interpolation of data takes minutes to operate

I have a data of 5068x59 and I want to interpolate each column with respect to column number 4. I have used the below code and its working fine. However, it takes sometime to operate. Therefore, is there any easier way to perform the below tasks

T=readmatrix('1.xlsx');
x = T(:,4) ; 

for b=1:width(T)
    
    for bb=1: length(x)
        
        T1(bb,b)=T(bb,b)+ bb*1E-11; %to make all values unique
    
    end

end

distanceQ = 1:max(T1(:,4)); % new distance vector with step size of 1m

for cc=1:width(T1)
speedQ = interp1(T1(:,4),T1(:,cc) , distanceQ);
filename = 'x)Data.xlsx';
writematrix(speedQ,filename,'WriteMode','append')
end

[KSSV]

What for is these lines?

for b=1:width(T)    
    for bb=1: length(x)        
        T1(bb,b)=T(bb,b)+ bb*1E-11; %to make all values unique  
    end
end

[Abdulkarim Almukdad]

I have repetitive numbers in my data for the same column like (1,1,1,1,1,1,2,3,4,5,5,etc) so I'm doing this loop to add very small values to each cell in the table. if I skip this step an error will occur stating that values must be unique.

[Mathieu NOE]

hi

First, you can avoid the first two nested loops . This is fairly simple to do (dummy example) :

% dummy data
T = [1;1;1;1;1;1;2;3;4;5;5]*[1 2 3 4]


%%%%% main code %%%%%%%%%%
[m,n]  = size(T);

offset = 1e-4*((1:m)'*ones(1,n));

T1 = T+offset

 T =
     1     2     3     4
     1     2     3     4
     1     2     3     4
     1     2     3     4
     1     2     3     4
     1     2     3     4
     2     4     6     8
     3     6     9    12
     4     8    12    16
     5    10    15    20
     5    10    15    20
T1 =
    1.0001    2.0001    3.0001    4.0001
    1.0002    2.0002    3.0002    4.0002
    1.0003    2.0003    3.0003    4.0003
    1.0004    2.0004    3.0004    4.0004
    1.0005    2.0005    3.0005    4.0005
    1.0006    2.0006    3.0006    4.0006
    2.0007    4.0007    6.0007    8.0007
    3.0008    6.0008    9.0008   12.0008
    4.0009    8.0009   12.0009   16.0009
    5.0010   10.0010   15.0010   20.0010
    5.0011   10.0011   15.0011   20.0011

Next , I am not sure if the "append" method to save your data in the excel file is taking long.

I would prefer to store all results first in an array or cell and then save it once.

is "speedQ" a vector of constant lenght ? if yes, there might be a faster solution.

You can use the profiler to see the bottlenecks in your code.

[Abdulkarim Almukdad]

Thanks a lot, that was helpfull. firstly I have changed the loop to the offest function and it works great, as for your what takes most of the time is indeed the writefile function so I have followed what you mentioned to store all the data in 1 variable then to save that variable once as shown below. as for your question, my code is intending to read several files and not all of the files have the same length of data but the difference is just between 1-5 columns. I just have 1 simple questions and I will be very thankfull if you can help me with it. how can I make the result from the interpolation appear in 1 column rather than in 1 row. without using additional transpose functions if possible.

T=readmatrix('1.xlsx');
x = T(:,4) ; 

[m,n]  = size(T);
offset = 1e-11*((1:m)'*ones(1,n));
T2 = T+offset;

distanceQ = 1:max(T2(:,4)); % new distance vector with step size of 1m
for cc=1:width(T2)
speedQ(cc,:)= interp1(T2(:,4),T2(:,cc) , distanceQ);

end

filename = 'xxx111)Data.xlsx';
writematrix(speedQ,filename,'WriteMode','append')

[Mathieu NOE]

hi

glad it helped you

I believe the ouput of the interp1 function is a row vector because distanceQ must also be a row vector

you must change the orientation of distanceQ , by putting distanceQ(:) instead of distanceQ in the interp1 function call

so this should give the correct orientation for speedQ:

speedQ(:,cc)= interp1(T2(:,4),T2(:,cc) , distanceQ(:));