The following code does what I need it to do, but since I am bound to have the same task in future codes, I would like to know what the best way to achieve the outcome is:
p_last = fill(NaN, (n,periods-1))
p_first = ones(n) * 0.5
p = hcat(p_first,p_last)
There are many ways like:
p = fill(NaN, n, periods)
p[:, 1] .= 0.5
or
p = [j == 1 ? 0.5 : NaN for i in 1:n, j in 1:periods]
or similarly:
p = [ifelse(j == 1, 0.5, NaN) for i in 1:n, j in 1:periods]
All of them have the advantage that they allocate only one matrix. The first one is a bit faster but requires two statements.
Related
I want to find the P(X<Y<Z) in r. For each value of z_i, I want to check whether it satisfies the conditions or not. I demonstrated the problem below. Here I used the ifelse function in r. I don't how to put multiple statements within ifelse. When I type ifelse(z[i]>y>x, 1, 0) I get errors. I want to know how to include this.
x = c(1,1)
y = c(2,2)
z = c(3,3)
value = NULL
n1 = length(x)
n2 = length(y)
n3 = length(z)
for(i in 1: length(z)){
value[i] = sum (ifelse(z[i]>y & z[i]> x & y > x, 1, 0))
}
value
The desired output should be 4 4. But the above code gives 2 2. Thanks in advance.
What I look for is basically an R-version of the answer to this question: Generating all permutation of numbers that sums up to N. First of all the answer uses java, which I have a really hard time reading. Second of all the code uses "deque", which I cant figure out a way to implement in R.
I have found several algorithms to do this, but they have all been written in programming languages using structures not available in R such as deques, heaps or list-comprehensions.
What I actually need is a way of finding all the vectors v of length N-1 where:
sum(v * 1:(N-1)) == N
and I think I can manage that myself if only I find a way of obtaining all the ordered integer partitions.
As an example for N = 4 all the ordered integer partitions using numbers 1 to N-1 are:
1+1+1+1
1+1+2
1+3
2+2
What I effectively need is output of the either form:
c(1,1,1,1)
c(1,1,2)
c(1,3)
c(2,2)
Or of the form:
c(4,0,0)
c(2,1,0)
c(1,0,1)
c(0,2,0)
since I should be able to convert the former format to the latter by myself. Any hint as to how to approach this problem using R would be greatly appreciated. The latter format is excactly the vectors v such that sum(v * 1:3) is 4.
EDIT:
My own attempt:
rek = function(mat, id1, id2){
if(id1 + id2 != length(mat) + 1){ #If next state not absorbing
mat[id1] = mat[id1] - 1
mat[id2] = mat[id2] - 1
mat[id1+id2] = mat[id1+id2] + 1
out = mat
id = which(mat > 0)
for(i in id){
for(j in id[id>=i]){
if(j == i & mat[i] == 1){
next
}
out = rbind(out, rek(mat,i,j))
}
}
return(out)
}
}
start = c(n, rep(0, n-2))
states = rbind(start, rek(start, 1, 1))
states = states[!duplicated(states), ] #only unique states.
This is incredibly inefficient. E. g. when n = 11, my states has over 120,000 rows prior to removing duplicates, which leaves only 55 rows.
EDIT 2:
Using the parts() function described below I came up with:
temp = partitions::parts(n)
temp = t(temp)
for(i in 1:length(temp[,1])){
row = temp[i,]
if(any(row>(n-1))){#if absorbing state
next
}
counts = plyr::count(row[row>0])
newrow = rep(0,n-1)
id = counts$x
numbs = counts$freq
newrow[id] = numbs
states = rbind(states, newrow)
}
states = states[-1,]#removing the first row, added manually
which excactly gives me the vectors v such that sum(v * 1:(N-1)) is N.
If anyone is interested, this is to be used within coalescent theory, as a way to describe the possible relations between N individuals omitting when all are related. As an example with N = 4:
(4, 0, 0) -- No individuals are related
(2, 1, 0) -- Two individuals are related, the rest are not
(0, 2, 0) -- The individuals are pair-wise related
(1, 0, 1) -- Three individuals are related, the other individual is not.
