Only required the mean value out rather then every single value in a data

Hi,

I am taking the mean of every 7 rows in each column unitl the end of the column. I have used the code below but the problem is whenever I am reshaping the data I am getting everything. what I want is only the mean values saved in a different folder rather then mean value along with the values that used for mean. I used the mod function to add the value at the end of 7 values I think that's why its triggering whole data together while I only want to see the mean data. Table is 1639 by 182 after calculating the mean I am getting 1880 by 182. I want only the mean rows which is around 241 in this case. File is attached

Code:

close all; clear all; clc;

% Read the table

t = readtable('Required_Average.xlsx', ...

'NumHeaderLines',1, ...

'PreserveVariableNames',true);

data = table2array(t);

% Making sure total number of rows filled with nan if not the multiple of 7

% in case the total number of rows is not a multiple of 7,

% add rows of NaNs to fill it out:

[N,M] = size(data);

n = mod(N,7);

if n ~= 0

data = [data; NaN(7-n,M)];

N = N+7-n;

end

% reshape data to have size 7 in the third dimension

data = permute(reshape(data.',[M 7 N/7]),[3 1 2]);

size(data)

% calculate the mean in the third dimension

% and append it to the end of data

% (using 'omitnan' to prevent NaNs from affecting the mean)

data(:,:,end+1) = mean(data,3,'omitnan');

%Reshape the data

data = reshape(permute(data,[3 1 2]),[],M)

Answer 1

[William Rose]

@muhammad choudhry,

I assume your desired output has N rows by 182 columns, where N=floor(1639/7)=234.

num=xlsread('Required_Average.xlsx');
[r,c]=size(num);
N=floor(r/7);
a=zeros(N,c);  %allocate array for the output
%row 1 of a() is mean of rows 1-7 of num()
%row 2 of a() is mean of rows 8-14 of num(), etc.
for i=1:N
    a(i,:)=mean(num(7*i-6:7*i,:));
end

Try it. Good luck.

[William Rose]

@muhammad choudhry, in the script I gave, array a() contains the result which is what I think you want. Array a() has 234 rows and 182 columns.

[muhammad choudhry]

short and precise! thanks

[William Rose]

@muhammad choudhry, you're welcome.

Answer 2

[Peter Perkins]

"7" sounds rather suspiciously like "weeks". If this were a timetable, it would be a one-liner with groupsummary. Moral: don't throw away information.

Dunno what the dates would be, but like this:

>> t = readtable("Required_Average.xlsx"); t = t(1:10,1:3)
t =
  10×3 table
     RZ1      RZ2        RZ3    
    _____    _____    __________
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
    12543    43424    3.2454e+05
>> tt = table2timetable(t,StartTime=datetime(2020,1,1),TimeStep=caldays(1))
tt =
  10×3 timetable
       Time         RZ1      RZ2        RZ3    
    ___________    _____    _____    __________
    01-Jan-2020    12543    43424    3.2454e+05
    02-Jan-2020    12543    43424    3.2454e+05
    03-Jan-2020    12543    43424    3.2454e+05
    04-Jan-2020    12543    43424    3.2454e+05
    05-Jan-2020    12543    43424    3.2454e+05
    06-Jan-2020    12543    43424    3.2454e+05
    07-Jan-2020    12543    43424    3.2454e+05
    08-Jan-2020    12543    43424    3.2454e+05
    09-Jan-2020    12543    43424    3.2454e+05
    10-Jan-2020    12543    43424    3.2454e+05
>> groupsummary(tt,"Time","week","mean")
ans =
  2×5 table
            week_Time             GroupCount    mean_RZ1    mean_RZ2     mean_RZ3 
    __________________________    __________    ________    ________    __________
    [29-Dec-2019, 05-Jan-2020)        4          12543       43424      3.2454e+05
    [05-Jan-2020, 12-Jan-2020)        6          12543       43424      3.2454e+05

[William Rose]

@Peter Perkins, nice!