how to loop in one time to get the max value minus the min value.

I have data in the workspace in 2 columns

year/ month / date hour minute seconds data

(1856x2)cell

output1 =

2015 12 19 02 40 0 ​​ 0

2015 12 19 02 50 0 ​​ 0

2015 12 19 03 00 0 ​​ 0

2015 12 19 03 10 0​​ 2.2

2015 12 19 03 20 0 ​​ 0

2015 12 19 03 30 0 ​​ 4.6

2015 12 19 03 40 0 ​​ ​​ 4.6

2015 12 19 03 50 0 ​​ 4.6

2015 12 19 04 00 0 ​​ 4.6

2015 12 19 04 10 0 ​​ 4.6

2015 12 19 04 20 0 ​​ 4.6

2015 12 19 04 30 0 ​​ 4.6

2015 12 19 04 40 0 ​​ ​​ 4.6

2015 12 19 04 50 0 ​​ 4.6

2015 12 19 05 00 0 ​​ 4.6

2015 12 19 05 10 0 ​​ 4.6

2015 12 19 05 20 0 ​​ 4.6

2015 12 19 05 30 0 ​​ 4.6

2015 12 19 05 40 0 ​​ ​​ 4.6

2015 12 19 05 50 0 ​​ 3

......

I want to reduce data in 1 hour (50 minutes - 00 minutes)

the last minute data was reduced by the initial minute in one hour

example:

2015 12 19 03 50 0 4.6

-

2015 12 19 03 00 0 0

obtained value =

2015 12 19 03 00 0 4.6

So that i get 1 data in one hour.

if i get a value (-) minus is considered zero

and I get data in one hour on each date in one year.

I tried the following data script and I felt wrong in making the program.

T = output1(:,1);
D = datenum(T,'dd-mm-yy HH:MM');
Days = day(D);
Months = month(D);
Hours = hour(D);
Minutes = minute(D);
C = output1(:,2);
M = cell2mat(C);
 max_minutes = zeros(23,2); min_days = zeros(23,2);
for i = 0:23
    if nonzeros(M(Days==i))
    [max_minutes(i,1) max_minutes(i,2)] = max(M(Days==i));
    Hours_temp = Hours(Days==i);
    Minutes_temp = Minutes(Days==i);
    max_hour(i) = Hours_temp(max_minutes(i,2));
    max_minute(i) = Minutes_temp(max_minutes(i,2));
    [min_days(i,1) min_days(i,2)] = min(M(Days==i));
    min_hour(i) = Hours_temp(min_days(i,2));
    min_minute(i) = Minutes_temp(min_days(i,2));
    else
        max_minutes(i,1:2) = NaN;
        min_days(i,1:2) = NaN;
   end
end

I thank everyone who helped me in resolving my difficulties.

I use matlab 2014a, which uses the old function so there are some programs whose functions are not readable in my matrix

Answer 1

[Andrei Bobrov]

[a,~,c] = unique(cdata{1}(:,1:4),'rows','stable');
out = [a,accumarray(c,cdata{2},[],@(x)max(x) - min(x))];

[Jan Risn]

oke thank you for the advice.. but input

out = [a, accumarray (c, cdata {2}, [ ], @ (x) max (x) - min (x))];

result

Second input VAL must be a one element vector for each row in SUBS, or a scalar.

Please help, thank you sir

[Andrei Bobrov]

Attach small part of your data as mat-file and read MATLAB help about accumarray.

[Jan Risn]

ok sir, I'm attaching the this mat.file ...

[Andrei Bobrov]

load('sample.mat')
time1 = datevec(cdata(:,1));
[a,~,c] = unique(time1(:,1:4),'rows','stable');
out = [cellstr(datestr([a,zeros(size(a,1),2)])),...
         num2cell(accumarray(c,cat(1,cdata{:,2}),[],@(x)max(x) - min(x)))];

[Andrei Bobrov]

last variant:

load('sample.mat')
time1 = datevec(cdata(:,1));
d = cat(1,cdata{:,2});
[a,~,c] = unique(time1(:,1:4),'rows','stable');
out = [cellstr(datestr([a,zeros(size(a,1),2)])),...
         num2cell(accumarray(c,(1:numel(d))',[],@(x)innerfun(d,x)))];

function out = innerfun(d,x)
    k = d(sort(x));
    out = max(0,k(end) - k(1));
end

[Jan Risn]

thank you very much sir, this is very helpful ..