Hope parts from package partitions could help
library(partitions)
N <- 4
res <- unique(lapply(asplit(parts(N),2),function(x) sort(x[x>0])))[-1]
which gives
> res
[[1]]
[1] 1 3
[[2]]
[1] 2 2
[[3]]
[1] 1 1 2
[[4]]
[1] 1 1 1 1
If you would like to write a custom base R function, here is a recursive version
f <- function(n, vhead = n, v = c()) {
if (n == 0) return(list(v))
unlist(lapply(seq_len(min(n, vhead)), function(k) f(n - k, k, c(k,v))), recursive = FALSE)
}
then we can run
res <- Filter(function(x) length(x)>1,f(N))
I am trying to set up a process using a while loop in order to have my code consistently sample among certain xd[i] before one particular xd[i] becomes equal to x.
I know it would be more efficient to put everything under one for loop (except for the while loop) but I am trying to create this step by step. Right now, I am stuck on the while loop part. I cannot run that part of the code without R crashing, or if it does not crash, it seems to continue sampling nonstop until I manually stop it. How can I change my while loop such that it samples over the xd vector until one of the elements of xd matches with x?
Thank you
reset1 = {
a = 0.3 #lower legal threshold
b = 0.9 #upper legal threshold
x = 0
theta = runif(1,min = a, max = b)
theta
A = 5 ## monetary value of harm from
maxw = 2*A
minw = 0
wbar = (maxw+minw)/2 ##average cost
wbar
xd = c(1,2,3)
w = c(1,2,3)
}
for (i in 1:length(xd)){w[i] = runif(1, min = 0, max = 2)} #trying to make it create a w for each person
##Drivers problem: pick the x that will minimize your cost
for(i in 1:length(xd)){xd[i] = min(c(1-(w[i]/(2*A)),((2+b)-sqrt(b^2-2*b+1+3*(w[i]/A)*(b-a)))/3,b))}
xd
for(i in 1:length(xd)){proba = function(xd){(xd-1)^2}}
proba(xd) #ith individual probability of getting in an accident given their xd[i]
proba(xd[c(1:3)])
probn = 1 - proba(xd) #probability of not getting in an accident given driveri's effort level
probn
while (any(x!=xd)) {x = sample(c(xd[c(1,2,3)],0,0,0),size = 1, replace = TRUE, prob = c(proba(xd), probn)) ###the x is selected based on which ever x resulted in an accident
}
show(x)
Perhaps
while(sum(xd!=x)==3){}
This loops runs as long as no element of xd equals x
Supposed I have an arbitrary function
foo = function(a,b) {a+b}
How can I iterate this function onto itself n times?
foo(foo(foo(foo(x, 1), 2), 3, 4)
I am looking at purrr:compose but it doesn't look hopeful for arbitrary n. purrr:reduce feels like it will come into play also... but I'm stumped.
Here is a pure purrr version, that is really functional, as you said reduce comes in handy here, since compose is just a function and functions are just elements you can reduce functions by composing them. To just fill one argument use partial.
foo_n <- reduce(map(1:n, ~partial(foo, b=.x)), compose)
You can also just append results of each foo(a,b) function into a numeric vector and then pick up the last result.
Let's x = 1 and bs are elements of 1:4:
x = 1
n = 4
out = vector("numeric")
steps = seq(1, 4, by = 1)
for( b in steps){
## initial value
if (length(out) == 0){
out = append(out, values = foo(x, b) )
}else{
out = append(out, values = foo( tail( out, 1), b) )
}
}
tail(out, 1)
I want to multiply and then sum the unique pairs of a vector, excluding pairs made of the same element, such that for c(1:4):
(1*2) + (1*3) + (1*4) + (2*3) + (2*4) + (3*4) == 35
The following code works for the example above:
x <- c(1:4)
bar <- NULL
for( i in 1:length(x)) { bar <- c( bar, i * c((i+1) : length(x)))}
sum(bar[ 1 : (length(bar) - 2)])
However, my actual data is a vector of rational numbers, not integers, so the (i+1) portion of the loop will not work. Is there a way to look at the next element of the set after i, e.g. j, so that I could write i * c((j : length(x))?
I understand that for loops are usually not the most efficient approach, but I could not think of how to accomplish this via apply etc. Examples of that would be welcome, too. Thanks for your help.
An alternative to a loop would be to use combn and multiply the combinations using the FUN argument. Then sum the result:
sum(combn(x = 1:4, m = 2, FUN = function(x) x[1] * x[2]))
# [1] 35
Even better to use prod in FUN, as suggested by #bgoldst:
sum(combn(x = 1:4, m = 2, FUN = prod))