topic Re: search 1D array indices problem in LabVIEW

search 1D array indices problem

midori2 — Tue, 29 May 2012 15:46:50 GMT

Hello All

I'm trying do my project in labview, but I'm getting stuck

I have 1D array consists of series of 1 and 0 like this --> 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1

but I'm interested on finding the indices of 1 which are only the start and end point of the series

for example: input is 1 1 1 0 0 0 1 1 1 1 0 0 1 1 1

the output should be 0 2 6 9 12 14

Thank you

Re: search 1D array indices problem

crossrulz — Tue, 29 May 2012 15:54:55 GMT

Sounds like you need a for loop with a couple of shift registers. One shift register should be used hold the previous value and another to keep the array indecies. Compare the previous value to the current one, if they are different add the index to the array using the build array function.

What have you tried so far? Give us some code and explain where you are stuck.

Re: search 1D array indices problem

midori2 — Tue, 29 May 2012 16:26:56 GMT

right now i just got the indices of 1's using while loop but i cannot find the indices of each 1's series

Re: search 1D array indices problem

crossrulz — Tue, 29 May 2012 17:04:18 GMT

I think the way you're searching is flawed. Like I said earlier, I would look for changes in the array values and store those indices. I was also playing with your original code and alternated what to search for (1 and 0). Either solution should work.

I decided to build upon what you already had. I used the remainder of (i+1)/2 to alternate between searching for 1 and 0. I will admit that this is not a final solution (I can't just give you an answer). You will need to account for when the new value is a 0 (subtract 1 from the index) and add the last index if it is a 1 and wasn't already included.

Re: search 1D array indices problem

GerdW — Tue, 29 May 2012 18:33:02 GMT

Hi,

two more ideas:

Re: search 1D array indices problem

ben64 — Tue, 29 May 2012 18:57:23 GMT

And one more.

Ben64

Re: search 1D array indices problem

crossrulz — Tue, 29 May 2012 19:08:24 GMT

ben64,

That's a great way to do it. It removes the extra logic I needed in order to get the last element and the need to subtract the index for transitions to 0. My only thing is that I would put the build array in a case structure to save some memory.

Re: search 1D array indices problem

crossrulz — Tue, 29 May 2012 19:09:56 GMT

@GerdW wrote:

Hi,

two more ideas:

Gerd, you might be having a little too much fun with those Pt-by-Pt VIs.... In all seriousness, I really need to look at them. They seem VERY useful.

Re: search 1D array indices problem

JW-JnJ — Tue, 29 May 2012 20:41:54 GMT

Ok, I know this is already solved, but I wanted to have fun.

A bit more obscure, but smaller.

Re: search 1D array indices problem

crossrulz — Tue, 29 May 2012 20:54:02 GMT

@JW-L3CE wrote:

Ok, I know this is already solved, but I wanted to have fun.

A bit more obscure, but smaller.

Yeah, I can tell I've rubbed off on you at least a little bit. That was my original train of thought. I abandoned it however because I was trying to help the OP think through the problem.

Re: search 1D array indices problem

JW-JnJ — Tue, 29 May 2012 21:08:55 GMT

@crossrulz wrote:

@JW-L3CE wrote:

Ok, I know this is already solved, but I wanted to have fun.

A bit more obscure, but smaller.

Yeah, I can tell I've rubbed off on you at least a little bit. That was my original train of thought. I abandoned it however because I was trying to help the OP think through the problem.

Good point, I should explain for the OP.

It works by subtracting the current value from the previous value.

If you get a 0->1 transition, the subtraction results in a 1 and the current index is added (the "1" case).

If you get a 1->0 transition, the subtraction results in a -1 and the previous index is added (the "-1" case).

The last 0 is added to the array so that you can catch a 1 at the end of the array. If the last element is a 1, it will reach the end and subtract 0-1, get a "-1" result, and report the previous index (the REAL last index, not the added one)

The shift register is initialized to 0 so you can catch 1's at the start of the array. This would subtract 1-0 and get a "1" case.

Re: search 1D array indices problem

ben64 — Tue, 29 May 2012 21:30:16 GMT

@JW-L3CE wrote:

@crossrulz wrote:

@JW-L3CE wrote:

Ok, I know this is already solved, but I wanted to have fun.

A bit more obscure, but smaller.

Yeah, I can tell I've rubbed off on you at least a little bit. That was my original train of thought. I abandoned it however because I was trying to help the OP think through the problem.

Good point, I should explain for the OP.

It works by subtracting the current value from the previous value.

If you get a 0->1 transition, the subtraction results in a 1 and the current index is added (the "1" case).

If you get a 1->0 transition, the subtraction results in a -1 and the previous index is added (the "-1" case).

The last 0 is added to the array so that you can catch a 1 at the end of the array. If the last element is a 1, it will reach the end and subtract 0-1, get a "-1" result, and report the previous index (the REAL last index, not the added one)

The shift register is initialized to 0 so you can catch 1's at the start of the array. This would subtract 1-0 and get a "1" case.

There is a small (minor) issue with this, if you have a single 1 its index will be in the output twice (for the 0-1 and 1-0 transitions). Note that it migt be ok, it depend on what the OP wants.

Ben64

Re: search 1D array indices problem

JW-JnJ — Tue, 29 May 2012 21:33:47 GMT

@ben64 wrote:

There is a small (minor) issue with this, if you have a single 1 its index will be in the output twice (for the 0-1 and 1-0 transitions). Note that it migt be ok, it depend on what the OP wants.

Ben64

Wow. Good eye. Kudos

Re: search 1D array indices problem

midori2 — Wed, 30 May 2012 02:46:31 GMT

Thanks all of you!!! that's very helpful!!!

Crossrulz .. now i'm trying to follow your solution , i've already subtracted the indices of o's by 1

could you give me the suggest after that about how to add the last column of index to array

Re: search 1D array indices problem

crossrulz — Wed, 30 May 2012 11:53:07 GMT

The simple way is to append a 0 to the array before running through the loop. This way you will see a transition from 1 to 0 (grab the previous index) or 0 to 0 (which you don't care about).