How to make a model.matrix in a specific way? - r

How would one produce a matrix that looks like this using model.matrix?
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 0 0 0 0
[2,] 1 2 0 0 0 0
[3,] 1 3 0 0 0 0
[4,] 0 0 1 1 0 0
[5,] 0 0 1 2 0 0
[6,] 0 0 1 3 0 0
[7,] 0 0 0 0 1 1
[8,] 0 0 0 0 1 2
[9,] 0 0 0 0 1 3
I produced the first matrix by
fit = lmer(Temp ~ 1 + (1 + Time|Id), data = Data)
getME(fit, name = c("Z"))
Where
Time = rep(1:3, 3)
And Id
Id = c(0L, cumsum(diff(Time) < 0))
This is as close as I could get.
id = rep(c("a","b","c"),each = 3)
Z = model.matrix(~0+id)
[,1] [,2] [,3]
1 1 0 0
2 1 0 0
3 1 0 0
4 0 1 0
5 0 1 0
6 0 1 0
7 0 0 1
8 0 0 1
9 0 0 1


i'm not sure I get what you try to do, but does this work for you?
val = rep(1:3,3)
z = model.matrix(~0+id+id:val)

Related

Set values withing a range equal to 1 and rest to 0

I have a 7000 X 7000 matrix in R. For example purpose I will use a smaller matrix as following:-
a <- matrix(c(0:-9, 1:-8, 2:-7, 3:-6, 4:-5, 5:-4, 6:-3, 7:-2, 8:-1, 9:0),
byrow = TRUE, ncol = 10, nrow = 10)
I want to create a new matrix which has values equal to 1 where the absolute values in matrix a are between the closed interval of 2 and 5. And rest all other values equal to zero.
This would make the following matrix:-
b <- matrix(c(0,0,1,1,1,1,0,0,0,0
0,0,0,1,1,1,1,0,0,0
1,0,0,0,1,1,1,1,0,0
1,1,0,0,0,1,1,1,1,0
1,1,1,0,0,0,1,1,1,1
1,1,1,1,0,0,0,1,1,1
0,1,1,1,1,0,0,0,1,1
0,0,1,1,1,1,0,0,0,1
0,0,0,1,1,1,1,0,0,0
0,0,0,0,1,1,1,1,0,0),
byrow = TRUE, ncol = 10, nrow = 10)
I can do this using for loop, but I just want to know if there is a much better and effcient solution to do this.
Thanks in advance.

You can just write down the comparison. It gives you a logical matrix and you can then use unary + to turn the result into an integer matrix.
+(abs(a) >= 2 & abs(a) <= 5)
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# [1,] 0 0 1 1 1 1 0 0 0 0
# [2,] 0 0 0 1 1 1 1 0 0 0
# [3,] 1 0 0 0 1 1 1 1 0 0
# [4,] 1 1 0 0 0 1 1 1 1 0
# [5,] 1 1 1 0 0 0 1 1 1 1
# [6,] 1 1 1 1 0 0 0 1 1 1
# [7,] 0 1 1 1 1 0 0 0 1 1
# [8,] 0 0 1 1 1 1 0 0 0 1
# [9,] 0 0 0 1 1 1 1 0 0 0
#[10,] 0 0 0 0 1 1 1 1 0 0

Perhaps you can try
> +((abs(a) - 2) * (abs(a) - 5) <= 0)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 1 1 1 0 0 0 0
[2,] 0 0 0 1 1 1 1 0 0 0
[3,] 1 0 0 0 1 1 1 1 0 0
[4,] 1 1 0 0 0 1 1 1 1 0
[5,] 1 1 1 0 0 0 1 1 1 1
[6,] 1 1 1 1 0 0 0 1 1 1
[7,] 0 1 1 1 1 0 0 0 1 1
[8,] 0 0 1 1 1 1 0 0 0 1
[9,] 0 0 0 1 1 1 1 0 0 0
[10,] 0 0 0 0 1 1 1 1 0 0

How to form the matrix of logical '1' and '0' using two vectors and logical operators in r?

Here is Matlab code to form the matrix of logical values of '0' and '1'
A=[1 2 3 4 5 6 7 8 9 10 ];
N = numel(A);
step = 2; % Set this to however many zeros you want to add each column
index = N:-step:1;
val = (1:N+step).' <= index;
Which result in
val=
1 1 1 1 1
1 1 1 1 1
1 1 1 1 0
1 1 1 1 0
1 1 1 0 0
1 1 1 0 0
1 1 0 0 0
1 1 0 0 0
1 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
How to do same task in r ,particularly val = (1:N+step).' <= indexthis step?

