wish of those help Since multiple entries in data[i] might contain the same value, I don't see how this could be vectorized simply since there can be race conditions. The point of vectorization is that each element is independent of the other elements, and so can be computed in parallel. But your algorithm doesn't allow that. "Vectorize" is not the same thing as "make go faster." What you seem to be building here is a histogram, and iOS has builtin, optimized support for that. You can create a singlechannel, singlerow image and use vImageHistogramCalculation_Planar8 like this: code :
void f1(unsigned char *data, unsigned int size) {
unsigned long A[256] = {0u};
vImage_Buffer src = { data, 1, size, size };
vImage_Error err = vImageHistogramCalculation_Planar8(&src, A, kvImageDoNotTile);
if (err != kvImageNoError) {
// error
}
...
}
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Refactoring/Vectorize a simple for loop (example data and loop included)
By : developer
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Here is the sample code , Aren't you just doing code :
sif < matrix(s, ncol=N, byrow=T)
sif < matrix(s[1:(round(length(s)/N)*N)], ncol=N, byrow=T)
> max(abs(sif[1:9999,1:100]sif2[1:9999,1:100]))
[1] 0
> sif[10000,]
[1] 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 ...
sif2[nrow(sif2), ] < nrow(sif2)
User System elapsed
Your way 57.831 14.056 71.525
R's way 0.004 0.002 0.006

Trying to vectorize a for loop in R
By : Rajesh
Date : March 29 2020, 07:55 AM
may help you . In general, try to avoid if when not needed. I think your desired output can be built as follows. code :
model < unlist(sapply(1:length(model.list), function(i) model.list[[i]][model.lookup[[i]]]))
model[model=='']<NA

Vectorize the following for loop
By : khodaeim
Date : March 29 2020, 07:55 AM
With these it helps Approach #1 The way I see it as, you would have a bottleneck with the cell2mat inside the loop. If you are running it for considerable number of loop iterations, this bottleneck would be felt. So, with this post I would try to reorganize the input array a so that the cell2mat would be used on it once outside the loop. That would be used to extract all elements and then regroup cell elements into another cell array, whose each cell would hold values corresponding to the cell2mat() extracted values at each iterations with the original code. Thus, each cell of such a reorganized cell array could be fed directly to replace cell2mat(a(a{elem})). code :
% Reorganize a to create another cell array in which each cell
% would have "cell2mat(a(a{elem}))" for iterator elem
lens = cellfun('length',a)
arr = [a{:}]
cumlens = cumsum(lens(arr))
grouped_cumlens = cumlens(cumsum(lens))
grouped_a = mat2cell(cell2mat(a(arr)),1,diff([0 grouped_cumlens]))
% Use grouped_a to perform the same operations but without cell2mat in loops
outx = 0;
for elem = i
outx = outx + sum(ismember(grouped_a{elem},b{elem}));
end
% Code unchanged from the previous version
lens = cellfun('length',a)
arr = [a{:}]
cumlens = cumsum(lens(arr))
grouped_cumlens = cumlens(cumsum(lens))
% Extract data into a regular array and decide start and stop indices
data = cell2mat(a(arr))
starts = [1 grouped_cumlens(1:end1)+1]
stops = grouped_cumlens
% Use extracted data with its start,stop indices for each iteration
outx = 0;
for elem = i
outx = outx + sum(ismember(data(starts(elem):stops(elem)),b{elem}));
end

How to vectorize this for loop
By : MDombrowski
Date : March 29 2020, 07:55 AM
wish helps you In this case you need to transform your subscripted indexes stored in x_cords and y_cords into linear indexes using the sub2indfunction and then you can directly assign the ones like this: code :
buffImg=zeros(100,100);
x=randperm(100);
y=randperm(100);
buffImg(sub2ind(size(buffImg),x,y))=1;
x=randperm(100);
y=randperm(100);
buffImg=zeros(100,100);
buffImg2=zeros(100,100);
for i = 1:length(x)
buffImg(x(i),y(i)) = 1;
end
buffImg2(sub2ind(size(buffImg),x,y))=1;
all(all(buffImg==buffImg2))

selective `for` loop for an R function: `for` loop Or `Vectorize`
By : 曹先胜
Date : March 29 2020, 07:55 AM
Does that help I highly appreciate the answer by our colleague @r2evans, but I think this problem might have a bit simpler solution (i.e., one line of if syntax): code :
ab = function(m, s, lo = Inf, hi = Inf){
loop = if(length(m) > 1) length(m) else length(s)
p = if(length(m) > 1) function(x) dnorm(x, m[i], s) else function(x) dnorm(x, m, s[i])
for(i in 1:loop){
f = function(x) p(x)/integrate(p, lo, hi)[[1]]
curve(f, 3, 3, add = i!= 1, col = i)
}
}
# Example of use:
ab(m = c(0, 1), 1)
#Example of use:
ab(m = 0, c(1, 2))

