I have a square matrix M with 25x25 dimension.
Then I want to create 25 matrices as follow:
the first matrix is matrix M without the first row and first column,
the second matrix is matrix M without the second row and second column, - ... so on until 25th matrix.
this little snippet will do:
lapply(1:25, function(i) M[-i, -i])
Related
I have a vector containing times in milliseconds looking like this;
vector <- c(667753, 671396, 675356, 679286, 683413, 687890, 691742,
695651, 700100, 704552, 708832, 713117, 717082, 720872, 725002, 729490,
733824, 738233, 742239, 746092, 750003, 754236, 867342, 870889, 873704,
876617, 879626, 882595, 885690, 888602, 891789, 894717, 897547, 900797,
903615, 906646, 909624, 912613, 915645, 918566, 921792, 924625, 927538,
930721, 933542)
Now i want to look into a large data frame with a lot of time columns and search for a single column that contains time values being closest (row-wise) to my vector time values.
The data.frame containing all the columns is of the same number of rows. So lets say my vector has 240 elements, then every column in the larger data.frame consists of 240 rows.
Any idia how to do this ?
You can calculate the euclidean distance from your vector and each column of the dataframe and then check which column has the smallest distance:
which.min(sapply(1:ncol(dataFrame), function(i) sqrt(sum((t(v)-dataFrame[,i])^2))))
The above returns the index of the column with the lowest distance.
Where dataFrame is the data frame containing columns of different times(so we compare each column to the vector v) and v is the vector.
The following is just the square root of the sum of squared distances (euclidean distance):
sqrt(sum((t(v)-dataFrame[,i])^2)))
You can also use the following as a distance measure:
abs(t(v)-dataFrame[,i])
EDIT
As Evan Friedland pointed out you can actually just use:
which.min(colSums(abs(v-dataFrame)))
or
which.min(sqrt(colSums((t(v)-dataFrame)^2)))
I want to multiply each column of matrix A by matrix C. For this I am using for loop as follows:
A=[ 0. 1. 2. 3;0. 1. 2. 3.]
C=[2 0;0 2].
for i=1:4
B(i)=C*A(:,i);
end
But no matrix B(i) is displaying.
The result of C*A(:,i) is a column matrix. To store all columns in a single matrix, you have to use the same notation you used to retrieve a single column from A. Therefore, you should write this in your loop:
B(:,i) = C * A(:,i);
I would like to create a (N*M)-Incidence Matrix for a bipartite graph (N=M=200).
However, the following restrictions have to be considered:
Each column i ( 1 , ... , 200 ) has a column sum of g = 10
each row has a Row sum of h = 10
no multiedges (The values in the incidence Matrix only take on the values [0:1]
So far I have
M <- 200; # number of rows
N <- 200; # number of colums
g <- 10
I <- matrix(sample(0:1, M*N, repl=T, prob= c(1-g/N,g/N)), M, N);
Does anybody has a solution?
Here's one way to do what you want. First the algorithm idea, then its implementation in R.
Two step Algorithm Idea
You want a matrix of 0's and 1's, with each row adding up to be 10, and each column adding up to be 10.
Step 1: First,create a trivial solution as follows:
The first 10 rows have 1's for the first 10 elements, then 190 zeros.
The second set of ten rows have 1's from the 11th to the 20th element and so on.
In other words, a feasible solution is to have a 200x200 matrix of all 0's, with dense matrices of 10x10 1's embedded diagonally, 20 times.
Step 2: Shuffle entire rows and entire columns.
In this shuffle, the rowSum and columnSums are maintained.
Implementation in R
I use a smaller matrix of 16x16 to demonstrate. In this case, let's say we want each row and each column to add up to 4. (This colsum has to be integer divisible of the larger square matrix dimension.)
n <- 4 #size of the smaller square
i <- c(1,1,1,1) #dense matrix of 1's
z <- c(0,0,0,0) #dense matrix of 0's
#create a feasible solution to start with:
m <- matrix(c(rep(c(i,z,z,z),n),
rep(c(z,i,z,z),n),
rep(c(z,z,i,z),n),
rep(c(z,z,z,i),n)), 16,16)
#shuffle (Run the two lines following as many times as you like)
m <- m[sample(16), ] #shuffle rows
m <- m[ ,sample(16)] #shuffle columns
#verify that the sum conditions are not violated
colSums(m); rowSums(m)
#solution
print(m)
Hope that helps you move forward with your bipartite igraph.
I have a (nxc+n+c) by 1 matrix. And I want to deselect the last n+c rows and convert the rest into a nxc matrix. Below is what I've tried, but it returns a matrix with every element the same in one row. I'm not sure why is this. Could someone help me out please?
tmp=x[1:n*c,]
Membership <- matrix(tmp, nrow=n, ncol=c)
You have a vector x of length n*c + n + c, when you do the extract, you put a comma in your code.
You should do tmp=x[1:(n*c)].
Notice the importance of parenthesis, since if you do tmp=x[1:n*c], it will take the range from 1 to n, multiply it by c - giving a new range and then extract based on this new range.
For example, you want to avoid:
(1:100)[1:5*5]
[1] 5 10 15 20 25
You can also do without messing up your head with indexing:
matrix(head(x, n*c), ncol=c)
I have a full matrix of numbers. On a computer, I can easily set with zeroes a row or a column. I would like to know how I can represent this operation symbolically in a mathematical expression.
For a n x n matrix A and with
e = ones(n)
e[k] = 0
matrix multiplication
A*diag(e)
zeros the k-th column and
diag(e)*A
zeros the k-th row