I have two arrays.
Using numpy.append we can merge two arrays.
How can we do same thing in R?
merge can not do that.
Python Output/Example:
a=np.array([1,2,3,4,5,5])
b=np.array([0,0,0,0,0,0])
np.append(a,b)
array([1, 2, 3, 4, 5, 5, 0, 0, 0, 0, 0, 0]) # this is what I want
x<-c(mat , (0.0) * (l - length(demeaned)))
mat is matrix (size is 20)
l - length(demeaned) is 10
i want at the end 30 size
The c-function concatenates its arguments. A vector can be a concatenation of numbers or of other verctors:
a = c(1,2,3,4,5,5)
b = c(0,0,0,0,0,0)
c(a,b)
[1] 1 2 3 4 5 5 0 0 0 0 0 0
At least for one-dimensional arrays like in your python-example this is equivalent to np.append
Adding to the previous answer, you can use rbind or cbind to create two-dimensional arrays (matrices) from simple arrays (vectors):
cbind(a,b)
# output
a b
[1,] 1 0
[2,] 2 0
[3,] 3 0
[4,] 4 0
[5,] 5 0
[6,] 5 0
or
rbind(a,b)
# output
[,1] [,2] [,3] [,4] [,5] [,6]
a 1 2 3 4 5 5
b 0 0 0 0 0 0
If you want to convert it back to vector, use as.vector. This
as.vector(rbind(a,b))
will give you a joined vector with alternating elements.
Also, note that c can flatten lists if you use the recursive=TRUE argument:
a <- list(1,list(1,2,list(3,4)))
b <- 10
c(a,b, recursive = TRUE)
# output
[1] 1 1 2 3 4 10
Finally, you can use rep to generate sequences of repeating numbers:
rep(0,10)
Related
Suppose I have a list of matrices. Suppose further I have found the smallest values by the column.
Here is my last question
I really need to know from which matrix each smallest value is selected. My original function is very complicated. Therefore, I provided a simple example. I have one idea and really do not know to implement it correctly in R.
My idea is:
Suppose that [i,j] is the elements of the matrix. Then,
if(d[[1]][i,j] < d[[2]][i,j]){
d[[1]][i,j] <– "x"
}else { d[[2]][i,j] <– "z"}
So, I would like to sign the name of the matrix that corresponds to each smallest value. Then, store the names in a separate matrix. So, then I can see the values in one matrix and their corresponding names (from where they come from) in another matrix
For example,
y <- c(3,2,4,5,6, 4,5,5,6,7)
x[lower.tri(x,diag=F)] <- y
> x
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 0 0
[2,] 3 0 0 0 0
[3,] 2 6 0 0 0
[4,] 4 4 5 0 0
[5,] 5 5 6 7 0
k <- c(1,4,5,2,5,-4,4,4,4,5)
z[lower.tri(z,diag=F)] <- k
> z
[,1] [,2] [,3] [,4] [,5]
[1,] 0 0 0 0 0
[2,] 1 0 0 0 0
[3,] 4 5 0 0 0
[4,] 5 -4 4 0 0
[5,] 2 4 4 5 0
d <- list(z, x)
Then:
do.call(pmin, d) (answered by #akrun)
Then, I will only get the matrix with smallest values. I would like to know where each value is come from?
Any idea or help, please?
You can use Map and do.call to create your own functions that will be applied element-wise to a list of inputs,
in your case a list of matrices.
pwhich.min <- function(...) {
which.min(c(...)) # which.min takes a single vector as input
}
di <- unlist(do.call(Map, c(list(f = pwhich.min), d)))
dim(di) <- dim(x) # take dimension from one of the inputs
di
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 1 1
[2,] 2 1 1 1 1
[3,] 1 2 1 1 1
[4,] 1 2 2 1 1
[5,] 2 2 2 2 1
EDIT:
To elaborate,
you could do something like Map(f = min, z, x) to apply min to each pair of values in z and x,
although in that case min already supports arbitrary amount of inputs through an ellipsis (...).
By contrast,
which.min only takes a single vector as input,
so you need a wrapper with an ellipsis that combines all values into a vector
(pwhich.min above).
Since you may want to have more than two matrices,
you can put them all in a list,
and use do.call to put each element in the list as a parameter to the function you specify in f.
Or another option would be to convert it to a 3D array and use apply with which.min
apply(array(unlist(d), c(5, 5, 2)), c(1, 2), which.min)
Or with pmap from purrr
library(purrr)
pmap_int(d, ~ which.min(c(...))) %>%
array(., dim(x))
I'm looking to generate all possible binary vectors of length n in R. What is the best way (preferably both computationally efficient and readable code) to do this?
n = 3
expand.grid(replicate(n, 0:1, simplify = FALSE))
# Var1 Var2 Var3
#1 0 0 0
#2 1 0 0
#3 0 1 0
#4 1 1 0
#5 0 0 1
#6 1 0 1
#7 0 1 1
#8 1 1 1
Inspired by this question generating all possible binary vectors of length n containing less than m 1s, I've extended this code to produce all possible combinations. It's not pretty, though.
