I have the following
cells = c(3, 5, 8, 6)
names(cells) = c('A', 'B', 'C', 'D')
I want to randomly choose half the cells, and assign then to a new vector m1, and the other half of the cells will be assigned to m2 = cells - m1. I'm new to R, and the sample function only allows me choose all the cells of one type, instead of an individual cell at a time, so I'm unsure of where to go from here.
My code currently looks like this
y = ceiling(sum(cells)/2)
for(i in 1:y){
z = sample(cells[cells>0], 1, replacement = FALSE, prob = NULL)
if(z == cells[1]){
cells[1] = cells[1] - 1
m1[1] = m1[1] + 1
}
if(z == cells[2]){
cells[2] = cells[2] - 1
m1[2] = m1[2] + 1
}
if(z == cells[3]){
cells[3] = cells[3] - 1
m1[3] = m1[3] + 1
}
if(z == cells[4]){
cells[4] = cells[4] - 1
m1[4] = m1[4] + 1
}
}
I know this is wrong as it only chooses the cell type randomly, instead of choosing each cell randomly. Any help would be appreciated.
Why not define cells = c(rep('A',3),rep('B',5), rep('C',8), rep('D',6)) and then
> cells
[1] "A" "A" "A" "B" "B" "B" "B" "B" "C" "C" "C" "C" "C" "C" "C" "C" "D" "D" "D"
[20] "D" "D" "D"
x <- sample(1:length(cells), ceiling(length(cells)/2))
m1 <- cells[x]
[1] "D" "B" "C" "A" "B" "A" "B" "D" "D" "C" "C"
m2 <- cells[setdiff(1:length(cells), x)]
[1] "A" "B" "B" "C" "C" "C" "C" "C" "D" "D" "D"
Or you could use parameter prob in sample function
x <- sample(letters[1:4], 22, replace = T, prob = c(3,5,8,6))
x
[1] "d" "c" "a" "c" "b" "a" "b" "b" "b" "c" "b" "c" "d" "c" "c" "b" "c" "b" "c" "d" "c" "b"
m1 <- x[1:11]; m2 <- x[12:22]
Instead of sampling from cells directly, sample indexes and use those to choose your elements.
cells = c(3, 5, 8, 6)
categories = c('A', 'B', 'C', 'D')
n = length(cells)
cells_long = unlist(sapply(1:n
, function(i){ rep(categories[i],cells[i]) }
))
n = length(cells_long)
ix = sample(1:n, floor(n/2))
m1 = cells_long[ix]
m2 = cells_long[-ix]
This is a useful trick for sampling from complex objects or collections of related objects as well.
Related
I am trying to reconstruct a Markov process from Shannons paper "A mathematical theory of communication". My question concerns figure 3 on page 8 and a corresponding sequence (message) from that Markov chain from page 5 section (B). I just wanted to check if I coded the right Markov chain to this figure from Shannons paper:
Here is my attempt:
install.packages("markovchain")
library(markovchain)
MessageABCDE = c("A", "B", "C", "D", "E")
MessageTransitionMatrix = matrix(c(.4,.1,.2,.2,.1,
.4,.1,.2,.2,.1,
.4,.1,.2,.2,.1,
.4,.1,.2,.2,.1,
.4,.1,.2,.2,.1),
nrow = 5,
byrow = TRUE,
dimname = list(MessageABCDE, MessageABCDE))
MCmessage = new("markovchain", states = MessageABCDE,
byrow = TRUE,
transitionMatrix = MessageTransitionMatrix,
name = "WritingMessage")
steadyStates(MCmessage)
markovchainSequence(n = 20, markovchain = MCmessage, t0 = "A")
My goal was to also create a sequence (message) from that chain. I am mostly uncertain about the transition matrix, where infered the probabilities had to be all the same in every row. I am happy with the output of markovchainSequence, but I am not 100% sure, if I am doing it right.
