I try to implement a example using R in Simulation (2006, 4ed., Elsevier) by Sheldon M. Ross, which wants to generate a random permutation and reads as follows:
Suppose we are interested in generating a permutation of the numbers 1,2,... ,n
which is such that all n! possible orderings are equally likely.
The following algorithm will accomplish this by
first choosing one of the numbers 1,2,... ,n at random;
and then putting that number in position n;
it then chooses at random one of the remaining n-1 numbers and puts that number in position n-1 ;
it then chooses at random one of the remaining n-2 numbers and puts it in position n-2 ;
and so on
Surely, we can achieve a random permutation of the numbers 1,2,... ,n easily by
sample(1:n, replace=FALSE)
For example
> set.seed(0); sample(1:5, replace=FALSE)
[1] 1 4 3 5 2
However, I want to get similar results manually according to the above algorithmic steps. Then I try
## write the function
my_perm = function(n){
x = 1:n # initialize
k = n # position n
out = NULL
while(k>0){
y = sample(x, size=1) # choose one of the numbers at random
out = c(y,out) # put the number in position
x = setdiff(x,out) # the remaining numbers
k = k-1 # and so on
}
out
}
## test the function
n = 5; set.seed(0); my_perm(n) # set.seed for reproducible
and have
[1] 2 2 4 5 1
which is obviously incorrect for there are two 2 . How can I fix the problem?
You have implemented the logic correctly but there is only one thing that you need to be aware which is related to R.
From ?sample
If x has length 1, is numeric (in the sense of is.numeric) and x >= 1, sampling via sample takes place from 1:x
So when the last number is remaining in x, let's say that number is 4, sampling would take place from 1:4 and return any 1 number from it.
For example,
set.seed(0)
sample(4, 1)
#[1] 2
So you need to adjust your function for that after which the code should work correctly.
my_perm = function(n){
x = 1:n # initialize
k = n # position n
out = NULL
while(k>1){ #Stop the while loop when k = 1
y = sample(x, size=1) # choose one of the numbers at random
out = c(y,out) # put the number in position
x = setdiff(x,out) # the remaining numbers
k = k-1 # and so on
}
out <- c(x, out) #Add the last number in the output vector.
out
}
## test the function
n = 5
set.seed(0)
my_perm(n)
#[1] 3 2 4 5 1
Sample size should longer than 1. You can break it by writing a condition ;
my_perm = function(n){
x = 1:n
k = n
out = NULL
while(k>0){
if(length(x)>1){
y = sample(x, size=1)
}else{
y = x
}
out = c(y,out)
x = setdiff(x,out)
k = k-1
}
out
}
n = 5; set.seed(0); my_perm(n)
[1] 3 2 4 5 1
Related
Let's say I have a population like {1,2,3, ..., 23} and I want to generate a sample so that the sample's mean equals 6.
I tried to use the sample function, using a custom probability vector, but it didn't work:
population <- c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23)
mean(population)
minimum <- min(population)
maximum <- max(population)
amplitude <- maximum - minimum
expected <- 6
n <- length(population)
prob.vector = rep(expected, each=n)
for(i in seq(1, n)) {
if(expected > population[i]) {
prob.vector[i] <- (i - minimum) / (expected - minimum)
} else {
prob.vector[i] <- (maximum - i) / (maximum - expected)
}
}
sample.size <- 5
sample <- sample(population, sample.size, prob = prob.vector)
mean(sample)
The mean of the sample is about the mean of the population (oscillates around 12), and I wanted it to be around 6.
A good sample would be:
{3,5,6,8,9}, mean=6.2
{2,3,4,8,9}, mean=5.6
The problem is different from sample integer values in R with specific mean because I have a specific population and I can't just generate arbitrary real numbers, they must be inside the population.
