How can i randomly divide a dataset(matrix) into k parts ??

I have a database and i want to randomly divide it into ka parts of equal size . if the database has n row each part will contain n/k randomly chosen row from the dataset .

Answer 1

[Mariem Harmassi]

function [idxo prtA]=randDivide(M,K) 

[n,m]=size(M);
np=(n-rem(n,K))/K;
B=M;
[c,idx]=sort(rand(n,1));
C=M(idx,:);
i=1;
j=1;
ptrA={};
idxo={};
n-mod(n,K)
while i<n-mod(n,K)
    prtA{j}=C(i:i+np-1,:);
    idxo{i}=idx(i:i+np-1,1);
    i=i+np;
    j=j+1;

end
prtA{j}=C(i:n,:);

end

this a my algo it works very well think u for ur answers

Answer 2

[Oleg Komarov]

Suppose you have the N by M matrix A. I would randomly permute positions from 1:N and then group them into k partitions. Follows the code.

% Sample inputs

N     = 100;
A     = rand(N,2);

% Number of partitions

k     = 6;

% Scatter row positions

pos   = randperm(N);

% Bin the positions into k partitions

edges = round(linspace(1,N+1,k+1));

Now you can "physically" partition A, or apply your code to the segments of without actually separating into blocks.

% Partition A

prtA  = cell(k,1);
for ii = 1:k
    idx      = edges(ii):edges(ii+1)-1;
    prtA{ii} = A(pos(idx),:); % or apply code to the selection of A
end

EDIT

You can also avoid the loop, but in that case you have to build a group index that points the row to which partition it belongs and then apply accumarray() to execute your code on the partitions.

[Mariem Harmassi]

i tried ur code and

N     = 20;
A     = rand(N,4);
% Number of partitions

k     = 3;
% Scatter row positions

pos   = randperm(N);
% Bin the positions into k partitions

edges = round(linspace(1,N+1,k+1));

% Partition A

prtA  = cell(k,1);
for ii = 1:k
    idx      = edges(ii):edges(ii+1)-1;
    prtA{ii} = A(pos(idx),:); % or apply code to the selection of A
end

>> prtA{1}

ans =  

      0.8449    0.0473    0.8380    0.9653
      0.8271    0.1157    0.2192    0.7255
      0.3530    0.7918    0.0578    0.7980
      0.3644    0.4857    0.3543    0.1227
      0.4233    0.5737    0.5924    0.1201
      0.6064    0.3261    0.5511    0.9597
      0.2981    0.2081    0.4948    0.5097

>> prtA{2}

ans =

      0.7283    0.3060    0.6703    0.6065
      0.5678    0.5613    0.3997    0.6631
      0.9188    0.7568    0.8532    0.2698
      0.0012    0.8892    0.3832    0.4133
      0.2181    0.3098    0.6159    0.0196
      0.1474    0.3711    0.1479    0.9500

>> prtA{3}

ans =

      0.8401    0.1154    0.2291    0.4997
      0.0225    0.8911    0.2773    0.6086
      0.6951    0.6617    0.1875    0.4398
      0.4731    0.3774    0.0707    0.7327
      0.3307    0.4178    0.1733    0.9329
      0.3416    0.0135    0.3195    0.2772
      0.3953    0.6725    0.6642    0.1881
  prtA = 

      [7x4 double]
      [6x4 double]
      [7x4 double]

how can i do to get the smaller partition at the end

[Oleg Komarov]

Why does it matter?