> z <- 3
> z <- rep(0, n)
> do.call(rbind, lapply(0:n, function(i) t(apply(combn(1:n,i), 2, function(k) {z[k]=1;z}))))
[,1] [,2] [,3]
[1,] 0 0 0
[2,] 1 0 0
[3,] 0 1 0
[4,] 0 0 1
[5,] 1 1 0
[6,] 1 0 1
[7,] 0 1 1
[8,] 1 1 1
What is it doing? Once we strip it back, the core of this one-liner is the following:
apply(combn(1:n,i), 2, function(k) {z[k]=1;z})
To understand this, let's step back one level further. The function combn(x,m) generates all possible combinations of x taken m at a time.
> combn(1:n, 1)
[,1] [,2] [,3]
[1,] 1 2 3
> combn(1:n, 2)
[,1] [,2] [,3]
[1,] 1 1 2
[2,] 2 3 3
> combn(1:n, 3)
[,1]
[1,] 1
[2,] 2
[3,] 3
For using apply(MARGIN=2), we pass in a column of this function at a time to our inner function function(k) {z[k]=1;z} which simply replaces all of the values at the indices k with 1. Since they were originally all 0, this gives us each possible binary vector.
The rest is just window dressing. combn gives us a wide, short matrix; we transpose it with t. lapply returns a list; we bind the matrices in each element of the list together with do.call(rbind, .).
You should define what is "the best way" (fastest? shortest code?, etc.).
One way is to use the package R.utils and the function intToBin for converting decimal numbers to binary numbers. See the example.
require(R.utils)
n <- 5
strsplit(intToBin(0:(2 ^ n - 1)), split = "")
I was wondering if is it possible to stringsplit each integer in a set of numbers and transform it into a transition matrix, e.g
data<-c(11,123,142,1423,1234,12)
What i would like to do is to split each integer in the data (considering only the first two elements in the dataset),first element will be 1,1 second element will be 1,2,3....and convert it into matrix e,g 1,1 will be 1 to 1, 1,2 will be 1 to 2 and 2,3 will be 2 to 3. generating the following matrix
1 2 3 4 5
1 1 1 0 0 0
2 0 0 1 0 0
3 0 0 0 0 0
4 0 0 0 0 0
5 0 0 0 0 0
My matrix will never go past 5x5. Below is what i have done which works but it's really really tedious.
data2<-as.matrix(as.character(data))
for(i in 1:nrow(data2)) {
values<-strsplit(data2,"")
}
values2<-t(sapply(values, '[', 1:max(sapply(values, length))))
values2[is.na(values2)]<-0
values3<-apply(values2,2,as.numeric)
from1to1<-0
from1to2<-0
from1to3<-0
from1to4<-0
from1to5<-0
from2to1<-0
from2to2<-0
from2to3<-0
from2to4<-0
...
from5to4<-0
from5to5<-0
for(i in 1:nrow(values3)){
for(j in 1:(ncol(values3)-1))
if (((values3[i,j]==1)&(values3[i,j+1]==1))){
from1to1<-from1to1 + 1
}else{
if (((values3[i,j]==1)&(values3[i,j+1]==2))){
from1to2<-from1to2 + 1
}else{
if (((values3[i,j]==1)&(values3[i,j+1]==3))){
from1to3<-from1to3 + 1
}else{
if (((values3[i,j]==1)&(values3[i,j+1]==4))){
from1to4<-from1to4 + 1
}else{
if (((values3[i,j]==1)&(values3[i,j+1]==5))){
from1to5<-from1to5 + 1
}else{
if (((values3[i,j]==1)&(values3[i,j+1]==1))){
from1to1<-from1to1 + 1
}else{.....continues through all other from2to1...from5to5``
I then place every single number into a 5x5 matrix.
This is obviously tedious and long and ridiculous. Is there anyway to shorten this? Any suggestions is appreciated.
Here's an option, presented here piped so as to be easy to follow:
library(magrittr) # for the pipe
# initialize a matrix of zeros
mat <- matrix(0, 5, 5)
# split each element into individual digits
strsplit(as.character(data), '') %>%
# turn list elements back to integers
lapply(as.integer) %>%
# make a 2 column matrix of each digit paired with the previous digit
lapply(function(x){matrix(c(x[-length(x)], x[-1]), ncol = 2)}) %>%
# reduce list to a single 2-column matrix
do.call(rbind, .) %>%
# for each row, add 1 to the element of mat they subset
apply(1, function(x){mat[x[1], x[2]] <<- mat[x[1], x[2]] + 1; x})
# output is the transpose of the matrix; the real results are stored in mat
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
## [1,] 1 1 2 1 4 1 4 2 1 2 3 1
## [2,] 1 2 3 4 2 4 2 3 2 3 4 2
mat
## [,1] [,2] [,3] [,4] [,5]
## [1,] 1 3 0 2 0
## [2,] 0 0 3 0 0
## [3,] 0 0 0 1 0
## [4,] 0 2 0 0 0
## [5,] 0 0 0 0 0
Alternately, if you'd like xtabs as suggested by alexis_laz, replace the last line with xtabs(formula = ~ .[,1] + .[,2]) instead of using mat.
