R : Faster way to repeat an array along a dimension? - r

Suppose i have an array z given by :
z = array(runif(100*50*200),c(100,50,200))
Is there a faster way to do :
dim(z) = c(100,50,1,200)
z = z[,,rep(1,300),]
Note that this is an exemple, the new dimension where i want to repeat the array is not always the 3rd, and the starting dimension of the array is not always 3.
profvis::profvis() only shows that the Garbage collector takes a certain time in the computation, but it does not show other internals..
It might be an allocation issue, although i'm not shure why it takes that kind of time. I have several of those very basic calls in my code, and 95% of my runtime is spend there.. So even if it's unavoidable, can you explain to me why is it so long ?

Related

Nondeterministic growth in Julia loop

I have a computation loop with adaptive time stepping and I need to store the results at each iteration. In other words, I do not know the vector size before the computation, so I can not preallocate a vector size to store the data. Right now, I build a vector using the push! function
function comp_loop()
clock = [0.0]
data = [0.0]
while run_time < model_time
# Calculate new timestep
timestep = rand(Float64) # Be sure to add Random
run_time += timestep
# Build vector
push!(clock,run_time)
push!(data,timestep)
end
end
Is there a more efficient way to go about this? Again, I know the optimal choice is to preallocate the vector, but I do not have that luxury available. Buffers are theoretically not an option either, as I don't now how large to make them. I'm looking for something more "optimal" on how to implement this in Julia (i.e. maybe some advanced application available in the language).
Theoretically, you can use a linked list such as the one from DataStructures.jl to get O(1) appending. And then optionally write that out to a Vector afterwards (probably in reverse order, though).
In practise, push!ing to a Vector is often efficient enough -- Vectors use a doubling strategy to manage their dynamic size, which leads to amortized constant time and the advantage of contiguous memory access.
So you could try the linked list, but be sure to benchmark whether it's worth the effort.
Now, the above is about time complexity. When you care about allocation, the argument is quite similar; with a vector, you are going to end up with memory proportional to the next power of two after your actual requirements. Most often, that can be considered amortized, too.

Parallel iteration over array with step size greater than 1

I'm working on a practice program for doing belief propagation stereo vision. The relevant aspect of that here is that I have a fairly long array representing every pixel in an image, and want to carry out an operation on every second entry in the array at each iteration of a for loop - first one half of the entries, and then at the next iteration the other half (this comes from an optimisation described by Felzenswalb & Huttenlocher in their 2006 paper 'Efficient belief propagation for early vision'.) So, you could see it as having an outer for loop which runs a number of times, and for each iteration of that loop I iterate over half of the entries in the array.
I would like to parallelise the operation of iterating over the array like this, since I believe it would be thread-safe to do so, and of course potentially faster. The operation involved updates values inside the data structures representing the neighbouring pixels, which are not themselves used in a given iteration of the outer loop. Originally I just iterated over the entire array in one go, which meant that it was fairly trivial to carry this out - all I needed to do was put .Parallel between Array and .iteri. Changing to operating on every second array entry is trickier, however.
To make the change from simply iterating over every entry, I from Array.iteri (fun i p -> ... to using for i in startIndex..2..(ArrayLength - 1) do, where startIndex is either 1 or 0 depending on which one I used last (controlled by toggling a boolean). This means though that I can't simply use the really nice .Parallel to make things run in parallel.
I haven't been able to find anything specific about how to implement a parallel for loop in .NET which has a step size greater than 1. The best I could find was a paragraph in an old MSDN document on parallel programming in .NET, but that paragraph only makes a vague statement about transforming an index inside a loop body. I do not understand what is meant there.
I looked at Parallel.For and Parallel.ForEach, as well as creating a custom partitioner, but none of those seemed to include options for changing the step size.
The other option that occurred to me was to use a sequence expression such as
let getOddOrEvenArrayEntries myarray oddOrEven =
seq {
let startingIndex =
if oddOrEven then
1
else
0
for i in startingIndex..2..(Array.length myarray- 1) do
yield (i, myarray.[i])
}
and then using PSeq.iteri from ParallelSeq, but I'm not sure whether it will work correctly with .NET Core 2.2. (Note that, currently at least, I need to know the index of the given element in the array, as it is used as the index into another array during the processing).
How can I go about iterating over every second element of an array in parallel? I.e. iterating over an array using a step size greater than 1?
You could try PSeq.mapi which provides not only a sequence item as a parameter but also the index of an item.
Here's a small example
let res = nums
|> PSeq.mapi(fun index item -> if index % 2 = 0 then item else item + 1)
You can also have a look over this sampling snippet. Just be sure to substitute Seq with PSeq

