I have a vector c(1,2,3,4,5,6,7,8,9,10,11). I want to draw a sample of N = 2, 5 times from this vector without replacement; I would like to save the results from each selection in a separate column.
The output could look like this:
structure(c(1, 2, 3, 9, 7, 10, 4, 8, 5, 6), dim = c(2L, 5L)).
matrix with two rows and five columns.
matrix(sample(x), nrow=2, ncol=5)
# [,1] [,2] [,3] [,4] [,5]
# [1,] 5 7 4 8 2
# [2,] 3 10 9 1 6
Something like the following:
set.seed(7*11*13)
x <- c(1,2,3,4,5,6,7,8,9,10,11)
result <-matrix(as.numeric(NA), 2, 5)
for (i in 1:5) result[,i] <-
sample(x, 2, replace=FALSE)
Related
I am trying to calculate the sum of the differences of all possible pairs of an array (sum of difference first/second + difference first/third, difference second/first + difference second/third, difference third/first + difference third/second). The original dataframe consists of multiple columns like the one below, still the colSum function doesn't work because of:
"Error in colSums(FC_TS1test1, dims = 1) : 'x' must be an array of at least two dimensions".
I appreciate your help!
array <- c(5, 3, 1)
The calculation behind it should be:
|(5-3)|+|(5-1)|=2+4=6
|(3-5)|+|(3-1)|=2+2=4
|(1-5)|+|(1-3)|=4+2=6
6+4+6=16
We can use sum + dist
> sum(dist(array))
[1] 840
We need combinations without repetition.
sum(combn(a, 2, diff))
# [1] -94
Or, with some packages:
colSums(matrixStats::rowDiffs(RcppAlgos::comboGeneral(a, 2, repetition=F)))
# [1] -94
or
sum(unlist(RcppAlgos::comboGeneral(a, 2, repetition=F, FUN=diff)))
# [1] -94
Two smaller examples demonstrate what the core of the code does.
combn(b, 2)
# [,1] [,2] [,3] [,4] [,5] [,6]
# [1,] 3 3 3 4 4 2
# [2,] 4 2 5 2 5 5
RcppAlgos::comboGeneral(b, 2, repetition=F)
# [,1] [,2]
# [1,] 3 4
# [2,] 3 2
# [3,] 3 5
# [4,] 4 2
# [5,] 4 5
# [6,] 2 5
Edit
For the absolute differences according to your recent edit we may define an anonymous function:
sum(combn(a, 2, function(x) abs(diff(x))))
# [1] 840
Data:
a <- c(3, 4, 2, 5, 4, 4, 1, 5, 5, 4, 1, 4, 7, 2, 1, 3, 5, 2, 4, 2,
7, 4, 4, 1, 4, 3, 4, 4, 2, 4, 1)
b <- c(3, 4, 2, 5)
Perhaps I am not using the right terms in my search, but I was wondering if anyone could point out an easy way of doing the following:
I have two matrices:
mat1 = matrix(1:12, 3)
mat2 = matrix(c(1, 2, 1, 2, 3, 2, 1, 2), 2, 4)
I want to multiply each of row in a certain column of mat1 by the first row of mat2. For example, column 3 of mat1 would become (7*3, 8*3, 9*3)=(21, 24, 27). After this, I want to add the second row of mat2 to each row in a certain column of mat1, so the column 3 would become (21+2, 24+2, 27+2) = (23, 26, 29).
You can try the code below
t(t(mat1)*mat2[1,]+mat2[2,])
such that
> t(t(mat1)*mat2[1,]+mat2[2,])
[,1] [,2] [,3] [,4]
[1,] 3 6 23 12
[2,] 4 7 26 13
[3,] 5 8 29 14
I have a nxp data frame and I want to convert it to an array with n matrices where each matrix is, let's say, ixj where i+j = p. Considering the following reprex:
library(tidyverse)
df <- tribble(
~x1, ~x2, ~x3, ~x4,
1, 2, 3, 4,
5, 6, 7, 8)
The desired result is an array with 2 matrices similar to the one produced by:
array(1:8, c(2,2,2))
Would anybody have an efficient method to obtain such results in high dimensional data frames?
With baseR and array:
array(t(df), c(2, 2, 2))
#, , 1
#
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
#
#, , 2
#
# [,1] [,2]
#[1,] 5 7
#[2,] 6 8
I'm a programming beginner and I'm not able to solve this problem:
I have a vector length 132 and two matrices A and B with the size of 132x24. I would like to take every single value of the vector and compare it rowwise with matrix A. If the value occurs in A I want to have the index of the column to go to matrix B and pick the value from the column with the same position (row and column indices) as in matrix A. The results should be given back as a vector with the same length of 132.
