How to find out if a curve passes through zero and goes from positive to negative value?

Hi.I have divided this curve:

into 8 region where the points of 1st 1/8 portion is in x=1,2nd portion in x=2 etc like this:

Now I want to find out if there is a code by which for x=1,2,3,4,5,6,7,8 if there is zero crossing and points changing from positive to negative value.Hope you understand.Thank you in advance.

My code is this:

j=diff(I);

figure

plot(j);

n = 8 ;

d=j';

a=length(j);

% check whether data needed to be appended to reshape into n

b = n * ceil(a / n)-a ;

%

d = [d NaN(1,b)] ;

iwant = reshape(d,[],n);

% plot

figure

plot(iwant','b*')

[KSSV]

find the roots....

[joynob ahmed]

fzero doesn't work here showing the error:

''FUN must be a function, a valid character vector expression, or an inline function object''

[joynob ahmed]

If I write this for the image

x0=[0 1];

if fzero(iwant',x0)==true

B1=1

else

B1=0

end

Same error is shown. Can't I do this for x=1,2,3.... in seperate code in the same way? Then my code will be easier for me.

Answer 1

[Image Analyst]

A single value cannot "change" sign or "cross the axis". Chances are that none of the y values where x = 0, 2, 4, ... are EXACTLY zero. So you need to look at x and find which ones are above the axis and which are below. You can then determine where the value went from - to + or vice versa.

y = rand(1, 20) - 0.5
s = sign(y)
% Find first index of the two where the crossing occurs.
positiveGoingIndexes = strfind(s, [-1, 1])
negativeGoingIndexes = strfind(s, [1, -1])

If you need more help, please attach "I" (poor name by the way) in a mat file with the paper clip icon.

[joynob ahmed]

there is an error: strfind works with one row.

I am attaching my file with the image.

[Image Analyst]

You gave no context for your question. As far as I knew you only had one row (vector). You didn't even explain what that curve was. It would help if you did a few simple things to improve your code writing with these three simple concepts (and this is what I tell everybody who writes code for me):

1. Put in lots of comments (yours had hardly any)
2. Use descriptive variable names. Yours had names like I,i,j,x,n,a,b,d. Non-descriptive, single letter variable names makes is harder for anyone to follow your code and know what it's doing.
3. Label your graphs. When I ran it, six figures came up all on top of one another and none of them was labeled and I don't know what they were. So I needed to go back to the code and that's where I encountered a confusing alphabet soup of non-descriptive variable names.

Again, I tell this to everybody, not just you, so don't feel like I'm picking on you. I'm just trying to make it easier for me to help you. If I get time later today, I may try to fix your code but if you do it first, that would be great. Then I can try to figure out what you actually want or need. Is this one of those ABCDE melanoma situations?

[joynob ahmed]

I am so sorry. I understood. I will edit and upload it again.

Yes,it is about melanoma detection.Here I am trying to detect border irregularity.

[joynob ahmed]

I have edited it. Please take a look.

As I have to count its value for detecting border irregularity,I have divided the curve into 8 region. Here each region will have value 1 if there is a zero crossing that means positive to negative going points.Otherwise the value will be zero. So the final value will be from 0 to 8.

This is my last part of code.I need your help badly as I am not so good in coding.

[Image Analyst]

None of the plots have any labels on them. And though you added a few comments the code still looks like alphabet soup. And I just see the image divided up into 4 sectors, not 8 sectors.

Let's try something else. Why don't you just explain in words what you want to measure for each of the 4 triangular zones?

[joynob ahmed]

Actually I have worked according to a paper.So it's a mess I know as I didn't understand that part of 4 sectors.

But as I required 8 regions,I have divided the border curve into 8 region. and in each region I want to find the irregularity which is defined by local maxima in main function (zero crossing from positive to negative in derivative).

Now as the points are divided into 8 parts,they are discontinuous (image in question). So I couldn't find the zero crossing or curve going from positive to negative value.I need to find out if there is minimum one positive to negative going points.then value 1 will be assigned to that region otherwise 0.Then by adding the value of all 8 region,result will be any value from 0-8

[Image Analyst]

I gave you the code in my answer. It gives you the indexes right before the line crosses the axis.

[joynob ahmed]

But it didn't work.I have showed you the error.

[Image Analyst]

Where did you show that error? I looked above and am not seeing it.

And I ran your latest attached code and it ran without error, though it did not have strfind() in it anywhere so it doesn't look like you took my suggestion.

And I still think looking for zero crossings of the derivative is not needed.

[joynob ahmed]

I have used it but it didn't give correct result. I have wrote this:

negativeGoingIndexes = strfind(region3, [1, -1])

where the image is :

But the result is:

negativeGoingIndexes =

[]

whereas you can see there are many negative going points.