You might also check out the permutations package, which from what I can tell seems to be for working with this kind of data, though it's somewhat high-level.
I am trying to solve the TSP (Traveling Salesman Problem) using the rbga.bin from the genealg package. I have matrix that stores the distances between the cities
like this:
[,1] [,2] [,3] [,4]
[1,] 0 2 10 4
[2,] 0 0 12 12
[3,] 0 0 0 5
[4,] 0 0 0 0
but I'm not able to code a proper evaluation function (even though I already saw some examples in documentation and on the web). I know a chromosome will be passed as a parameter to the evaluation function, but I don't know what operations to do to return a proper value.
Basically, you are asking how to evaluate the length of a path given the endpoints on that path and a distance matrix (for path 1-3-2-4, you want d13+d32+d24+d41). You can do this with matrix indexing and the sum function. Let's consider your distance matrix and solution 1-3-2-4 (since the TSP is typically posed in a symmetric form, I've made it symmetric):
(d <- matrix(c(0, 2, 10, 4, 2, 0, 12, 12, 10, 12, 0, 5, 4, 12, 5, 0), nrow=4))
# [,1] [,2] [,3] [,4]
# [1,] 0 2 10 4
# [2,] 2 0 12 12
# [3,] 10 12 0 5
# [4,] 4 12 5 0
sln <- c(1, 3, 2, 4)
Now you can grab matrix indexes from your solution and pull the distances, summing them for your final evaluation:
(idx <- cbind(sln, c(tail(sln, -1), sln[1])))
# [1,] 1 3
# [2,] 3 2
# [3,] 2 4
# [4,] 4 1
d[idx]
# [1] 10 12 12 4
sum(d[idx])
# [1] 38
This question already has answers here:
How to replace non-diagonal elements in a matrix?
(6 answers)
Closed 9 years ago.
Okay, I asked this question earlier but I got bashed (deservedly) for not specifying anything and showing no sign of previous attempt. So let me try again..
I'm using R, and I have a 463✕463 matrix. What I would like to do is to replace all elements other than the diagonal ones (X11, X22, X33,...,Xjj) with zero.
E.g. I want:
[1 4 5
2 3 5
3 9 8]
to be:
[1 0 0
0 3 0
0 0 8]
When I use the diag() function, it simply gives me a column vector of the diagonal values. I imagine I can use the replace() function somehow combined with a "if not diagonal" logic...but I am lost.
And yes, as some here have guessed, I am probably much younger than many people here and am completely new at this...so please put me in the right direction. Really appreciate all your help!
In R, the diag method has two functions.
It returns the diagonal of a matrix. I.e.
m <- matrix(1:9, ncol=3)
m
# [,1] [,2] [,3]
# [1,] 1 4 7
# [2,] 2 5 8
# [3,] 3 6 9
diag(m)
# [1] 1 5 9
It can construct a diagonal matrix.
diag(1:3)
# [,1] [,2] [,3]
# [1,] 1 0 0
# [2,] 0 2 0
# [3,] 0 0 3
So in your case, extract the diagonal from your existing matrix and supply it to diag:
diag(diag(m))
# [,1] [,2] [,3]
# [1,] 1 0 0
# [2,] 0 5 0
# [3,] 0 0 9
using outer
You can use the following to compute a logical matrix which describes the non-diagonal entries of a n×n matrix:
outer(1:n, 1:n, function(i,j) i!=j)
Applied to your example:
> m <- matrix(c(1,2,3,4,3,9,5,5,8),ncol=3)
> m
[,1] [,2] [,3]
[1,] 1 4 5
[2,] 2 3 5
[3,] 3 9 8
> m[outer(1:3, 1:3, function(i,j) i!=j)] <- 0
> m
[,1] [,2] [,3]
[1,] 1 0 0
[2,] 0 3 0
[3,] 0 0 8
using triangles
A possible alternative would be combining the two triangles on either side of the diagonal. In this case, you use the matrix m itself as input to determine the size.
upper.tri(m) | lower.tri(m)
Applied to your use case:
> m[upper.tri(m) | lower.tri(m)] <- 0
It seems you already got this answer in response to your original post…
m[ col(m)==row(m) ] <- 0
> m <- matrix(1:9, 3)
> m[ col(m)==row(m) ]
[1] 1 5 9
> m[ col(m)!=row(m) ] <- 0
> m
[,1] [,2] [,3]
[1,] 1 0 0
[2,] 0 5 0
[3,] 0 0 9