The plot of the probability vector:
You can try this:
m = local({b=combn(1:23,5);
d = colMeans(b);
e = b[,d>5.5 &d<6.5];
function()sample(e[,sample(ncol(e),1)])})
m()
[1] 8 5 6 9 3
m()
[1] 6 4 5 3 13
breakdown:
b=combn(1:23,5) # combine the numbers into 5
d = colMeans(b) # find all the means
e = b[,d>5.5 &d<6.5] # select only the means that are within a 0.5 range of 6
sample(e[,sample(ncol(e),1)]) # sample the values the you need
I would like to randomly assign positive integers to G groups, such that they sum up to V.
For example, if G = 3 and V = 21, valid results may be (7, 7, 7), (10, 6, 5), etc.
Is there a straightforward way to do this?
Editor's notice (from 李哲源):
If values are not restricted to integers, the problem is simple and has been addressed in Choosing n numbers with fixed sum.
For integers, there is a previous Q & A: Generate N random integers that sum to M in R but it appears more complicated and is hard to follow. The loop based solution over there is also not satisfying.
non-negative integers
Let n be sample size:
x <- rmultinom(n, V, rep.int(1 / G, G))
is a G x n matrix, where each column is a multinomial sample that sums up to V.
By passing rep.int(1 / G, G) to argument prob I assume that each group has equal probability of "success".
positive integers
As Gregor mentions, a multinomial sample can contain 0. If such samples are undesired, they should be rejected. As a result, we sample from a truncated multinomial distribution.
In How to generate target number of samples from a distribution under a rejection criterion I suggested an "over-sampling" approach to achieve "vectorization" for a truncated sampling. Simply put, Knowing the acceptance probability we can estimate the expected number of trials M to see the first "success" (non-zero). We first sample say 1.25 * M samples, then there will be at least one "success" in these samples. We randomly return one as the output.
The following function implements this idea to generate truncated multinomial samples without 0.
positive_rmultinom <- function (n, V, prob) {
## input validation
G <- length(prob)
if (G > V) stop("'G > V' causes 0 in a sample for sure!")
if (any(prob < 0)) stop("'prob' can not contain negative values!")
## normalization
sum_prob <- sum(prob)
if (sum_prob != 1) prob <- prob / sum_prob
## minimal probability
min_prob <- min(prob)
## expected number of trials to get a "success" on the group with min_prob
M <- round(1.25 * 1 / min_prob)
## sampling
N <- n * M
x <- rmultinom(N, V, prob)
keep <- which(colSums(x == 0) == 0)
x[, sample(keep, n)]
}
Now let's try
V <- 76
prob <- c(53, 13, 9, 1)
Directly using rmultinom to draw samples can occasionally result in ones with 0:
## number of samples that contain 0 in 1000 trials
sum(colSums(rmultinom(1000, V, prob) == 0) > 0)
#[1] 355 ## or some other value greater than 0
But there is no such issue by using positive_rmultinom:
## number of samples that contain 0 in 1000 trials
sum(colSums(positive_rmultinom(1000, V, prob) == 0) > 0)
#[1] 0
Probably a less expensive way, but this seems to work.
G <- 3
V <- 21
m <- data.frame(matrix(rep(1:V,G),V,G))
tmp <- expand.grid(m) # all possibilities
out <- tmp[which(rowSums(tmp) == V),] # pluck those that sum to 'V'
out[sample(1:nrow(out),1),] # randomly select a column
Not sure how to do with runif
I figured out what I believe to be a much simpler solution. You first generate random integers from your minimum to maximum range, count them up and then make a vector of the counts (including zeros).
Note that this solution may include zeros even if the minimum value is greater than zero.
Hope this helps future r people with this problem :)
rand.vect.with.total <- function(min, max, total) {
# generate random numbers
x <- sample(min:max, total, replace=TRUE)
# count numbers
sum.x <- table(x)
# convert count to index position
out = vector()
for (i in 1:length(min:max)) {
out[i] <- sum.x[as.character(i)]
}
out[is.na(out)] <- 0
return(out)
}
rand.vect.with.total(0, 3, 5)
# [1] 3 1 1 0
rand.vect.with.total(1, 5, 10)
#[1] 4 1 3 0 2
Note, I also posted this here Generate N random integers that sum to M in R, but this answer is relevant to both questions.