Here is my console output for markovchainSequence:
> markovchainSequence(n = 20, markovchain = MCmessage, t0 = "A")
[1] "D" "E" "A" "D" "A" "A" "B" "D" "E" "C" "A" "A" "E" "C" "C" "D" "D" "D"
[19] "A" "C"
Looks fine. It's maybe odd because with fully independent states like this there isn't any need for a Markov chain. One could equally well use
tokens <- c("A", "B", "C", "D", "E")
probs <- c(0.4, 0.1, 0.2, 0.2, 0.1)
sample(tokens, size=20, replace=TRUE, prob=probs)
## [1] "A" "B" "A" "B" "D" "B" "C" "D" "A" "D" "C" "E" "A" "A" "C" "E" "C" "D" "C" "C"
Will likely make more sense once there is a variety of conditional probabilities.
I want to randomize/shuffle a vector. Some of the vector elements are identical. After shuffling, identical elements should have a minimum distance of three (i.e. two other elements should be between identical elements).
Consider the following example vector in R:
x <- rep(LETTERS[1:5], 3) # Create example vector
x
# [1] "A" "B" "C" "D" "E" "A" "B" "C" "D" "E" "A" "B" "C" "D" "E"
If I shuffle my vector using the sample function, some of the identical elements may be too close together. For instance, if I use the following R code, the element "C" appears directly after each other at positions 5 and 6:
set.seed(53135)
sample(x) # sample() function puts same elements too close
# [1] "B" "A" "E" "D" "C" "C" "E" "A" "B" "C" "D" "E" "A" "D" "B"
How could I ensure that identical elements have a minimum distance of three?
So basically we need to conditionally sample one element from the x vector that have not been selected in the min.dist-1 runs. Using purrr's reduce we can achieve this:
min.dist <- 2
reduce(integer(length(x)-1), ~c(.x, sample(x[!x %in% tail(.x, min.dist)], 1)), .init=sample(x,1))
[1] "A" "E" "D" "B" "A" "D" "E" "C" "D" "A" "C" "E" "B" "A" "E"
Bundled in a function
shuffle <- function(x, min.dist=2){
stopifnot(min.dist < length(unique(x)))
reduce(integer(length(x)-1), ~c(.x, sample(x[!x %in% tail(.x, min.dist)], 1)), .init=sample(x,1))
}
> shuffle(x, 3)
[1] "A" "C" "B" "D" "E" "A" "B" "C" "E" "D" "A" "B" "C" "E" "A"
> shuffle(x, 3)
[1] "A" "B" "D" "E" "C" "A" "B" "D" "E" "C" "A" "D" "E" "C" "A"
> shuffle(x, 4)
[1] "C" "E" "D" "A" "B" "C" "E" "D" "A" "B" "C" "E" "D" "A" "B"
> shuffle(x, 4)
[1] "A" "B" "D" "E" "C" "A" "B" "D" "E" "C" "A" "B" "D" "E" "C"
> shuffle(x, 2)
[1] "E" "A" "D" "E" "B" "D" "A" "E" "C" "D" "A" "E" "C" "A" "B"
> shuffle(x, 2)
[1] "B" "A" "D" "C" "B" "A" "E" "B" "A" "E" "B" "C" "D" "A" "E"
after #27ϕ9 comment:
shuffle <- function(x, min.dist=2){
stopifnot(min.dist < length(unique(x)))
reduce(integer(length(x)-1), ~ c(.x, sample(x[!x %in% tail(.x, min.dist) &( x %in% names(t <- table(x[x%in%.x]) > table(.x))[t] | !x %in% .x)], 1)), .init=sample(x,1))
}
> table(shuffle(rep(LETTERS[1:5], 3),2))
A B C D E
3 3 3 3 3
> table(shuffle(rep(LETTERS[1:5], 3),2))
Error in sample.int(length(x), size, replace, prob) :
invalid first argument
UPDATE
After some trial and error, looking at the fact that not always you're gonna have enough elements to space out the min.dist I came up with a solution this code is the most explained from the ones above :
shuffle <- function(x, min.dist=2){
stopifnot(min.dist < length(unique(x)))
reduce(integer(length(x)-1), function(.x, ...){
# whether the value is in the tail of the aggregated vector
in.tail <- x %in% tail(.x, min.dist)
# whether a value still hasn't reached the max frequency
freq.got <- x %in% names(t<-table(x[x%in%.x]) > table(.x))[t]
# whether a value isn't in the aggregated vector
yet <- !x %in% .x
# the if is there basically to account for the cases when we don't have enough vars to space out the vectors
c(.x, if(any((!in.tail & freq.got) | yet )) sample(x[(!in.tail & freq.got) | yet ], 1) else x[which(freq.got)[1]] )
}, .init=sample(x,1))
}
now running the table(shuffle(rep(LETTERS[1:5], 3),2)) will always return 3 for all vars and we can say with some certainty that in the vector the variables are spaced with a minimum distance of 2. the only way to guarantee that no elements are duplicated is by using min.dist=length(unique(x))-1 otherwise there will be instances where at maximum r < min.dist elements are not min.dist distanced from their last occurrences, and if such elements exist they're going to be in the length(x) + 1 - 1:min.dist subset of the resulting vector.