One option is
i <- seq_len(ncol(m1))
sapply(rev(i), function(.i) {
m1[,.i][sequence(.i *2)] <- 1
m1[,.i]
})
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
Or vectorize it
i1 <- rep(i, rev(2*i))
m1[cbind(ave(i1, i1, FUN = seq_along), i1)] <- 1
m1
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
Or another option without creating a matrix beforehand
n <- 5
i1 <- seq(10, 2, by = -2)
r1 <- c(rbind(i1, rev(i1)))
matrix(rep(rep(c(1, 0), n), r1), ncol = n)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 1 1 1 1 1
# [2,] 1 1 1 1 1
# [3,] 1 1 1 1 0
# [4,] 1 1 1 1 0
# [5,] 1 1 1 0 0
# [6,] 1 1 1 0 0
# [7,] 1 1 0 0 0
# [8,] 1 1 0 0 0
# [9,] 1 0 0 0 0
#[10,] 1 0 0 0 0
#[11,] 0 0 0 0 0
#[12,] 0 0 0 0 0
data
m1 <- matrix(0, 12, 5)

R: Matrix Combination with specific number of values

I want to make all combinations of my Matrix.
Ex. a binary 5 X 5 matrix where I only have two 1 rows (see below)
Com 1:
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
Com 2:
1 0 1 0 0
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
1 1 0 0 0
.
.
.
Com ?:
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
0 0 0 1 1
I tried using Combination package in R, but couldn't find a solution.

Using RcppAlgos (I am the author) we can accomplish this with 2 calls. It's quite fast as well:
library(tictoc)
library(RcppAlgos)
tic("RcppAlgos solution")
## First we generate the permutations of the multiset c(1, 1, 0, 0, 0)
binPerms <- permuteGeneral(1:0, 5, freqs = c(2, 3))
## Now we generate the permutations with repetition choose 5
## and select the rows from binPerms above
allMatrices <- permuteGeneral(1:nrow(binPerms), 5,
repetition = TRUE,
FUN = function(x) {
binPerms[x, ]
})
toc()
RcppAlgos solution: 0.108 sec elapsed
Here is the output:
allMatrices[1:3]
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 0 0 0
[2,] 1 1 0 0 0
[3,] 1 1 0 0 0
[4,] 1 1 0 0 0
[5,] 1 1 0 0 0
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 0 0 0
[2,] 1 1 0 0 0
[3,] 1 1 0 0 0
[4,] 1 1 0 0 0
[5,] 1 0 1 0 0
[[3]]
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 0 0 0
[2,] 1 1 0 0 0
[3,] 1 1 0 0 0
[4,] 1 1 0 0 0
[5,] 1 0 0 1 0
len <- length(allMatrices)
len
[1] 100000
allMatrices[(len - 2):len]
[[1]]
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 1 1
[2,] 0 0 0 1 1
[3,] 0 0 0 1 1
[4,] 0 0 0 1 1
[5,] 0 0 1 1 0
[[2]]
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 1 1
[2,] 0 0 0 1 1
[3,] 0 0 0 1 1
[4,] 0 0 0 1 1
[5,] 0 0 1 0 1
[[3]]
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 1 1
[2,] 0 0 0 1 1
[3,] 0 0 0 1 1
[4,] 0 0 0 1 1
[5,] 0 0 0 1 1

The code I've written below worked for me. A list of 100,000 5x5 matrices. Each of the rows has two places set to 1.
n <- 5 # No of columns
k <- 2 # No. of ones
m <- 5 # No of rows in matrix
nck <- combn(1:n,k,simplify = F)
possible_rows <-lapply(nck,function(x){
arr <- numeric(n)
arr[x] <- 1
matrix(arr,nrow=1)
})
mat_list <- possible_rows
for(i in 1:(m-1)){
list_of_lists <- lapply(mat_list,function(x){
lapply(possible_rows,function(y){
rbind(x,y)
})
})
mat_list <- Reduce(c,list_of_lists)
print(c(i,length(mat_list)))
}

How to convert matrix elements from 0|1 to 1|0 in R?

I have a graph that converted to matrix.
g = sample_k_regular(10,3)
m =get.adjacency(g)
I want to select randomly some elements and convert to 0|1.(if it is 0 to become 1 and if it is 1 to become 0).
How to do this work?