How to do this? Do I need a for loop or are there some smart ways to work with packages?
Unfortunately I can not give example data.
Thank you for your help!
# vector v contains values that I want to compare with matrix A
> v
[1] 5 1 10 1 7
# every single value of v occurs in every row of A only once
# I want to have the position of this value in matrix A
> A
[,1] [,2] [,3] [,4]
[1,] 5 7 4 1
[2,] 14 1 3 3
[3,] 13 3 1 10
[4,] 2 1 5 8
[5,] 13 2 5 7
# the position in matrix A equals the position in matrix B
# now the values of B have to be returned as a vector
> B
[,1] [,2] [,3] [,4]
[1,] 6 3 4 3
[2,] 5 2 5 5
[3,] 4 6 3 1
[4,] 3 6 1 5
[5,] 2 4 6 3
# vector with fitting values of B
> x
[1] 6 2 1 6 3
v <- c(5, 1, 10, 1, 7)
A <- matrix(c(
5, 7, 4, 1,
14, 1, 3, 3,
13, 3, 1, 10,
2, 1, 5, 8,
13, 2, 5, 7), 5, byrow = TRUE)
B <- matrix(c(
6, 3, 4, 3,
5, 2, 5, 5,
4, 6, 3, 1,
3, 6, 1, 5,
2, 4, 6, 3), 5, byrow = TRUE)
myfun <- function(i) which(v[i]==A[i,])
ii <- 1:length(v)
B[cbind(ii, sapply(ii, myfun))]
The function myfun() is quick'n'dirty.
To test if your data are ok you can calculate how often the value v[i] is found in the row A[i,]
countv <- function(i) sum(v[i]==A[i,])
all(sapply(ii, countv)==1) ### should be TRUE
If you get FALSE then inspect:
which(sapply(ii, countv)!=1)
Alright, I'm not sure how you pictured your output, but I've got something that comes near.
Example data:
x <- 1:132
set.seed(123)
A <- matrix(sample(1:1000, size = 132*24, replace = TRUE), nrow = 132, ncol = 24)
B <- matrix(rnorm(132*24), nrow = 132, ncol = 24)
Now we check for every value of vector x if and where it occurs in every row of matrix A:
x.vs.A <- sapply(x, function(x){
apply(A, 1, function(y) {
match(x, y)
})
})
This gives us a matrix x.vs.A with 132 rows (the rows of A) and 132 columns (the values of x). Within the cells of this matrix, we will find either NA, if the combination of one value of x and one row of A was unsuccessful, or the column position within A of the FIRST match of the value of x.
And now we extract the rowwise position and bind them together with the cell value, depiting the second (column) dimension of the matched value. Thus we create for every value of x a matrix of row/column position of matches in matrix A:
x.in.A <- apply(x.vs.A, 2, function(x) cbind(which(!is.na(x)), x[!is.na(x)]))
Example:
> x.in.A[[1]]
[,1] [,2]
[1,] 12 17
[2,] 42 17
[3,] 73 12
[4,] 123 21
This would show that the first value in vector x can be found in A[12, 17], in A[42, 17] and so on.
Now access these values in B, returning vectors for each value of x, and bind them to the matrices in the list:
x.in.B <- lapply(x.in.A, function(x){
apply(x, 1, function(y){
B[y[1], y[2]]
})
})
x.in.AB <- mapply(function(x, y) cbind(x, y),
x.in.A, x.in.B)
> x.in.AB[[1]]
y
[1,] 12 17 -0.2492526
[2,] 42 17 -0.7985330
[3,] 73 12 0.1253824
[4,] 123 21 -0.9704919
I have a problem I think best can be solved using cominatorics.
Lets say you have values 4 values (2,5,6,7). I would like to get all vectors where I pick out 3 of them, that is I would like a matrix with (2,5,6),(2,5,7),(5,6,7). I would like to do this with a general vector. How do I do it?
x <- c(2, 5, 6, 7)
combn(x, 3)
gives
> combn(x, 3)
[,1] [,2] [,3] [,4]
[1,] 2 2 2 5
[2,] 5 5 6 6
[3,] 6 7 7 7
I have created a package iterpc which is able to solve most of the combination/permutation related problems.
library(iterpc)
x <- c(2, 5, 6, 7)
# combinations
getall(iterpc(4, 3, label=x))
# combinations with replacement
getall(iterpc(4, 3, replace=TRUE, label=x))
# permutations
getall(iterpc(4, 3, label=x, ordered=TRUE))