Just to be completely certain using a loop to check whether tail of the output vector has unique values: (remove the print statement I used it just for demonstration purposes)
shuffler <- function(x, min.dist=2){
while(!length(unique(print(tail(l<-shuffle(x, min.dist=min.dist), min.dist+1))))==min.dist+1){}
l
}
table(print(shuffler(rep(LETTERS[1:5], 3),2)))
[1] "A" "B" "C" "E" "B" "C" "D" "A" "C" "D" "A" "E" "B" "D" "E"
A B C D E
3 3 3 3 3
table(print(shuffler(rep(LETTERS[1:5], 3),2)))
[1] "D" "C" "C"
[1] "C" "C" "E"
[1] "C" "A" "C"
[1] "D" "B" "D"
[1] "B" "E" "D"
[1] "C" "A" "E" "D" "A" "B" "C" "E" "A" "B" "D" "C" "B" "E" "D"
A B C D E
3 3 3 3 3
Update:
shuffler <- function(x, min.dist=2){
while(any(unlist(lapply(unique(tl<-tail(l<-shuffle(x, min.dist=min.dist), 2*min.dist)), function(x) diff(which(tl==x))<=min.dist)))){}
l
}
this new version does a rigorous test on whether the elements in the tail of the vector are min.distanced, the previous version works for min.dist=2, however this new version does better testing.
If your data is large, then it may be (way) faster to rely on probability to do that kind of task.
Here's an example:
prob_shuffler = function(x, min.dist = 2){
n = length(x)
res = sample(x)
OK = FALSE
# We loop until we have a solution
while(!OK){
OK = TRUE
for(i in 1:min.dist){
# We check if identical elements are 'i' steps away
pblm = res[1:(n-i)] == res[-(1:i)]
if(any(pblm)){
if(sum(pblm) >= (n - i)/2){
# back to square 1
res = sample(x)
} else {
# we pair each identical element with
# an extra one
extra = sample(which(!pblm), sum(pblm))
id_reshuffle = c(which(pblm), extra)
res[id_reshuffle] = sample(res[id_reshuffle])
}
# We recheck from the beginning
OK = FALSE
break
}
}
}
res
}
Even though the while loop looks scary, in practice convergence is fast. Of course, the lower the probability to have two characters at min.dist away, the faster the convergence.
The current solutions by #Abdessabour Mtk and #Carles Sans Fuentes work but, depending on the size of the input data, quickly become prohibitively slow. Here's a benchmark:
library(microbenchmark)
x = rep(c(letters, LETTERS), 10)
length(x)
#> [1] 520
microbenchmark(prob_shuffler(x, 1), shuffler_am(x, 1), shuffler_csf(x, 1), times = 10)
#> Unit: microseconds
#> expr min lq mean median uq max neval
#> prob_shuffler(x, 1) 87.001 111.501 155.071 131.801 192.401 264.401 10
#> shuffler_am(x, 1) 17218.100 18041.900 20324.301 18740.351 22296.301 26495.200 10
#> shuffler_csf(x, 1) 86771.401 88550.501 118185.581 95582.001 98781.601 341826.701 10
microbenchmark(prob_shuffler(x, 2), shuffler_am(x, 2), shuffler_csf(x, 2), times = 10)
#> Unit: microseconds
#> expr min lq mean median uq max neval
#> prob_shuffler(x, 2) 140.1 195.201 236.3312 245.252 263.202 354.101 10
#> shuffler_am(x, 2) 18886.2 19526.901 22967.6409 21021.151 26758.800 29133.400 10
#> shuffler_csf(x, 2) 86078.1 92209.901 97151.0609 97612.251 99850.101 107981.401 10
microbenchmark(prob_shuffler(x, 3), shuffler_am(x, 3), shuffler_csf(x, 3), times = 10)
#> Unit: microseconds
#> expr min lq mean median uq max neval
#> prob_shuffler(x, 3) 318.001 450.402 631.5312 573.352 782.2 1070.401 10
#> shuffler_am(x, 3) 19003.501 19622.300 23314.4808 20784.551 28281.5 32885.101 10
#> shuffler_csf(x, 3) 87692.701 96152.202 101233.5411 100925.201 108034.7 113814.901 10
We can remark two things: a) in all logic, the speed of prob_shuffler depends on min.dist while the other methods not so much, b) prob_shuffler is about 100-fold faster for just 520 observations (and it scales).
