Counting the number of elements in an array based on a set criteria

[wei.jian]

Hi there,

I'm working on a code that would enable me to count the number of elements
that fulfil a particular type of criteria. Here's what I've but it seems
that the array.concat function throws my code off if I've this code embedded
in a loop. I.e. data from other channels are also concatenated. Can someone
suggest an alternate method for me to make this work? In the e.g. below,
I've set values that are equal or above 20 to be identified for counting.

a = newArray(10, 20, 30, 40, 50, 40, 30, 20, 10);

for (i=0; i<a.length; i++) {
                if (a[i]>=20) {
                X = Array.concat(X, a[i]);
                Y = Array.slice(X, 1, X.length);
                }
        }
        Array.show("Results", Y);
        print(Y.length/scale);

Many thanks.
wj




--
Sent from: http://imagej.1557.x6.nabble.com/

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Herbie]

Good day wj,

I'm far from sure what you would like to achieve with your macro code
but this

        X = Array.concat(X, a[i]);

wont work!

(X as an argument is not known to the function. It is not defined.)

Please tell us what _exactly_ you expect as a result and BTW variable
scale is not defined as well.

Regards

Herbie


::::::::::::::::::::::::::::::::::::::
Am 04.09.17 um 04:35 schrieb wei.jian:

> Hi there,
>
> I'm working on a code that would enable me to count the number of elements
> that fulfil a particular type of criteria. Here's what I've but it seems
> that the array.concat function throws my code off if I've this code embedded
> in a loop. I.e. data from other channels are also concatenated. Can someone
> suggest an alternate method for me to make this work? In the e.g. below,
> I've set values that are equal or above 20 to be identified for counting.
>
> a = newArray(10, 20, 30, 40, 50, 40, 30, 20, 10);
>
> for (i=0; i<a.length; i++) {
> if (a[i]>=20) {
> X = Array.concat(X, a[i]);
> Y = Array.slice(X, 1, X.length);
> }
> }
> Array.show("Results", Y);
> print(Y.length/scale);
>
> Many thanks.
> wj
>
>
>
>
> --
> Sent from: http://imagej.1557.x6.nabble.com/
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[wei.jian]

Hi Herbie,

Thanks for your reply. This was written as a function so you are right about
the 'scale' in the code. Below is the updated one. X as an argument serves
as a filler for me to pick out values that are equal or greater than 20. In
this eg. X is (0, 20, 30, 40, 50, 40, 30, 20). Next, I used the Array.slice
function to remove the filler. Thus Y gives (20, 30, 40, 50, 40, 30, 20) and
Y.length gives 7 as there are 7 elements based on my criteria.

a = newArray(10, 20, 30, 40, 50, 40, 30, 20, 10);
 
for (i=0; i<a.length; i++) {
 if (a[i]>=20) {
 X = Array.concat(X, a[i]);
 Y = Array.slice(X, 1, X.length);
 }
 }
 Array.show("Results", Y);
 print(Y.length);

I understand that my approach in isolating elements is clunky so are there
alternatives that I can use to pick out elements based on my criteria?

Thanks.
wj



--
Sent from: http://imagej.1557.x6.nabble.com/

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html

[Herbie]

Well wj,

perhaps this is what you are looking for:

///////////////////////////////////////////////////
a = newArray( 10, 20, 30, 40, 50, 40, 30, 20, 10 );
Y = newArray(a.length);
j = 0;
for (i=0; i<a.length; i++) {
        if  (a[i]>=20 ) {
                Y[j] = a[i];
                j++;
        }
}
Y = Array.trim( Y,  j );
Array.show( "Results", Y);
///////////////////////////////////////////////////

Regards

Herbie

:::::::::::::::::::::::::::::::::::::
Am 04.09.17 um 16:57 schrieb wei.jian:

> Hi Herbie,
>
> Thanks for your reply. This was written as a function so you are right about
> the 'scale' in the code. Below is the updated one. X as an argument serves
> as a filler for me to pick out values that are equal or greater than 20. In
> this eg. X is (0, 20, 30, 40, 50, 40, 30, 20). Next, I used the Array.slice
> function to remove the filler. Thus Y gives (20, 30, 40, 50, 40, 30, 20) and
> Y.length gives 7 as there are 7 elements based on my criteria.
>
> a = newArray(10, 20, 30, 40, 50, 40, 30, 20, 10);
>  
> for (i=0; i<a.length; i++) {
>   if (a[i]>=20) {
>   X = Array.concat(X, a[i]);
>   Y = Array.slice(X, 1, X.length);
>   }
>   }
>   Array.show("Results", Y);
>   print(Y.length);
>
> I understand that my approach in isolating elements is clunky so are there
> alternatives that I can use to pick out elements based on my criteria?
>
> Thanks.
> wj
>
>
>
> --
> Sent from: http://imagej.1557.x6.nabble.com/
>
> --
> ImageJ mailing list: http://imagej.nih.gov/ij/list.html
>

--
ImageJ mailing list: http://imagej.nih.gov/ij/list.html