how can I count up the number from a specific interval?

HI,I am pretty new with Mablab and I need some help.

I have created a vector d with a random function and I need to count up the number of instances where di> 8.

How can I do it?

Answer 1

[Nadir Altinbas]

hello for instance interval 8 < x > 999

x = 8 : 0.5: 999;

it increases as 0.5 up to 999 example

ok

-nadir altinbas

Answer 2

[ME]

Assuming you want the number of elements greater than 8 rather than the sum of the elements greater than 8 then you can simply use:

   numel(d(d>8))

Answer 3

[Steven Lord]

Generate a logical mask that is true where your data is in your specific interval and false where it is not.

x = 0:10
mask = x > 8

Count the number of nonzero (non-false) values in your mask using the nnz function.

nnz(mask) % 2, corresponding to x = 9 and x = 10

You can use the logical indexing technique shown by ME if you need to extract the elements of x that are in your interval. But if you're just trying to count them, that's extra work you don't have to do.

elementsGreaterThan8 = x(mask)

[ME]

I’ve just compared the time taken for each of our approaches. This approach is very slightly quicker - by about 0.0001s for an array with 100 random elements. That is as long as you miss out the nnz(mask) step which doesn’t seem to be necessary. If you include that step then this approach takes about 0.0001s longer for the same 100 random elements.

[Steven Lord]

Try this benchmarking script (with local functions) comparing your approach and mine. You should define a value n before running it. If you don't it'll start off at 0. I increase n after each run so we get different random numbers each run, to simulate a more realistic real-world application.

For sufficiently large vector sizes, I'm seeing MEsol take about 10 times as long as SLsol (they each return the same count.) Extracting about half the elements in the x vector into a temporary array simply to count how many elements it has with numel requires memory copying which slows down MEsol. For small enough vector sizes the time to perform that extra memory copying seems negligible but for larger sizes it can make a difference.

if ~exist('n', 'var')
    n = 0;
end

disp("Seed " + n + " for Mersenne Twister")
rng(n, 'twister');
x = randn(1, 1e7);

M = timeit(@() MEsol(x));
disp("Time for MEsol: " + M)
countME = MEsol(x);

S = timeit(@() SLsol(x));
disp("Time for SLsol: " + S)
countSL = SLsol(x);

disp("" + newline)
disp("MESol returned " + countME + newline + ...
     "SLSol returned " + countSL + newline);
n = n + 1;

function n = MEsol(x)
n = numel(x(x > 0));
end

function n = SLsol(x)
mask = x > 0;
n = nnz(mask);
end

[ME]

I agreed that your approach was faster! Yes, for the 100 elements I used (i.e. a small example) the time difference was small. I thought it went without saying that this difference would scale up as the number of elements increased but it’s great to get a better handle on these differences.