I am very new to Matlab, and I feel completely overwhelmed by the use of arrays. What is the most efficient implementation of the following C++ code in Matlab?
A = std::vector<double>();
for (int i = 0; i < 100; i++) {
if (complicatedBoolFunction(i)) {
A.push_back(i);
}
}
Edit: By efficiency I mean to use as little resources as possible to grow the array A - that is, to avoid copy-pasting it into temporary memory
You can do this 2 ways
Pre-allocating for the maximum size, and removing unused elements. This has the advantage of pre-allocating memory in case the condition is often met...
A = NaN(100,1)
for ii = 0:99
if rand > 0.5 % some condition
A(ii+1) = ii; % some value
end
end
A(isnan(A)) = []; % remove unused elements
Appending to the array. This avoids making A way too large if appending is unlikely...
A = []; % empty array
for ii = 0:99
if rand > 0.5 % some condition
A(end+1, 1) = ii; % some value. Equivalent to 'A = [A; ii];'
end
end
A better, and more Matlab-esque way of doing this would be to vectorise your conditional function. This way you avoid looping and allocation issues...
ii = 0:99;
A = ii(rand(100, 1) > 0.5);
You can use any Boolean function you like as an indexing array, as long as it returns a logical array with the same number of elements as the array you're indexing (ii here) or integer indices of the elements to choose.
The most efficient implementation of such C++ code would be
i = 0:99;
A = i(complicatedBoolFunction(i));
Anyway you can grow an array with concatenation, which is (or was) usually not recommended, like the following
A = [];
for i = 0:99
if (complicatedBoolFunction(i))
A = [A i];
end
end
or much more efficiently like this:
A = [];
for i = 0:99
if (complicatedBoolFunction(i))
A(end + 1) = i;
end
end
Related
How can I delete all-zero pages from a 3D matrix in a loop?
I have come up with the following code, though it is not 'entirely' correct, if at all. I am using MATLAB 2019b.
%pseudo data
x = zeros(3,2,2);
y = ones(3,2,2);
positions = 2:4;
y(positions) = 0;
xy = cat(3,x,y); %this is a 3x2x4 array; (:,:,1) and (:,:,2) are all zeros,
% (:,:,3) is ones and zeros, and (:,:,4) is all ones
%my aim is to delete the arrays that are entirely zeros i.e. xy(:,:,1) and xy(:,:,2),
%and this is what I have come up with; it doesn't delete the arrays but instead,
%all the ones.
for ii = 1:size(xy,3)
for idx = find(xy(:,:,ii) == 0)
xy(:,:,ii) = strcmp(xy, []);
end
end
Use any to find indices of the slices with at least one non-zero value. Use these indices to extract the required result.
idx = any(any(xy)); % idx = any(xy,[1 2]); for >=R2018b
xy = xy(:,:,idx);
I am unsure what you'd expect your code to do, especially given you're comparing strings in all-numerical arrays. Here's a piece of code which does what you desire:
x = zeros(3,2,2);
y = ones(3,2,2);
positions = 2:4;
y(positions) = 0;
xy = cat(3,x,y);
idx = ones(size(xy,3),1,'logical'); % initialise catching array
for ii = 1:size(xy,3)
if sum(nnz(xy(:,:,ii)),'all')==0 % If the third dimension is all zeros
idx(ii)= false; % exclude it
end
end
xy = xy(:,:,idx); % reindex to get rid of all-zero pages
The trick here is that sum(xy(:,:,ii),'all')==0 is zero iff all elements on the given page (third dimension) are zero. In that case, exclude it from idx. Then, in the last row, simply re-index using logical indexing to retain only pages whit at least one non-zero element.
You can do it even faster, without a loop, using sum(a,[1 2]), i.e. the vectorial-dimension sum:
idx = sum(nnz(xy),[1 2])~=0;
xy = xy(:,:,idx);
I have this problem: I want to iterate on the elements of an array that satisfy a certain condition.
The first thing I though is to use the filter method on the original array and then iterates over the resulting elements. But I had some memory usage problem with that, i.e. java heap space.
When a filter is applied on an array, it will instantiate a new array? So it will copy each element?
Is it better to use this approach:
array.filter(<condition>).foreach{ element =>
do something
}
Or the following one?
for(i <- array.indices if <condition>){
do something
}
Moreover, I wrote these two tests:
With for
val size = 10000000
val elements = Array.ofDim[Double](size)
for (i <- elements.indices) {
elements.update(i, math.random)
}
var cont = 0
val n = 0.5
while(true){
cont = 0
for (j <- elements.indices if elements(j) < n) {
cont += 1
}
println(cont)
}
with filter
val size = 10000000
val elements = Array.ofDim[Double](size)
for (i <- elements.indices) {
elements.update(i, math.random)
}
val n = 0.5
while(true){
val valid = elements.filter(x => x < n)
println(valid.size)
}
and I checked the memory usage with VisualVM, it seem that the first solution uses less memory than the second one.
