How to locate certain pixels and picture them in RGB image

Question

Dear all:
I have a code, as the attached one in LV 8.2, and have problems for this. I hope someone can kindly help me.  I have a 2-D data set, just like the random numbers generated for a 512x500 array in the code. I need to locate all pixels with values higher than a threshold, say 0.8 in the code, and all pixels with values lower than another certain threshold, say 0.2 in the code. The purpose is to picture the array with gray level color, and give different colors for groups of higher or lower values.
In my code, I got the following problems. First, when I count the number of pixels in a certain group, the value, for "large" and "small" indicators, are negative. What mistake did I make?
Second, I plan to duplicate the initial array (data set) and locate all the pixels with the expected values, assigning the zero value to each of them. Then, combine three arrays into one 512x500x3. I used Build Array in the code but got 3x512x500. What mistake did I make here?
Third, I plan to picture the data set in an intensity graph (I hope I did it right). Since the initial data is a 2-D array, it is supposed to be black-and-white in gray level, but I always got blue-and-white. Sigh, what silly mistake did I make for this?
Fourth, as you can see that eventually I am supposed to obtain a 3-D array. Each pixel without the expected values will have 3 the same values as the initial one. Each pixel with the expected values will have 2 the same values as the initial one and 1 zero. If I can present this array in a RGB image, the pixels without the expected values are supposed to have gray color, and the pixels with the expected values are supposed to have color different than gray. Is it possible to draw such a picture in LV? And is it possible to save such a picture?
I really appreciate for anyone who can help me figure out my mistakes and correct my code. Thank you very much in advance.
Attachments:
LocatePixels.vi ‏24 KB

Answer 1

Hi ccyang,
This is Stephen in Applications Engineering.  I've started taking a look at your questions and I wanted to let you know that I've found a solution to your first issue.  The 512x500 array that you created has approximately 250,000 values.  Since the random number generator is uniform, we can expect ~50,000 to be greater than 0.8.  50,000 is larger than the value that an I16 integer can hold.  An I16 integer type is the default output for the Boolean to (0,1) function you are using.  Therefore, add a "To Quad Integer" conversion function to your block diagram after the Boolean converter but before the sum.  I'll be back soon with the solutions to the other two questions.  Just wanted to let you know I was working on a solution.
Stephen Meserve
Applications Engineering
National Instruments
Stephen Meserve
National Instruments