Image processing for a set of images but showing "Array indices must be positive integers or logical values" for another set of images?

I made an image processing script which can work for a set of images, but when I try to process another set of images with the same script, it shows errors saying "Array indices must be positive integers or logical values".

Could anyone help me with this please?

Array indices must be positive integers or logical values.

Error in training (line 16)
mean5=mean(double(A3(:)<1500))

This is my code:

%read images
A=imread('r30jun97.giant.04--1---2.tif');
 
%increase intensity
A1 = A*1000;
 
%mask
mask = A1>=2000;
A2 = uint16(mask).*A1;
 
%median filter to remove noise
meanFilter = ones(3)/9;
A3 = imfilter(A2,meanFilter);
 
%statistics
mean5=mean(double(A3(:)<1500));
sd5=sqrt(var(double(A3(:)<1500)));

Answer 1

[Walter Roberson]

Guessing

mask = A3<1500;
mean5 = mean(A3(mask));
sd5 = std(A3(mask)) ;

However you should check whether you accidentally created a variable named mean

[Yuhong Jin]

Thank you for replying. I solved this problem by empty my workspace, but I do not know why that worked?

[Walter Roberson]

If you accidentally had a variable named mean in your workspace, then

mean(double(A3(:)<1500))

would be a request to calculate the logical vector A3(:)<1500, and convert the logical values to double (0 and 1) values, and use those double precision 0 and 1 values to index into the variable mean which would be an error because 0 is not a valid index.

Whereas if you did not have a variable named mean in your workspace, it would be a request to calculate the logical vector A3(:)<1500, and convert the logical values to double (0 and 1) values, total the 0 and 1 values, and divide by the number of values. Note that if that were your intention, to find the fraction of values A3 < 1500, then you could skip the double() and just use

mean(A3(:)<1500)

[Yuhong Jin]

My intention is to calculate the mean and standard deviation in a specified range of A3. Should I define another variable or I can do it with mean?

[Walter Roberson]

In that case the code I posted above would work. Or you might prefer to write it as

A3_5 = A3(A3<1500);
mean5 = mean(A3_5);
sd5 = std(A3_5);