Find function misbehaves when Indexing into array

I have a function which searches an array for the first value which is below a given tolerance. First, it creates a logic array using the input and a "less than" condition. Then it finds the index of the first instance of "True" in the LogicArray using the find function. If no "True" value could be found, it sets the index to be the last value in the array (in this case, 125). Then it uses the found index value to return a corresponding array value.

function [FindIndex, RxComplete] = IndexCheck(x,y)
Tolerance = 100;
LogicArray = x < Tolerance;
FindIndex = find(LogicArray,1,'first');
if isempty(FindIndex)
    FindIndex = 125;
end
RxComplete = y(FindIndex,1);
end

The function is then applied to a cell array using cellfun.

[Index_output, Array_output] = cellfun(@IndexCheck, x, y, 'UniformOutput',false);

The code works for the above application, but it mesbehaves when I try to only search certain elements of my input array. For example, I want the function to skip the first 5 elements of the array so I index into LogicArray with (5:end). This should tell the find function to only start searching for a "True" after the first 5 elements.

function [FindIndex, RxComplete] = IndexCheck(x,y)
Tolerance = 100;
LogicArray = x < Tolerance;
FindIndex = find(LogicArray(5:end),1,'first');
if isempty(FindIndex)
    FindIndex = 125;
end
RxComplete = y(FindIndex,1);
end

But when I include this indexing condition for FindIndex, the function no long works and returns index and array values that do not satisfy the Tolerance condition. (i.e index values which correspond to an array value that is greater than 100).

I can trim the data to remove the first five elements, but this feels like a bandaid solution which doesn't address the issue. I'm hoping someone can help me understand why this is happening.

Answer 1

[Matt J]

Perhaps this is what you meant,

LogicArray = x < Tolerance;

LogicArray(1:4)=false;

FindIndex = find(LogicArray,1,'first');

[Michael]

Ah thank you! This works beautifully. I feel silly for not thinking of this. Do you have any idea why find is not able to "ignore" elements in an array as I was attempting to do?

[Matt J]

Indexing into LogicArray just gives back a shorter array, which is then given to find(). Find() has no way of knowing what operations were done to produce the shorter array that it receives as input.

In particular, it has no way of knowing the relationship between LogicArray(5:end) and LogicArray alone, nor that you want it to return indices into the latter, rather than the former.

[Stephen23]

"Do you have any idea why find is not able to "ignore" elements in an array as I was attempting to do?"

This has nothing to do with FIND. This is a very basic concept of MATLAB. The simple answer to your question is that the elements are not there, so they cannot be ignored (or indeed have anything else done with them).

Lets try a very simple example, just ilke what you are doing. Only for clarity on two lines:

V = [1,1,0,0,0,1]

V = 1×6

1 1 0 0 0 1

W = V(3:end)

W = 1×4

0 0 0 1

find(W)

ans = 4

What is the input to FIND? The input to FIND is the vector [0,0,0,1]. Is the output correct? (hint: yes).

FIND cannot know anything about where that vector came from, e.g. how it was defined, if it was originally a sub-vector of a larger vector or array. Vectors do not have memory like that. All FIND knows about is the input it was provided with (as it should).

Answer 2

[Steven Lord]

%{
if isempty(FindIndex)
    FindIndex = 125;
end
%}

Rather than assuming that your x array (and your y array) has exactly 125 rows (which if you pass in only a subset of your data may not be a valid assumption) I would compute this value using either the size function or (since you want the index of the last row) the height function. [For columns the corresponding function is width or you can still use size.]

x = ones(3, 4)

x = 3×4

1 1 1 1 1 1 1 1 1 1 1 1

size(x, 1)

ans = 3

height(x)

ans = 3

size(x, 2)

ans = 4

width(x)

ans = 4

[Michael]

Thank you for this tip, I was also looking for a better way to define that element.