Of course if the probability to have two identical characters at min.dist away is extremely high, then the recursive methods should be faster. But in most practical cases, the probability method is faster.
I hope this answer works fine for you. It is done with base R, but it works. I leave the printing if you want to check line by line:
x <- rep(LETTERS[1:5], 3) # Create example vector
shuffle <- function(x, min_dist=3){
#init variables
result<-c() # result vector
count<-0
vec_use<-x
vec_keep<-c()
for(i in 1:length(x)){
# print(paste0("iteration =", i))
if (count>min_dist){
valback<-vec_keep[1]
# print(paste0("value to be returned:", valback))
ntimes_valback<-(table(vec_keep)[valback])
vec_use<- c(vec_use,rep(valback,ntimes_valback))
# print(paste0("vec_use after giving back valbak =", valback))
# print(paste0(vec_use,","))
vec_keep <- vec_keep[!vec_keep %in% valback]
# print(paste0("vec_keep after removing valback =", valback))
# print(paste0(vec_keep,","))
}
val<-sample(vec_use,1)
# print(paste0("val = ",val))#remove value
vec_keep<- c(vec_keep,x[x %in% val])
vec_keep<-vec_keep[1:(length(vec_keep)-1)]#removing 1 letter
# print(paste0("vec_keep ="))
# print(paste0(vec_keep,","))
vec_use <- vec_use[!vec_use %in% val]
# print(paste0("vec_use ="))
# print(paste0(vec_use,","))
result[i]<-val
count<-count+1
}
return(result)
}
shuffle(x)
"C" "D" "B" "E" "C" "A" "B" "D" "E" "A" "C" "D" "B" "E" "C"
I need to generate a random sample with a multivariate normal distribution using seed(12346) with 100 columns and 5000 rows.
So far I have got this:
set.seed(12346)
Preg1 <- data.frame(MASS::mvrnorm(n=5000,mu=c(0,0,0),Sigma = diag(3)))
The above gives me three columns, how can I get 100?
I cannot figure out how to get the vector of mu with 100 zeros without typing them in and the Sigma would then be Sigma = diag(100)
You can use mu = rep(0, 100). The rep function is used to repeat values.
set.seed(12346)
ncol = 100
Preg1<-data.frame(mvrnorm(n = 5000, mu = rep(0, ncol), Sigma = diag(ncol)))
dim(Preg1)
# [1] 5000 100
The rep function is quite useful, it can be used in various ways that aren't applicable here but are good to know about:
rep(c("A", "B", "C"), times = 3)
# [1] "A" "B" "C" "A" "B" "C" "A" "B" "C"
rep(c("A", "B", "C"), times = 1:3)
# [1] "A" "B" "B" "C" "C" "C"
rep(c("A", "B", "C"), each = 3)
# [1] "A" "A" "A" "B" "B" "B" "C" "C" "C"
In this particular case, because your Sigma is an identity matrix, each column is actually independent. So it would be equivalent to generate each column (or even each draw) independently, which we could do either of these ways:
x = replicate(n = ncol, rnorm(5000))
dim(x)
# [1] 5000 100
z = matrix(rnorm(5000 * ncol), ncol = ncol)
dim(z)
# [1] 5000 100
I'm learning to use R and I'm trying to extract cont, p0 and pf variables from nested loop as 3 different vectors from this code.
