Comments on: Find Largest Positive Value Next to Zero

By: OysterEngineer

OysterEngineer — Tue, 27 Oct 2009 23:30:18 +0000

Loren,

Your solution is clear to follow. i.e., the code demonstrates what it is doing.

The solutions that are shown from Steve, Doug & Jos are not clear. Certainly it can be fun to study them. But, I would never use them in real code.

I really think that Djykstra was correct in saying that the code should demonstrate that it is correct.

By: Akiel

Akiel — Sun, 19 Jul 2009 19:04:41 +0000

How about this:
max(A(logical(conv(~A+0, [1 0 1], ’same’))))
I haven’t looked at any of the other solutions, so maybe someone’s already done this.

By: Brett Shoelson

Brett Shoelson — Fri, 17 Jul 2009 19:24:36 +0000

Hi Loren,
Late to the game–have been on vacation–but wanted to throw my solution into the mix. Cheers!
Brett

First, pad B, then use |BSXFUN|:

C = [-Inf B -Inf];
max(max(C(bsxfun(@plus,find(~C),[-1,1]'))))

By: Thijs

Thijs — Mon, 06 Jul 2009 07:14:02 +0000

My option would be the following:

max(A([false A(1:end-1)==0]|[A(2:end)==0 false]))

or the identical, but slightly more compact:

max(A([0 A(1:end-1)==0]|[A(2:end)==0 0]))

pretty similar to what already has been posted though.

By: Loren

Loren — Fri, 03 Jul 2009 12:04:04 +0000

Jos-

You’re right about the title!

Ben-

Thanks for another solution.

–Loren

By: Jos

Jos — Fri, 03 Jul 2009 09:31:17 +0000

Again an interesting blog! However, the title of this blog is somewhat misleading. Shouldn’t the word “positive” be removed?

- jos

By: Ben

Ben — Fri, 03 Jul 2009 07:48:45 +0000

I think Mike’s answer is probably the most intuitive, but I would have answered it like Jos only using “|” instead of “any”.

fb = @(X)max(X([1,X(1:end-1)]==0 | [X(2:end),1]==0))

An even shorter code is

fb2 = @(X)max(X([1,X(1:end-1)].*[X(2:end),1]==0))

By: Pär Lansåker

Pär Lansåker — Thu, 02 Jul 2009 18:53:12 +0000

It is often better to use an obvious path for understanding. I would use a combinationd of diff(A) (find the largest, not forgetting the negative left hand side) and A==0 (adjecent to 0).

By: Mike Koets

Mike Koets — Thu, 02 Jul 2009 16:23:41 +0000

If we do want to allow answers of zero (I assumed we did not want that), then we can use this code

max(c(intersect(1:length(c),[find(c==0)+1 find(c==0)-1])))

Now the intersect just removes out of bounds results.

By: Loren

Loren — Thu, 02 Jul 2009 15:06:01 +0000

Matt-

Your solution is certainly viable as well!

–Loren