Calculate values inside a nested loop with different steps and store them in the same variable.

I have a matrix (x) and want to calculate where its values are above certain thresholds and store the results in a new matrix (C). To be more specific, I need to check whether the condition sum(x>i) is met in 18 different values of i (classes). My difficulty is that I want a different step so for the first ten values step=1, while the remaining 8 while have step=5. I tried a nested loop but the result was not the expected. Any ideas how to proceed?

for i=1:10
     C(i)= sum(x>i);  
for j=15:5:50
    C(j)= sum(x>j);
end 
end

Answer 1

[Stephen23]

Assuming that SUM() returns a scalar:

V = [1:10,15:5:50];
C = V; % preallocate
for k = 1:numel(V) % loop over indices, not your data values
    C(k) = sum(x>V(k));
end

Note that in MATLAB it is almost always easier to iterate over indices, rather than over data values.

Or alternatively:

C = arrayfun(@(v)sum(x>v), V)

Answer 2

[Jiri Hajek]

Hi, you can use any special vector of indexes (natual numbers) like this:

j = [1:10,15:5:50];
for i=j
     C(i)= sum(x>i);
end

[Ancalagon8]

Thank you! What I noticed is that the C vector has 50 values and not 18 as expected. Do you know why? I mean that according to the steps, I expected values 18 values in total (1,2,3,4,5,6,7,8,9,10,15,20,25,30,35,40,45 and 50).

[Stephen23]

"I expected values 18 values in total (1,2,3,4,5,6,7,8,9,10,15,20,25,30,35,40,45 and 50)."

V = [1:10,15:5:50]

V = 1×18

1 2 3 4 5 6 7 8 9 10 15 20 25 30 35 40 45 50

[Ancalagon8]

Strange, your answer makes absolute sense, yet I still receive 50 values.

[Stephen23]

"Strange, your answer makes absolute sense, yet I still receive 50 values."

There is nothing strange going on at all: you are using the values of j as indices into C. The largest value of j is 50, so the largest index into C is 50, so C will be expanded to have 50 elements. The unassigned elements will have value zero, the default when arrays are expanded like that.

It might not be what you want (your request is totally unclear to me), but there is nothing strange going on here.

[Ancalagon8]

So V is 1X18 or 1X50?

Let me explain better: I want to check whether the condition sum(x>i) is met in 18 different values of i (classes). One option that it worked for me was to use one variable (C) for i =1:10, a second one (D) for j=15:5:50 and finally combine C and D into a new one (E) with 18 values. My initial question was how to combine this into a nested loop, where the steps are 1 and 5.

[Stephen23]

"So V is 1X18 or 1X50? "

As far as I can tell, the basic issue is that both your question and this answer use data/parameter values as indices, whereas you really should be iterating over actual indices. You are confusing code and data. But you have already accepted this answer as answering your question, which tells everyone that you are happy with the results you get.

[Ancalagon8]

@Stephen23 your answer does not help me understand why I receive 50 values instead of 18, but thank you for your time.

[Steven Lord]

Let's take a smaller example, one where we can view the vector without scrolling.

indicesToAssignTo = [1 3 5]

indicesToAssignTo = 1×3

1 3 5

for whichInd = indicesToAssignTo
    x(whichInd) = whichInd.^2
end

x = 1

x = 1×3

1 0 9

x = 1×5

1 0 9 0 25

When I run this you see that MATLAB displays x three times, once for each assignment to x inside the loop.

In the first iteration x is a 1-by-1 because we assign to element number 1.

In the second iteration x is a 1-by-3 because we assign to element number 3. MATLAB can't just leave element 2 empty, so it fills it with a default fill value of 0.

In the third iteration x is a 1-by-5 because we assign to element number 5. Again MATLAB can't just leave element 4 empty, so it fills it with 0.

In your code your vector has 50 elements because you assign to element number 50. MATLAB can't leave the elements to which you don't assign empty so it fills them with 0.

[Ancalagon8]

One option is to use nonzeros, but what if a value that I need is zero?

[Stephen23]

"your answer does not help me understand why I receive 50 values instead of 18"

Basically you are doing this:

A(50) = pi

A = 1×50

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

How many elements do you expect an array to have, when you assign something to its 50th element?

"One option is to use nonzeros, but whati if a value that I need is zero?"

As I wrote earlier, just actual indices for indexing, not your data values.