v<-c("a","b","c","d","e","f","g","h")
n<-length(v)
mv<-5
a<-n-(mv-1)
cont<-0
for (num in (a-1):0){
for (i in 0:num){
cont<-cont+1
p0<-v[a-num]
pf<-v[n-i]
}
}
The expected result should be:
> print(cont)
[1] 1 2 3 4 5 6 7 8 9 10
> print (p0)
[1] "a" "a" "a" "a" "b" "b" "b" "c" "c" "d"
> print (pf)
[1] "h" "g" "f" "e" "h" "g" "f" "h" "g" "h"
I would keep cont as an index variable and store the other variables in vectors.
v<-c("a","b","c","d","e","f","g","h")
n<-length(v)
mv<-5
a<-n-(mv-1)
cont = 0
cont_stored = vector();
p0 = vector();
pf = vector();
for (num in (a-1):0){
for (i in 0:num){
cont <- cont+1
cat("cont = ", cont, "\n"); ## useful function for printing stuff out in loops
cont_stored[cont] = cont;
p0[cont] = v[a-num]
pf[cont] = v[n-i]
}
}
cont_stored
p0
pf
You can do this without explicit for loop :
v <- c("a","b","c","d","e","f","g","h")
n <- length(v)
mv <- 5
a <- n-(mv-1)
cont <- 0
p0 <- rep(v[1:a], a:1)
pf <- v[unlist(sapply((n-a + 1):n, function(x) n:x))]
p0
# [1] "a" "a" "a" "a" "b" "b" "b" "c" "c" "d"
pf
# [1] "h" "g" "f" "e" "h" "g" "f" "h" "g" "h"
If you need cont you could use p0 or pf with seq_along.
cont <- seq_along(p0)
cont
#[1] 1 2 3 4 5 6 7 8 9 10
I am writing a program that (as a part of it) automatically creates dendrograms from an input dataset.
For each node/split I want to extract all the labels that are under that node and the location of that node on the dendrogram plot (for further plotting purposes).
So, let's say my data looks like this:
> Ltrs <- data.frame("A" = c(3,1), "B" = c(1,1), "C" = c(2,4), "D" = c(6,6))
> dend <- as.dendrogram(hclust(dist(t(Ltrs))))
> plot(dend)
The dendrogram
Now I can extract the location of the splits/nodes:
> library(dendextend)
> nodes <- get_nodes_xy(dend)
> nodes <- nodes[nodes[,2] != 0, ]
> nodes
[,1] [,2]
[1,] 1.875 7.071068
[2,] 2.750 3.162278
[3,] 3.500 2.000000
Now I want to get all the labels under a node, for each node (/row from the 'nodes' variable).
This should look something like this:
$`1`
[1] "D" "C" "B" "A"
$`2`
[1] "C" "B" "A"
$`3 `
[1] "B" "A"
Can anybody help me out? Thanks in advance :)
How about something like this?
library(tidyverse)
library(dendextend)
Ltrs <- data.frame("A" = c(3,1), "B" = c(1,1), "C" = c(2,4), "D" = c(6,6))
dend <- as.dendrogram(hclust(dist(t(Ltrs))))
accumulator <- list();
myleaves <- function(anode){
if(!is.list(anode))return(attr(anode,"label"))
accumulator[[length(accumulator)+1]] <<- (reduce(lapply(anode,myleaves),c))
}
myleaves(dend);
ret <- rev(accumulator); #generation was depth first, so root was found last.
Better test this. I am not very trustworthy. In particular, I really hope the list ret is in an order that makes sense, otherwise it's going to be a pain associating the entries with the correct nodes! Good luck.
Function partition_leaves() extracts all leaf labels per each node and makes a list ordered in the same fashion as get_nodes_xy() output. With your example,
Ltrs <- data.frame("A" = c(3,1), "B" = c(1,1), "C" = c(2,4), "D" = c(6,6))
dend <- as.dendrogram(hclust(dist(t(Ltrs))))
plot(dend)
partition_leaves(dend)
yields:
[[1]]
[1] "D" "C" "A" "B"
[[2]]
[1] "D"
[[3]]
[1] "C" "A" "B"
[[4]]
[1] "C"
[[5]]
[1] "A" "B"
[[6]]
[1] "A"
[[7]]
[1] "B"
filtering list by vector length will give output similar to the desired one.