You can make sample of n elements (10 in example) and change it
m1=as.matrix(m)
m1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 1 0 0 0 0 1 0
[2,] 0 0 0 0 1 0 0 0 1 1
[3,] 1 0 0 0 0 0 0 1 0 1
[4,] 1 0 0 0 1 0 1 0 0 0
[5,] 0 1 0 1 0 1 0 0 0 0
[6,] 0 0 0 0 1 0 1 0 0 1
[7,] 0 0 0 1 0 1 0 1 0 0
[8,] 0 0 1 0 0 0 1 0 1 0
[9,] 1 1 0 0 0 0 0 1 0 0
[10,] 0 1 1 0 0 1 0 0 0 0
set.seed(1)
ss=sample(length(m1),size = 10)
m1[ss]=1-m1[ss]
m1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 1 1 0 0 0 0 1 0
[2,] 0 0 0 0 1 0 1 0 1 1
[3,] 1 0 0 0 0 0 0 1 0 1
[4,] 1 0 0 0 1 0 1 0 0 0
[5,] 0 1 0 1 0 1 0 0 0 0
[6,] 1 0 0 0 1 0 1 0 1 1
[7,] 0 0 1 0 0 0 0 1 0 1
[8,] 0 0 1 0 0 1 1 0 1 0
[9,] 1 1 0 0 0 0 0 1 1 0
[10,] 0 0 1 0 0 1 0 0 0 0
For exclude diagonal as #ZheyuanLi told
you can calculate diad position and exlude it from data for sample :
m1=as.matrix(m)
m1
set.seed(1)
m_l=1:length(m1)
m_l=m_l[-which(diag(1,nrow = nrow(m1))==1)]
ss=sample(m_l,size = 10)
m1[ss]=1-m1[ss]
m1
For big matrix beter use seq.int than diag
n=1000
Unit: microseconds
expr min lq mean median uq max neval
{ which(diag(1, nrow = n) == 1) } 8976.718 9422.967 14397.44991 10489.0520 16001.550 190959.200 100
{ seq(1, by = n + 1, length = n) } 12.941 17.404 37.90449 31.9075 56.004 83.448 100
{ seq.int(1, by = n + 1, length = n) } 5.355 6.248 8.90736 7.1405 12.272 16.512 100
{ 1 + { (1:n) - 1 } * (1 + n) } 5.355 6.248 9.77758 8.9255 11.826 25.437 100

How to transform a item set matrix in R

How to transform a matrix like
A 1 2 3
B 3 6 9
c 5 6 9
D 1 2 4
into form like:
1 2 3 4 5 6 7 8 9
1 0 2 1 1 0 0 0 0 0
2 0 0 1 1 0 0 0 0 0
3 0 0 0 0 0 1 0 0 1
4 0 0 0 0 0 0 0 0 0
5 0 0 0 0 0 1 0 0 1
6 0 0 0 0 0 0 0 0 2
7 0 0 0 0 0 0 0 0 0
8 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0
I have some implement for it ,but it use the for loop
I wonder if there has some inner function in R (for example "apply")
add:
Sorry for the confusion.The first matrix just mean items sets, every set of items come out pairs ,for example the first set is "1 2 3" , and will become (1,2),(1,3),(2,3), correspond the second matrix.
and another question :
If the matrix is very large (10000000*10000000)and is sparse
should I use sparse matrix or big.matrix?
Thanks!

Removing the row names from M gives this:
m <- matrix(c(1,3,5,1,2,6,6,2,3,9,9,4), nrow=4)
> m
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 3 6 9
## [3,] 5 6 9
## [4,] 1 2 4
# The indicies that you want to increment in x, but some are repeated
# combn() is used to compute the combinations of columns
indices <- matrix(t(m[,combn(1:3,2)]),,2,byrow=TRUE)
# Count repeated rows
ones <- rep(1,nrow(indices))
cnt <- aggregate(ones, by=as.data.frame(indices), FUN=sum)
# Set each value to the appropriate count
x <- matrix(0, 9, 9)
x[as.matrix(cnt[,1:2])] <- cnt[,3]
x
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
## [1,] 0 2 1 1 0 0 0 0 0
## [2,] 0 0 1 1 0 0 0 0 0
## [3,] 0 0 0 0 0 1 0 0 1
## [4,] 0 0 0 0 0 0 0 0 0
## [5,] 0 0 0 0 0 1 0 0 1
## [6,] 0 0 0 0 0 0 0 0 2
## [7,] 0 0 0 0 0 0 0 0 0
## [8,] 0 0 0 0 0 0 0 0 0
## [9,] 0 0 0 0 0 0 0 0 0

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