How to verify number of function evaluations when profiling R code

When profiling R code with Rprof-type functions we get the time spent in function alone and the time spent in function and callees. However, as far as I know we don't get the number of times a given function was evaluated.
For example, assume I wants to compare two integration functions:
integrate_1(myfunc, from = -Inf, to = Inf)
integrate_2(myfunc, from = -Inf, to Inf)
I could easily see how much time each function takes and where this time was spent, but I don't know how to check how many times myfunc had to be evaluated in each of the integrate functions.
Thanks,
One way of implementing Joran's counter method is to use the trace function.
For example, first we set the counter to zero. (Assigned in the global environment, for convenience.)
count <- 0
Then set up the trace. Here we set it on the identity function (that just returns the value that you input to it).
trace("identity", quote(count <<- count + 1), print = FALSE)
Now whenever identity is called, the value of count is incremented. print = FALSE just stops a message being printed to the console when the function is called.
Let's call the function a few times and inspect the count:
for(i in seq_len(123)) identity(1)
count
## [1] 123
Rprof works by sampling the call stack on a timer. It does not count calls.
It records the sampled call stacks in a file, and though it does not record line numbers where calls occur, those samples are still useful for seeing what causes time to be spent.
For example, if you happen to look at M random samples, and you see a pattern like A calling B calling C on N of them, then you know the program spends roughly fraction N/M of its time doing that (assuming N > 1).
If you see such a thing, and you can think of a way to avoid even part of it, you will save a substantial fraction of the total time.
Rprof comes with a summarization tool that gives you the kind of numbers you mentioned, but I don't find those numbers useful anyway.
I would much rather get a real sense of what's happening.

modifying an element of a list in-place in J, can it be done?