Is there a method to generate random integers in R such that any two consecutive numbers are different? It is probably along the lines of x[k+1] != x[k] but I can't work out how to put it all together.
Not sure if there is a function available for that. Maybe this function can do what you want:
# n = number of elements
# sample_from = draw random numbers from this range
random_non_consecutive <- function(n=10,sample_from = seq(1,5))
{
y=c()
while(length(y)!=n)
{
y= c(y,sample(sample_from,n-length(y),replace=T))
y=y[!c(FALSE, diff(y) == 0)]
}
return(y)
}
Example:
random_non_consecutive(20,c(2,4,6,8))
[1] 6 4 6 2 6 4 2 8 4 2 6 2 8 2 8 2 8 4 8 6
Hope this helps.
The function above has a long worst-case runtime. We can keep that worst-case more constant with for example the following implementation:
# n = number of elements
# sample_from = draw random numbers from this range
random_non_consecutive <- function(n=10,sample_from = seq(1,5))
{
y= rep(NA, n)
prev=-1 # change this if -1 is in your range, to e.g. max(sample_from)+1
for(i in seq(n)){
y[i]=sample(setdiff(sample_from,prev),1)
prev = y[i]
}
return(y)
}
Another approach is to over-sample and remove the disqualifying ones as follows:
# assumptions
n <- 5 # population size
sample_size <- 1000
# answer
mu <- sample_size * 1/n
vr <- sample_size * 1/n * (1 - 1/n)
addl_draws <- round(mu + vr, 0)
index <- seq(1:n)
sample_index <- sample(index, sample_size + addl_draws, replace = TRUE)
qualified_sample_index <- sample_index[which(diff(sample_index) != 0)]
qualified_sample_index <- qualified_sample_index[1:sample_size]
# In the very unlikely event the number of qualified samples < sample size,
# NA's will fill the vector. This will print those N/A's
print(which(is.na(qualified_sample_index) == TRUE))
I am looking for a general purpose algorithm to identify short numeric series from lists with a max length of a few hundred numbers. This will be used to identify series of masses from mass spectrometry (ms1) data.
For instance, given the following list, I would like to identify that 3 of these numbers fit the series N + 1, N +2, etc.
426.24 <= N
427.24 <= N + 1/x
371.10
428.24 <= N + 2/x
851.47
451.16
The series are all of the format: N, N+1/x, N+2/x, N+3/x, N+4/x, etc, where x is an integer (in the example x=1). I think this constraint makes the problem very tractable. Any suggestions for a quick/efficient way to tackle this in R?
This routine will generate series using x from 1 to 10 (you could increase it). And will check how many are contained in the original list of numbers.
N = c(426.24,427.24,371.1,428.24,851.24,451.16)
N0 = N[1]
x = list(1,2,3,4,5,6,7,8,9,10)
L = 20
Series = lapply(x, function(x){seq(from = N0, by = 1/x,length.out = L)})
countCoincidences = lapply(Series, function(x){sum(x %in% N)})
Result:
unlist(countCoincidences)
[1] 3 3 3 3 3 3 3 3 3 2
As you can see, using x = 1 will have 3 coincidences. The same goes for all x until x=9. Here you have to decide which x is the one you want.