This is the memory used by the first solution
This is the memory used by the second solution
The for expression use the withFilter method rather than filter, which avoids creating the intermediate Array. So either use the for version or use withFilter rather than filter.
var amazon = activeSpreadsheet.getSheetByName('Amazon');
var lastRow1 = amazon.getLastRow();
var array1 = amazon.getRange('A2:A' + lastRow1).getValues();
var source = activeSpreadsheet.getSheetByName('ProductDetails');
var lastRow2 = source.getLastRow();
var array2 = source.getRange('B2:B' + lastRow2).getValues();
n = 2;
x = 0; // match
z = 0; // non-match
for (var i = 0; i < array2.length; i++){
for (var j = 0; j < array1.length; j++){
if (array2[i] !== array1[j]){
z = z + 1;
}
else {
x = 9999
}
}
newsheet.getRange([n],[5]).setValue(z);
newsheet.getRange([n],[6]).setValue(x);
if (z > x) {
newsheet.getRange([n],[1]).setValue(array2[i]);
n == n++;
z = 0;
x = 0;
}
else if (z < x) {
z = 0;
x = 0;
}
}
My project is written in GAS (google app scripts) which is essentially, for all intents and purposes JS with variation in libraries.
Basically I am grabbing an element in the array2 and passing it through a loop to match to array1. For every time it does not match it adds 1, and when it matches (should only match once if it has any matches) it stores an arbitrary large number (larger than length of array1) and compares them.
As you can see I've written out to display these values and I always get z = 5183 (length of array1) and x = 0 (meaning no non-matches found). Therefore even if something exists in array 2 and 1 it will always write it to the cell.
What should happen is if there is a match, z= 5182 and x= 9999 (or arbitrary large number) and since 5182 < 9999 it doesn't do anything.
Is my scope wrong? Or am I not writing the If/Else correctly? Or is it something else?
Your code performs a strict comparison between the elements of two Arrays. That's fine, in general. However, for those specific Arrays those elements are also Arrays, which means strict (in)equality is checking to see if those are the exact same array object in memory. See this question for more information.
You probably wanted to do a value-based comparison, which means you need to compare the specific element of that inner array (i.e., index again). if (array2[i][0] !== array1[j][0]) {...} will check the 1st element of the inner array.
Looking at the instantiation of array1 and array2, we see that these are indeed 2D arrays from a single-column Ranges, so there will be only 1 element in each inner array. You can reduce the level of indexing necessary by flattening these arrays when you read them:
const array1 = sheet.getRange(...).getValues().map(function (row) { return row[0]; });
const array2 = ...;
I'm also not sure why you are passing in arrays to Sheet#getRange - you should pass in 1-4 arguments in manners consistent with the method signatures detailed in the Apps Script documentation.
Note that there are much better algorithms for checking if a value exists in a given array - you re-scan all of the 2nd array for every value of the first array. You should conduct thorough research into this topic before asking a new question on how to improve your algorithm.
Lastly, you should implement the best practice of using batch methods - you currently call setValue in a loop. Consider storing the results to write in an array, and then writing with Range#setValues once your loop has completed. There are many questions available for you to review on this topic.
Is there a way to grow a 3D array in the third dimension using the end index in a loop in Matlab?
In 2D it can be done like
a = [];
for x = y
a(end + 1, :) = f(x);
end
But in 3D the same thing will not work as a(1,1,end) will try to index a(1,1,1) the first iteration (not a(1,1,0) as one might expect). So I can't do
im = [];
for x = y
im(:, :, end + 1) = g(x);
end
It seems the end of a in third dimension is handled a bit differently than in the first two:
>> a = [];
>> a(end,end,end) = 1
Attempted to access a(0,0,1); index must be a positive integer or logical.
Am I missing something about how end indexing works here?
What you're asking...
If you know the size of g(x), initialize im to an empty 3d-array:
im = zeros(n, m, 0); %instead of im = [];
I think your code should work now.
A better way...
Another note, resizing arrays each iteration is expensive! This doesn't really matter if the array is small, but for huge matrices, there can be a big performance hit.
I'd initialize to:
im = zeros(n, m, length(y));
And then index appropriately. For example:
i = 1;
for x = y
im(:, :, i) = g(x);
i = i + 1;
end
This way you're not assigning new memory and copying over the whole matrix im each time it gets resized!
I wan to declare an array in MATLAB without specifying the size, rather like std::vector in C++, and then I want to "push" elements to the array. How can I declare this array and push to it?
Altough the answer of Paul R is correct, it is a very bad practice to let an array grow in Matlab without pre-allocation. Note that even std::vector has the option to reserve() memory to avoid repeated re-allocations of memory.
You might want to consider pre-allocating a certain amount of memeory and then resize to fit the actual needed size.
You can read more on pre-allocation here.
You can just define an empty array like this:
A = [];
To "push" a column element:
A = [ A 42 ];
To "push" a row element:
A = [ A ; 42 ];
As Shai pointed out, pushing elements onto a vector is not a good approach in MATLAB. I'm assuming you're doing this in a loop. In that case, this would be better approach:
A = NaN(max_row, 1);
it = 0;
while condition
it = it + 1;
A(it) = value;
end
A = A(1:it);
If you don't know the maximum dimension, you may try something like this:
stack_size = 100;
A = NaN(stack_size,1);
it = 0;
while some_condition
it = it + 1;
if mod(it, stack_size) == 0
A = [A; NaN(stack_size,1)];
end
A(it) = value;
end
A = A(1:it);