I have been playing with an implementation of lookandsay (OEIS A005150) in J. I have made two versions, both very simple, using while. type control structures. One recurs, the other loops. Because I am compulsive, I started running comparative timing on the versions.
look and say is the sequence 1 11 21 1211 111221 that s, one one, two ones, etc.
For early elements of the list (up to around 20) the looping version wins, but only by a tiny amount. Timings around 30 cause the recursive version to win, by a large enough amount that the recursive version might be preferred if the stack space were adequate to support it. I looked at why, and I believe that it has to do with handling intermediate results. The 30th number in the sequence has 5808 digits. (32nd number, 9898 digits, 34th, 16774.)
When you are doing the problem with recursion, you can hold the intermediate results in the recursive call, and the unstacking at the end builds the results so that there is minimal handling of the results.
In the list version, you need a variable to hold the result. Every loop iteration causes you to need to add two elements to the result.
The problem, as I see it, is that I can't find any way in J to modify an extant array without completely reassigning it. So I am saying
try. o =. o,e,(0&{y) catch. o =. e,(0&{y) end.
to put an element into o where o might not have a value when we start. That may be notably slower than
o =. i.0
.
.
.
o =. (,o),e,(0&{y)
The point is that the result gets the wrong shape without the ravels, or so it seems. It is inheriting a shape from i.0 somehow.
But even functions like } amend don't modify a list, they return a list that has a modification made to it, and if you want to save the list you need to assign it. As the size of the assigned list increases (as you walk the the number from the beginning to the end making the next number) the assignment seems to take more time and more time. This assignment is really the only thing I can see that would make element 32, 9898 digits, take less time in the recursive version while element 20 (408 digits) takes less time in the loopy version.
The recursive version builds the return with:
e,(0&{y),(,lookandsay e }. y)
The above line is both the return line from the function and the recursion, so the whole return vector gets built at once as the call gets to the end of the string and everything unstacks.
In APL I thought that one could say something on the order of:
a[1+rho a] <- new element
But when I try this in NARS2000 I find that it causes an index error. I don't have access to any other APL, I might be remembering this idiom from APL Plus, I doubt it worked this way in APL\360 or APL\1130. I might be misremembering it completely.
I can find no way to do that in J. It might be that there is no way to do that, but the next thought is to pre-allocate an array that could hold results, and to change individual entries. I see no way to do that either - that is, J does not seem to support the APL idiom:
a<- iota 5
a[3] <- -1
Is this one of those side effect things that is disallowed because of language purity?
Does the interpreter recognize a=. a,foo or some of its variants as a thing that it should fastpath to a[>:#a]=.foo internally?
This is the recursive version, just for the heck of it. I have tried a bunch of different versions and I believe that the longer the program, the slower, and generally, the more complex, the slower. Generally, the program can be chained so that if you want the nth number you can do lookandsay^: n ] y. I have tried a number of optimizations, but the problem I have is that I can't tell what environment I am sending my output into. If I could tell that I was sending it to the next iteration of the program I would send it as an array of digits rather than as a big number.
I also suspect that if I could figure out how to make a tacit version of the code, it would run faster, based on my finding that when I add something to the code that should make it shorter, it runs longer.
lookandsay=: 3 : 0
if. 0 = # ,y do. return. end. NB. return on empty argument
if. 1 ~: ##$ y do. NB. convert rank 0 argument to list of digits
y =. (10&#.^:_1) x: y
f =. 1
assert. 1 = ##$ y NB. the converted argument must be rank 1
else.
NB. yw =. y
f =. 0
end.
NB. e should be a count of the digits that match the leading digit.
e=.+/*./\y=0&{y
if. f do.
o=. e,(0&{y),(,lookandsay e }. y)
assert. e = 0&{ o
10&#. x: o
return.
else.
e,(0&{y),(,lookandsay e }. y)
return.
end.
)
I was interested in the characteristics of the numbers produced. I found that if you start with a 1, the numerals never get higher than 3. If you start with a numeral higher than 3, it will survive as a singleton, and you can also get a number into the generated numbers by starting with something like 888888888 which will generate a number with one 9 in it and a single 8 at the end of the number. But other than the singletons, no digit gets higher than 3.
Edit:
I did some more measuring. I had originally written the program to accept either a vector or a scalar, the idea being that internally I'd work with a vector. I had thought about passing a vector from one layer of code to the other, and I still might using a left argument to control code. With I pass the top level a vector the code runs enormously faster, so my guess is that most of the cpu is being eaten by converting very long numbers from vectors to digits. The recursive routine always passes down a vector when it recurs which might be why it is almost as fast as the loop.
That does not change my question.
I have an answer for this which I can't post for three hours. I will post it then, please don't do a ton of research to answer it.
assignments like
arr=. 'z' 15} arr
are executed in place. (See JWiki article for other supported in-place operations)
Interpreter determines that only small portion of arr is updated and does not create entire new list to reassign.
What happens in your case is not that array is being reassigned, but that it grows many times in small increments, causing memory allocation and reallocation.
If you preallocate (by assigning it some large chunk of data), then you can modify it with } without too much penalty.
After I asked this question, to be honest, I lost track of this web site.
Yes, the answer is that the language has no form that means "update in place, but if you use two forms
x =: x , most anything
or
x =: most anything } x
then the interpreter recognizes those as special and does update in place unless it can't. There are a number of other specials recognized by the interpreter, like:
199(1000&|#^)199
That combined operation is modular exponentiation. It never calculates the whole exponentiation, as
199(1000&|^)199
would - that just ends as _ without the #.
So it is worth reading the article on specials. I will mark someone else's answer up.
The link that sverre provided above ( http://www.jsoftware.com/jwiki/Essays/In-Place%20Operations ) shows the various operations that support modifying an existing array rather than creating a new one. They include:
myarray=: myarray,'blah'
If you are interested in a tacit version of the lookandsay sequence see this submission to RosettaCode:
las=: ,#((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)#]^:(1+i.#[)
5 las 1
11 21 1211 111221 312211

MPI_ALLREDUCE PROBLEM

I have a code which has a 2D local array (cval).This local array is being calculated by every processor and in the end I call MPI_ALLREDUCE to sum this local array to a global array(gns).
This local array has different sizes for different processors.The way I do a all reduce is as follows
k = n2spmax- n2spmin + 1 ! an arbitrary big value
do i = nmin, nmax
call MPI_ALLREDUCE(cval(i,:),gns(i,:),k,MPI_DOUBLE_PRECISION,MPI_SUM,MPI_COMM_WORLD,ierr)
end do
Is this the correct way of writing it.I am not sure about it ?
No, you can't do it this way. MPI_Allreduce requires that all of the processes in the communicator are contributing the same amount of data. That's why there's a single count argument.
To give more guidance on what is the right way to do this, we'll need a bit more clarity on what you're trying to do. Is the idea that you're calculating gns(i,j) = the sum over all ranks of cval(i,j), but not all ranks have all the cval(i,j)s?

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