# Get Smart, Be Logical!1

Posted by Loren Shure,

Recently a customer asked me an interesting question about a good way to compute something. Before I sat down to tackle it, I happened to mention the problem to my pal, Sean. And he quickly answered, "max logical". It took me just a moment to realize what he was saying, and it is the answer to the problem. But when I heard it, I heard a person's name "Max Logical", and though I didn't know who that was, I immediately thought of Maxwell Smart, from the 60s tv comedy in the US, called "Get Smart". By General Artists Corporation-GAC-management. [Public domain], via Wikimedia Commons And you can see the precursor to today's mobile phones!

### Contents

#### The Problem

So what's the problem? The customer posed it by way of an example. Given a 3-D array A, generated for this example from a matrix, find the first plane in which some condition is true.
A=[1 2; 3 4]
M = cat(3, A, A+1, A+2, A+3);

A =
1     2
3     4

So the function should work like this: IDX = someFunction(M >=5);
IDX = [NaN 4; 3 2]

IDX =
NaN     4
3     2

And, the customer wanted this to work for arrays of higher dimensions as well.

#### The Solution

As I said, I mentioned this to Sean. But I had thought for about 2 seconds about it before and I realized I could do a find on the entire array, as if it were a vector, get the locations and sort them into their proper locations and find the first one in the correct last dimension. And then Sean said "Max Logical". Here's what a great solution looks like.
catdim = 3;
szA = size(A);
[mv,midx] = max(M >= 5,[],catdim);
IDX = NaN(szA);
IDX(mv) = midx(mv)

IDX =
NaN     4
3     2

Why/how did this work? Well, I use the condition I want met into function max, and it is logical input, meaning only 0s and 1s. And max returns the index where the first condition is met, and I do this along the final dimension. Then create an array of NaN values, and use logical indexing to update this index matrix. Voila!

#### Higher Dimensions

Now let's see if it makes sense in higher dimensions. For simplicity, I again will provide a base array in 1 dimension less than I am ultimately looking, so I can construct something for which I can reasonably easily check the result.
A = randi(17, [3 2 1 3]);
catdim = ndims(A)+1
M = cat(catdim, A, A+1, A+2, A+3, A+4);

catdim =
5

See a slice of M
sliceOfM = M(:,:,:,1)

sliceOfM =
12     2
13     4
8    16

Apply the max, logical indexing to M next.
szA = size(A)
[mv,midx] = max(M >= 13,[],catdim);
IDX = NaN(szA);
IDX(mv) = midx(mv)

szA =
3     2     1     3
IDX(:,:,1,1) =
2   NaN
1   NaN
NaN     1
IDX(:,:,1,2) =
NaN     1
1   NaN
4   NaN
IDX(:,:,1,3) =
NaN     1
1     1
NaN     1


#### Have You Embraced Logical Indexing?

I have made several posts over the years on indexing that readers have found helpful. Do you have any interesting logical stories to share? Post them here.

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Published with MATLAB® R2017b
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