max and min values in an array

thank you! Found the solution! Thank you!

3 Comments

findpeaks() ?
So as I understand you want to find the max value for each 16 rows ???
NO NO NO!!!!,Why did you delete all the contents of the question and the comments?, it's a terrible thing to do . Others may also benefit from the question.

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 Accepted Answer

madhan ravi
madhan ravi on 7 Apr 2019
Edited: madhan ravi on 7 Apr 2019
See if this does what you want , first we split into 16 separate rows each and then we conquer in the third dimension:
[m,n]=size(A); % where A is your matrix of size 256 X 40K
parts = 16;
AA = permute(reshape(A.',n,m/parts,[]),[2,1,3]);
Max = max(AA,[],[1,2]); % max(max(AA)) for versions <= 2016b
[r,~]=find(AA == Max) % r represents rows

6 Comments

Anh Dao
Anh Dao on 7 Apr 2019
Edited: Anh Dao on 7 Apr 2019
r,c seems to be wrong, I tried, and there seems to be problem with the split , the row appeared to be all 1, can you please check?
Could you illustrate with a short example? c gives you the linear index unlike r, what do you mean the split is wrong?
>> A = rand(8,2)
[m,n]=size(A);
parts = 2;
AA = permute(reshape(A.',n,m/parts,[]),[2,1,3]);
Max = max(AA,[],[1,2]);
[r,~]=find(AA == Max)
A =
0.6944 0.9400
0.3253 0.6584
0.4368 0.9652
0.7557 0.0311
0.4930 0.1461
0.7457 0.5300
0.8067 0.8428
0.9376 0.0922
r =
3
4
>>
I've got this error Product of known dimensions, 40000, not divisible into total number of elements, 15.
256 is divisible by 16 where did you get 15 and why are you bothered with columns now?? We are focused on rows ?
[r,c,p]=ind2sub(size(AA),find(AA==Max)) % r represents rows , c represents columns & p represents pages
Post the code that your trying.
Thank you, so I have a matrix A equal 256x40000
it was a mistake that I put in last time that i put parts = 15, sorry about that, so r should be correct right, what's the difference in the code you posted?
[m,n]=size(A)
parts = 16;
AA = permute(reshape(DoM.',n,m/parts,[]),[2,1,3]);
Max = max(AA,[],[1,2]);
[r,~]=find(AA == Max)
replace
[r,~]=find(AA == Max)
with
[r,c,p]=ind2sub(size(AA),find(AA==Max)) % r represents rows , c represents columns & p represents pages

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