Anyways, after the loop:

% Index smaller as last
[~,idx] = sort(diff(edges),'descend');
prtA    = prtA(idx);

[Mariem Harmassi]

Yes i tried it but prtA =

    [17x2 double]
    [17x2 double]
    [17x2 double]
    [17x2 double]
    [16x2 double]
    [16x2 double]
and what i want is all the result that i expect is
prtA = 

    [17x2 double]
    [17x2 double]
    [17x2 double]
    [17x2 double]
    [17x2 double]
    [15x2 double]
all the partition with the same size and the rest in the last partition 
HOw to do that

[Oleg Komarov]

Adapting to your requests, I build edges in a slightly different way then:

% Sample inputs scrambling
N     = 100;
A     = rand(N,2);
k     = 6;
pos   = randperm(N);

% Edges
edges = 1:round(N/k):N+1;
if numel(edges) < k+1
    edges = [edges N+1];
end

% partition
prtA  = cell(k,1);
for ii = 1:k
    idx      = edges(ii):edges(ii+1)-1;
    prtA{ii} = A(pos(idx),:); 
end

Answer 3

[Azzi Abdelmalek]

A=rand(210,4);[n,m]=size(A);
np=20;B=A;
[c,idx]=sort(rand(n,1));
C=A(idx,:);
idnan=mod(np-rem(n,np),np)
C=[C ;nan(idnan,m)];
[n,m]=size(C);
for k=1:n/np
  ind=(k-1)*np+1:k*np
res(:,:,k)=C(ind,:)
end
idxo=reshape([idx ;nan(idnan,1)],np,1,n/np)  % your original index

[Mariem Harmassi]

hi think for the answer , how can i save the original index ??

[Azzi Abdelmalek]

the original index is idxo

check the updated code

[Mariem Harmassi]

it works well , but i dont understand the role of each line A=rand(200010,4); %create the matrix [n,m]=size(A);%get the matrix size np=20;%define the submatrix size B=A;%just keep the original matrix [c,idx]=sort(rand(n,1)); %sort the matrix randomly /or sort a randomly genratematrix of n row and A column C=A(idx,:); %the matrix A randomly sorted idnan=mod(np-rem(n,np),np) %whats the role of thsi component ? do u add to the sorted matrix some rows in order to prevent class with less than np C=[C ;nan(idnan,m)]; %adjust the C with adding NaN rows just to keep the final submatrix with %np rows [n,m]=size(C);%get the new size res=reshape(C,[np,m,n/np]) % your submatrix : n/np matrix with np rows m columns idxo=reshape([idx ;nan(idnan,1)],np,1,n/np) % your original index Is it raght ?

[Azzi Abdelmalek]

 idnan=mod(np-rem(n,np),np)

% if you have 205 data and np=20, we have to complete with 15 nan value, will be ignored, it help us to reshape 220/20 like you said

[Mariem Harmassi]

 i tried the code , but i found the nan in every partition and in every column  how can i group them in the last partition and let them in the same rows? 
A=rand(5,4);[n,m]=size(A);
np=2;B=A;
[c,idx]=sort(rand(n,1));
C=A(idx,:);
idnan=mod(np-rem(n,np),np)
C=[C ;nan(idnan,m)];
[n,m]=size(C);
res=reshape(C,[np,m,n/np])     % your submatrix
idxo=reshape([idx ;nan(idnan,1)],np,1,n/np);  % your original index

idnan =

     1

res(:,:,1) =

    0.9571    0.0020    0.2607    0.5391
    0.3271    0.3492       NaN    0.9611

res(:,:,2) =

    0.8100    0.4447    0.3409    0.4193
    0.6083       NaN    0.2305    0.8073

res(:,:,3) =

    0.5130    0.7416    0.5576    0.3956
       NaN    0.4363    0.2633       NaN

[Azzi Abdelmalek]

%instead of reshape use

for k=1:n/np
  ind=(k-1)*np+1:k*np
res(:,:,k)=C(ind,:)
end

[Azzi Abdelmalek]

Because reshape is working column after column

[Mariem Harmassi]

Yes i see that , :((( so itsn't useful for my problem because i need to keep the initial configuration the matrix is dataset of objects so i can t change the column into rows what reshape do a column is descriptor.

[Azzi Abdelmalek]

look at the updated code