1 1
# [6,] 2 1 2 1
Maybe using lapply:
lapply(1:3, function(i) a)
Benchmarking:
library(microbenchmark)
a <- matrix(c(2,2,1,1,1,2,1,2,1,1,2,1,1,1,1,1,1,2,2,2,1,2,1,1),6)
microbenchmark(
replicate(3, a, simplify = FALSE),
rep(list(a), 3),
lapply(1:3, function(i) a),
times = 10000)
# Unit: nanoseconds
# expr min lq mean median uq max neval cld
# replicate(3, a, simplify = FALSE) 5987 7127 8044.9371 7413 7983 1150921 10000 c
# rep(list(a), 3) 0 285 396.2066 285 570 15395 10000 a
# lapply(1:3, function(i) a) 1995 2566 3013.6994 2851 3136 1290902 10000 b
I have a variable that is a list.
[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 1 1 1 1 1 1 4
[2,] 1 1 1 1 1 1 1 4
[3,] 1 1 1 1 1 1 1 4
[4,] 1 1 1 1 1 1 1 4
[5,] 1 1 1 1 1 1 1 4
[6,] 1 1 1 1 1 1 1 4
[7,] 1 1 1 1 1 1 1 4
[8,] 1 1 1 1 1 1 1 4
[9,] 1 1 1 1 1 1 1 4
[10,] 1 1 1 1 1 1 1 4
[11,] 1 1 1 1 1 1 1 4
[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 1 1 1 1 1 1 3
[2,] 1 1 1 1 1 1 1 4
[3,] 1 1 1 1 1 1 1 4
[4,] 1 1 1 1 1 1 1 3
[5,] 1 1 1 1 1 1 1 4
[6,] 1 1 1 1 1 1 1 4
[7,] 1 1 1 1 1 1 1 4
[8,] 1 1 1 1 1 1 1 4
[9,] 1 1 1 1 1 1 1 4
[10,] 1 1 1 1 1 1 1 4
[11,] 1 1 1 1 1 1 1 4
I need an empty variable which is same format as this variable.
My function will do some computation and will be put in the same location as this variable. However in the new variable I will not have 8th column.
Replicate your data
df = list(matrix(rep(1, 88), ncol = 8), matrix(rep(1, 88), ncol = 8))
Remove the 8th column from all sublist
new_df = lapply(df, function(x) x[,-8])
Remove the 8th column from all sublist and replace all the 1's with NA
new_df = lapply(df, function(x) replace(x[,-8], x[,-8] == 1, NA))
So, my goal is to take an input vector and to make an output matrix of different counters. So every time a value appears in my inputs, I want to find that counter and iterate it by 1. I understand that I'm not good at explaining this, so I illustrated a simple version below. However, I want to make 2 changes which I will enumerate after the example so that it makes sense.
nums = c(1,2,3,4,5,1,2,4,3,5)
unis = unique(nums)
counter = matrix(NA, nrow = length(nums), ncol = length(unis))
colnames(counter) = unis
for (i in 1:length(nums)){
temp = nums[i]
if (i == 1){
counter[1,] = 0
counter[1,temp] = 1
} else {
counter[i,] = counter[i-1,]
counter[i,temp] = counter[i-1,temp]+1
}
}
counter
which outputs
> counter
1 2 3 4 5
[1,] 1 0 0 0 0
[2,] 1 1 0 0 0
[3,] 1 1 1 0 0
[4,] 1 1 1 1 0
[5,] 1 1 1 1 1
[6,] 2 1 1 1 1
[7,] 2 2 1 1 1
[8,] 2 2 1 2 1
[9,] 2 2 2 2 1
[10,] 2 2 2 2 2
The 2 modifications. 1) Since the real data is much larger, I would want to do this using apply or however people who know R better than me says it should be done. 2) Whereas the input is a vector where each element is only an element, how could this be generalized if an element of a vector was a tuple? For example (if nums was a tuple of 4 and 5, then it would iterate both in that step and the last line of the output would then be 2,2,2,3,2)
Thanks and if you don't understand please ask questions and I'll try to clarify
Using the Matrix package (which ships with a standard installation of R)
nums <- c(1,2,3,4,5,1,2,4,3,5)
apply(Matrix::sparseMatrix(i=seq_along(nums), j=nums), 2, cumsum)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 0 0 0 0
# [2,] 1 1 0 0 0
# [3,] 1 1 1 0 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 1 1
# [7,] 2 2 1 1 1
# [8,] 2 2 1 2 1
# [9,] 2 2 2 2 1
# [10,] 2 2 2 2 2
Note that this behaves a bit differently in a couple of ways from thelatemail's suggested solution. Which behavior you prefer will depend on what you are using this for.
Here's a small example that illustrates the differences:
nums <- c(5,2,1,1)
# My suggestion
apply(Matrix::sparseMatrix(i=seq_along(nums), j=nums), 2, cumsum)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 0 0 0 0 1
# [2,] 0 1 0 0 1
# [3,] 1 1 0 0 1
# [4,] 2 1 0 0 1
# #thelatemail's suggestion
sapply(unique(nums), function(x) cumsum(nums==x) )
# [,1] [,2] [,3]
# [1,] 1 0 0
# [2,] 1 1 0
# [3,] 1 1 1
# [4,] 1 1 2
For your second question, you could do something like this:
nums <- list(1,2,3,4,5,1,2,4,3,c(4,5))
ii <- rep(seq_along(nums), times=lengths(nums)) ## lengths() is in R>=3.2.0
jj <- unlist(nums)
apply(Matrix::sparseMatrix(i=ii, j=jj), 2, cumsum)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 0 0 0 0
# [2,] 1 1 0 0 0
# [3,] 1 1 1 0 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 1 1
# [6,] 2 1 1 1 1
# [7,] 2 2 1 1 1
# [8,] 2 2 1 2 1
# [9,] 2 2 2 2 1
# [10,] 2 2 2 3 2
For your first query, you can get there with something like:
sapply(unique(nums), function(x) cumsum(nums==x) )
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 0 0 0 0
# [2,] 1 1 0 0 0
# [3,] 1 1 1 0 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 1 1
# [6,] 2 1 1 1 1
# [7,] 2 2 1 1 1
# [8,] 2 2 1 2 1
# [9,] 2 2 2 2 1
#[10,] 2 2 2 2 2
Another idea:
do.call(rbind, Reduce("+", lapply(nums, tabulate, max(unlist(nums))), accumulate = TRUE))
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 0 0 0 0
# [2,] 1 1 0 0 0
# [3,] 1 1 1 0 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 1 1
# [6,] 2 1 1 1 1
# [7,] 2 2 1 1 1
# [8,] 2 2 1 2 1
# [9,] 2 2 2 2 1
#[10,] 2 2 2 2 2
And generally:
x = list(1, 3, 6, c(6, 3), 2, c(4, 6, 1), c(1, 2), 3)
do.call(rbind, Reduce("+", lapply(x, tabulate, max(unlist(x))), accumulate = TRUE))
# [,1] [,2] [,3] [,4] [,5] [,6]
#[1,] 1 0 0 0 0 0
#[2,] 1 0 1 0 0 0
#[3,] 1 0 1 0 0 1
#[4,] 1 0 2 0 0 2
#[5,] 1 1 2 0 0 2
#[6,] 2 1 2 1 0 3
#[7,] 3 2 2 1 0 3
#[8,] 3 2 3 1 0 3