picking maximum difference across entries of row vectors, Part 2

I would like to generalize the previous question ( https://www.mathworks.com/matlabcentral/answers/404240-picking-maximum-difference-across-entries-of-row-vectors )

Now A is a table with the first column has the date data. (not a variale)

Please advise how to proceed in this general case.

for each row, take a difference between every possible pair of two elements in the row and returns the maximum difference

In a simpler case where A is a matrix, if A= [1 3 6 7 10], then B=[9].

[Stephen23]

Original question:

https://www.mathworks.com/matlabcentral/answers/404240-picking-maximum-difference-across-entries-of-row-vectors

Answer 1

[Matt J]

 result = max(A{:,2:end},[],2)-min(A{:,2:end},[],2)

[Süleyman burak Dogaroglu]

Hii

your answer help me soo much but i have to change it a little bit. such as;

p=[max(A(1:end,:),[],2)-min(A(1:end,:),[],2)];
result=p';

and i think you guessed what i need to find. so as a result could you please explain me that how these

max(A(1:end,:),[],2) and min(A(1:end,:) definitions work.

[Matt J]

Explain beyond what is in the documentation for max()?

[Süleyman burak Dogaroglu]

:) I m sorry i just want to understant the sequance behind the equation. also what's the meaning of ':' character and i replace it at the end of the inside bracet but the result didn't change. why could it be?

[Matt J]

Well, if A is a matrix then max(A,[],2) will give the row-wise maximum of A and similarly for min(). So, if your goal is to obtain a row vector containing the differences between the row-wise max and min of a matrix A, then it would simply be,

p = ( max(A,[],2)-min(A,[],2) ).'

You don't need the indexing A(1:end,:) when A is a matrix. My original answer was intended for when A is a table.

[Süleyman burak Dogaroglu]

thanks so much. now it is clear to me.

Answer 2

[Aakash Deep]

Hello,

In order to find out the maximum difference between two elements in a row vector, you can first sort it and then take the difference between the last and the first element.

vec = sort(A);
B = vec(end)-vec(1);

Hope this helps :)

[alpedhuez]

The point is that they are in a table, not in a matrix.

[Stephen23]

@alpedhuez: If you post a follow-up question please put a link to the original question. It helps us to keep track of what information or answers you have already been given, and what extra clarifications or data you have given us about the question.

[alpedhuez]

Thank you for letting me know.