How to sort a matrix's columns by using indexes

I wrote some codes to sort an array (a) descendingly. I want to use the indexes (indA) to sort the data matrix's columns.

close all;
clc
load lung.mat
data=lung; 
[n,m]=size(data);
l=1;
t=1;
data1=[];
data2=[];
for i=1:n
    if data(i,m)==1
        data1(l,:)=data(i,:);
        l=l+1;
    else
        data2(t,:)=data(i,:); 
        t=t+1;
    end
end
if t>l
    data1(l:t-1,:)=0;
else
    data2(t:l-1,:)=0;
end     
for i=1: m                       
   thisCol1=data1(:,i);      
   thisCol2=data2(:,i);
   a(i)=fHammingDist(thisCol1,thisCol2);
end
[A,indA]=sort(a,'descend'); 

I'll be very greatfull to have your opinions how to sort a matrix's columns by using an array of indexes.

Thank you

[Bruno Luong]

Are you sure to compute the distance of the last column which seems to contain special values

... 
if data(i,m)==1
  ...
end
...
for i=1: m                       
    ...
   a(i)=fHammingDist(thisCol1,thisCol2);
end

Answer 1

[Bruno Luong]

Short answer

data = data(:,indA);

Long answer

load lung.mat
data=lung; 

b = data(:,end) == 1;
data1 = data(b,:);
data2 = data(~b,:);
n1 = size(data1,1);
n2 = size(data2,1);
if n1 < n2
    data1(n2,1) = 0;
elseif n2 < n1
    data2(n1,1) = 0;
end
a = arrayfun(@(j) fHammingDist(data1(:,j),data2(:,j)), 1:size(data,2));
[A,indA] = sort(a,'descend'); 
data1 = data1(:,indA);
data2 = data2(:,indA);
data = data(:,indA);

Answer 2

[dpb]

See

doc sortrows

Not clear which array it is you wish sorted but mayhaps can accomplish directly -- altho the auxiliary array and distances may be needed to have been computed, not knowing exactly the problem trying to solve.

But, given the index vector you've built, simply

A=A(indA,:);

if we presume this mystery array is 'A' will rearrange rows in that order.

ERRATUM:

As Bruno points out, I mixed metaphors/switched horses midstream...

A=A(:,indA);

to sort the columns instead, of course.

[Bruno Luong]

Why do you rearrange rows with sorted column indexes?

[dpb]

Why? 'Cuz I didn't read the Q? carefully enough... :)

Good catch, Bruno!