How to read values for the range between each maximum and each minimum from a large uneven data file?

Hi,

I have a data file with 11000 rows and 4 columns. I need to read all values (in all columns) for data between each maximum and each minimum of column 3. Integrate column 3 and column 4, and calculate the mean of column 2, for each range between the maximum and minimum of column 3. Do I make sense?? I am really unsure about how to start. I am using Matlab2015a. I have attached what I want to do as image. For rudimentary case I can do like shown on the right in figure. But it would take a decade to solve. Need help.

[KSSV]

You want to do this by splitting at every sixth row?

[Madan Kumar]

No, the data is uneven. some times it's 5 rows, sometimes 6 rows and so on..

Answer 1

[per isakson]

"range between the maximum and minimum of column 3" with a bit of guessing my interpretation is implemented in ixb and ixe

%%
fid = fopen('axn.txt','r');
cac = textscan( fid, '%f%f%f%f', 'CollectOutput',true );
num = cac{1};
[~] = fclose( fid );
%%
ixb = [ 1; find(diff(num(:,3))>0)+1 ];  % beginning of section
ixe = [ ixb(2:end)+1; size(num,1) ];    % end of section
%%
len = length( ixb );
result = nan( len, 2 ); 
for jj = 1 : len
    result(jj,1) = mean(num(ixb(jj):ixe(jj),2));
    result(jj,2) = trapz(num(ixb(jj):ixe(jj),3),num(ixb(jj):ixe(jj),4));
end

outputs

>> result
result =
       51.494      -10.433
       51.472      -23.332
       51.479     -0.95127
       51.463      -29.865
>> 

[Madan Kumar]

Thank you. But it doesnot give the correct result. I tried checking manually. I think it might be due to the loop jj=1:len. This loop should run from maximum to minimum of column 3, .i.e. from row 1 to 5, and do the calculation. Next run should be from next maximum to next minimum of column 3, i.e. from row 6 to row 11, ...and so on.

[per isakson]

Is this better?

>> result
result =
       51.492      -24.937
       51.463      -29.865
       51.472      -24.485
       51.463      -29.865
>> 

My mistake. Replace

ixe = [ ixb(2:end)+1; size(num,1) ];

by

ixe = [ ixb(2:end)-1; size(num,1) ];

with a minus one, not plus.

Now the beginings and ends of the "sections" agree with my intention

>> ixb', ixe'
ans =
     1     6    12    18
ans =
     5    11    17    23
>> 

[Madan Kumar]

Thank you so much. It works.