Since you're looking for an arithmetic sequence, the difference k is constant. Thus, you can loop over the vector and subtract each value from the sequence. If you have a sequence, subtracting the second term from the vector will result in values of -k, 0, and k, so you can find the sequence by looking for matches between vector - value and its opposite, value - vector:
x <- c(426.24, 427.24, 371.1, 428.24, 851.47, 451.16)
unique(lapply(x, function(y){
s <- (x - y) %in% (y - x);
if(sum(s) > 1){x[s]}
}))
# [[1]]
# NULL
#
# [[2]]
# [1] 426.24 427.24 428.24
I am filling a 10x10 martix (mat) randomly until sum(mat) == 100
I wrote the following.... (i = 2 for another reason not specified here but i kept it at 2 to be consistent with my actual code)
mat <- matrix(rep(0, 100), nrow = 10)
mat[1,] <- c(0,0,0,0,0,0,0,0,0,1)
mat[2,] <- c(0,0,0,0,0,0,0,0,1,0)
mat[3,] <- c(0,0,0,0,0,0,0,1,0,0)
mat[4,] <- c(0,0,0,0,0,0,1,0,0,0)
mat[5,] <- c(0,0,0,0,0,1,0,0,0,0)
mat[6,] <- c(0,0,0,0,1,0,0,0,0,0)
mat[7,] <- c(0,0,0,1,0,0,0,0,0,0)
mat[8,] <- c(0,0,1,0,0,0,0,0,0,0)
mat[9,] <- c(0,1,0,0,0,0,0,0,0,0)
mat[10,] <- c(1,0,0,0,0,0,0,0,0,0)
i <- 2
set.seed(129)
while( sum(mat) < 100 ) {
# pick random cell
rnum <- sample( which(mat < 1), 1 )
mat[rnum] <- 1
##
print(paste0("i =", i))
print(paste0("rnum =", rnum))
print(sum(mat))
i = i + 1
}
For some reason when sum(mat) == 99 there are several steps extra...I would assume that once i = 91 the while would stop but it continues past this. Can somone explain what I have done wrong...
If I change the while condition to
while( sum(mat) < 100 & length(which(mat < 1)) > 0 )
the issue remains..
Your problem is equivalent to randomly ordering the indices of a matrix that are equal to 0. You can do this in one line with sample(which(mat < 1)). I suppose if you wanted to get exactly the same sort of output, you might try something like:
set.seed(144)
idx <- sample(which(mat < 1))
for (i in seq_along(idx)) {
print(paste0("i =", i))
print(paste0("rnum =", idx[i]))
print(sum(mat)+i)
}
# [1] "i =1"
# [1] "rnum =5"
# [1] 11
# [1] "i =2"
# [1] "rnum =70"
# [1] 12
# ...
See ?sample
Arguments:
x: Either a vector of one or more elements from which to choose,
or a positive integer. See ‘Details.’
...
If ‘x’ has length 1, is numeric (in the sense of ‘is.numeric’) and
‘x >= 1’, sampling _via_ ‘sample’ takes place from ‘1:x’. _Note_
that this convenience feature may lead to undesired behaviour when
‘x’ is of varying length in calls such as ‘sample(x)’. See the
examples.
In other words, if x in sample(x) is of length 1, sample returns a random number from 1:x. This happens towards the end of your loop, where there is just one 0 left in your matrix and one index is returned by which(mat < 1).
The iteration repeats on level 99 because sample() behaves very differently when the first parameter is a vector of length 1 and when it is greater than 1. When it is length 1, it assumes you a random number from 1 to that number. When it has length >1, then you get a random number from that vector.
Compare
sample(c(99,100),1)
and
sample(c(100),1)
Of course, this is an inefficient way of filling your matrix. As #josilber pointed out, a single call to sample could do everything you need.
The issue comes from how sample and which do the sampling when you have only a single '0' value left.
For example, do this:
mat <- matrix(rep(1, 100), nrow = 10)
Now you have a matrix of all 1's. Now lets make two numbers 0:
mat[15]<-0
mat[18]<-0
and then sample
sample(which(mat<1))
[1] 18 15
by adding a size=1 argument you get one or the other
now lets try this:
mat[18]<-1
sample(which(mat<1))
[1] 3 13 8 2 4 14 11 9 10 5 15 7 1 12 6
Oops, you did not get [1] 15 . Instead what happens in only a single integer (15 in this case) is passed tosample. When you do sample(x) and x is an integer, it gives you a sample from 1:x with the integers in random order.