max value every n interval

In this problem, I have a set of data collected every 2sec. This data set divided into several subset that contain 4 data's in each interval.Next, I am interested to find the maximum value in each intervals and the particular time related to that particular max value. For example, in the data set as shown in the figure attached together, I know the maximum value and time associated for each interval was [ 3,6,6,10,23,7] and [ 7,15,23,25,37,41] respectively.

Following this, the syntax below was developed and successfully did what it was intended.

However, this syntax look messy,lengthy and seem not computationally efficient. May I know if there are other faster and compact approach replacing the proposed syntax.

%%%%% This was syntax to solve the above problem.

n=4;
s.Data_to_find  = [1 2 1 3 2 4 5 6 2 3 5 6 10 3 4 2 4 5 23 6 7];
s.time          = 1:2:42;

%% Ensure empty space fill with something

z_data_to_find  = nan (mod (-numel(s.Data_to_find),n),1);
z_time          = nan (mod (-numel(s.time),n),1);

%% New matrix will nan fill empty space

a_Data_to_find  = [s.Data_to_find(:);z_data_to_find];
a_time          = [s.time(:);z_time];

%% Reshape the matrix so that the time and data go from left to right for every interval

b_Data_to_find  =(reshape(a_Data_to_find,n,[]))';
b_time          =(reshape(a_time,n,[]))';

%% Find the maximum value at each interval and output the column coordinate for each M row

[max_each_row_b_Data_to_find,column_cordinate] =  max (b_Data_to_find,[],2);

%%%% This for loop to get the full coordinate of the maximum data value in each of the interval

for i =1:6
      idx = column_cordinate (i,:);
      location (i,:) = [i idx]
end

%% Accessing the time

for i = 1:6
    zzz = location (i,:)
    actualtime(i,:) = b_time ((zzz (1,1)),(zzz (1,2)))
    end

%% The result

actualtime =

     7
    15
    23
    25
    37
    41

Answer 1

[dpb]

Got the right idea, just can be a little more parsimonious in the implementation...

>> [mx,ix]=max(reshape([Data nan(1,mod(-length(Data),n))],n,[]))
mx =
   3     6     6    10    23     7
>> t=reshape([t nan(1,mod(-length(Data),n))],n,[]);  % same "trick" for the time vector
>> tmx=zeros(1,length(mx));                          % preallocate
>> for i=1:length(ix),tmx=t(ix(i),i);end
>> tmx
tmx =
   7    15    23    25    37    41
>>

I'm sure you could use cumprod and accumarray with an offset of n-ix by column and do a direct lookup/accumulation of the time but the loop is "deadahead" and with preallocation will probably be faster anyways...

PS: I like your thinking of using the negative sign with mod to get the right number directly...good job! :)

[balandong]

Thanks dbp for making the code more compact and removing the two for-loops

Following are the complete code base on dpb approach.

n=4;
Data  = [1 2 1 3 2 4 5 6 2 3 5 6 10 3 4 2 4 5 23 6 7];
t          = 1:2:42;

[mx,ix]=max(reshape([Data nan(1,mod(-length(Data),n))],n,[]));

t=reshape([t nan(1,mod(-length(Data),n))],n,[]);
tmx=zeros(1,length(mx));  
for i=1:length(ix),tmx(i)=t(ix(i),i);end