reorder data in a matrix

Question

barath manoharan 2023-2-25

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https://ww2.mathworks.cn/matlabcentral/answers/1918770-reorder-data-in-a-matrix

评论： barath manoharan 2023-2-26

采纳的回答： Stephen23

i have a 7*1 double matrix and need to reorder data to satisfy some condition.Thank you in advance.

example:

Let A = [10010 ; 11000 ; 01100 ; 01011 ; 10111 ; 11010 ; 01111]

output :

B = [10010 ; 11010 ; 11000 ; 01100 ; 01111 ; 01011 ; 10111]

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barath manoharan 2023-2-26

@Stephen23 sorry for the inconvenience, for example

A = [10010 ; 10111 ; 10011 ; 01011]

i want to reorder the above A into a set of elements B lets say

B = [10010 ; 10011 ; 10111 ; 01011]

here let me take first "two" elements

10010

10011 - as you can see only the last bit of both elements are different so will it as 1

10111 - only 1 bit difference (3rd bit different) so will take it as 1

01011 - 3 bits different so 3

so total bit changes are 1 + 1 + 3 = 5. (reordering is done to get total as least as possible)

but now my primary goal is to reorder the A set into B set as i have mentioned in question and given below. A set is shown as a 7 * 1 double in my matlab. so please provide a solution keeping in mind this condition.

example:

Let A = [10010 ; 11000 ; 01100 ; 01011 ; 10111 ; 11010 ; 01111]

output :

B = [10010 ; 11010 ; 11000 ; 01100 ; 01111 ; 01011 ; 10111].

thank you in advance.

Image Analyst 2023-2-26

This looks like a homework problem. Is it? If so, ask your instructor or read the link below to get started:

How do I get help on homework questions on MATLAB Answers? - MATLAB Answers - MATLAB Central

Obviously we can't give you the full solution because you're not allowed to turn in our code as your own.

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Answer 1

Stephen23 2023-2-26

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1918770-reorder-data-in-a-matrix#answer_1180135

编辑：Stephen23 2023-2-26

在 MATLAB Online 中打开

This code finds all permutations with the minimum absolute difference as you specified here:

A = [1,0,0,1,0; 1,1,0,0,0; 0,1,1,0,0; 0,1,0,1,1; 1,0,1,1,1; 1,1,0,1,0; 0,1,1,1,1]
A = 7×5
     1     0     0     1     0
     1     1     0     0     0
     0     1     1     0     0
     0     1     0     1     1
     1     0     1     1     1
     1     1     0     1     0
     0     1     1     1     1
P = perms(1:size(A,1));
N = Inf;
C = {};
for k = 1:size(P,1)
    X = P(k,:);
    M = sum(vecnorm(diff(A(X,:),1,1),1,1));
    if M==N
        C{end+1} = X;
    elseif M<N
        N = M;
        C = {X};
    end
end
display(C) % mimimum permutations
C = 1×4 cell array
    {[5 1 6 2 3 7 4]}    {[4 7 5 1 6 2 3]}    {[4 7 3 2 6 1 5]}    {[3 2 6 1 5 7 4]}
display(N) % total absolute difference
N = 9
for k = 1:numel(C)
    A(C{k},:) % lets take a look at them
end
ans = 7×5
     1     0     1     1     1
     1     0     0     1     0
     1     1     0     1     0
     1     1     0     0     0
     0     1     1     0     0
     0     1     1     1     1
     0     1     0     1     1
ans = 7×5
     0     1     0     1     1
     0     1     1     1     1
     1     0     1     1     1
     1     0     0     1     0
     1     1     0     1     0
     1     1     0     0     0
     0     1     1     0     0
ans = 7×5
     0     1     0     1     1
     0     1     1     1     1
     0     1     1     0     0
     1     1     0     0     0
     1     1     0     1     0
     1     0     0     1     0
     1     0     1     1     1
ans = 7×5
     0     1     1     0     0
     1     1     0     0     0
     1     1     0     1     0
     1     0     0     1     0
     1     0     1     1     1
     0     1     1     1     1
     0     1     0     1     1

Your example output does not have the minimum according to the definition you gave. Its total is actually:

B = [1,0,0,1,0; 1,1,0,1,0; 1,1,0,0,0; 0,1,1,0,0; 0,1,1,1,1; 0,1,0,1,1; 1,0,1,1,1]
B = 7×5
     1     0     0     1     0
     1     1     0     1     0
     1     1     0     0     0
     0     1     1     0     0
     0     1     1     1     1
     0     1     0     1     1
     1     0     1     1     1
sum(vecnorm(diff(B,1,1),1,1))
ans = 10

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Stephen23 2023-2-26

在 MATLAB Online 中打开

In contrast your first example here does provide (one of) the correct result:

A = [1,0,0,1,0; 1,0,1,1,1; 1,0,0,1,1; 0,1,0,1,1];
P = perms(1:size(A,1));
N = Inf;
C = {};
for k = 1:size(P,1)
    X = P(k,:);
    M = sum(vecnorm(diff(A(X,:),1,1),1,1));
    if M==N
        C{end+1} = X;
    elseif M<N
        N = M;
        C = {X};
    end
end
N
N = 5
C
C = 1×8 cell array
    {[4 3 2 1]}    {[4 3 1 2]}    {[4 2 3 1]}    {[4 1 3 2]}    {[2 3 1 4]}    {[2 1 3 4]}    {[1 3 2 4]}    {[1 2 3 4]}
B = [1,0,0,1,0; 1,0,0,1,1; 1,0,1,1,1; 0,1,0,1,1];
sum(vecnorm(diff(B,1,1),1,1))
ans = 5

It happens to be the 7th permutation found with that total:

Z = cellfun(@(x)isequal(B,A(x,:)),C)
Z = 1×8 logical array
   0   0   0   0   0   0   1   0
isequal(B,A(C{7},:))
ans = logical
   1

barath manoharan 2023-2-26

@Stephen23 thank you for your response. really helpful

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Answer 2

Askic V 2023-2-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1918770-reorder-data-in-a-matrix#answer_1179795

编辑：Askic V 2023-2-25

在 MATLAB Online 中打开

It seems to me that you want this logic implemented:

A =  ["10010" ; "11000" ; "11010" ; "01100" ; "01011"]
A = 5×1 string array
    "10010"
    "11000"
    "11010"
    "01100"
    "01011"
N = numel(A);
B = A;
for i = N:-2:2
    B([i,i-1]) = B([i-1,i]);
end
B
B = 5×1 string array
    "10010"
    "11010"
    "11000"
    "01011"
    "01100"

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barath manoharan 2023-2-26

@Askic V thank you for your response and by mistake previously i have sent the wrong data and can you please solve the edited one above. thank you in advance

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