How can I check if the columns of my matrix have value in common?

Hi, I have a basic problem.

I have a matrix 7x21 double and I want to compare each column with the others. I want to understand if a value in the first column is repeated in the following 20 columns. Either a value from the sixth column in all five first or subsequent columns.

Answer 1

[Walter Roberson]

I am not certain what you are asking

Either a value from the sixth column in all five first or subsequent columns.

M = randi(2, 10, 8)

M = 10×8

2 1 1 2 1 2 2 1 1 2 1 1 1 1 2 1 1 2 1 2 1 2 1 1 1 2 1 2 1 1 2 1 1 1 1 2 2 2 2 1 2 2 2 1 2 1 1 1 2 2 2 2 2 2 1 1 2 2 2 2 2 2 2 1 2 2 2 1 2 1 1 1 2 2 2 2 1 1 2 2

sum(M(:,6) == M(:,1:5), 2) == 5

ans = 10×1 logical array
   0
   0
   0
   0
   0
   0
   1
   1
   0
   0

[HelpAStudent]

I have a matrix 7x21 double. I need to compare each element of one column with each element of the remaining columns. By compare I mean if the elements are equal or almost equal (I have an uncertainty range in my data sheet). The output should tell me wich elements are equal

[Walter Roberson]

format long g
tol = 1e-15;
M = randi(2, 7, 21) + rand(7,21)*1e-14

M = 7×21

2.0000 2.0000 1.0000 1.0000 1.0000 1.0000 1.0000 2.0000 2.0000 1.0000 1.0000 2.0000 1.0000 1.0000 2.0000 2.0000 1.0000 2.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 1.0000 2.0000 1.0000 2.0000 1.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 1.0000 2.0000 2.0000 2.0000 1.0000 2.0000 2.0000 1.0000 1.0000 2.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 1.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 2.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 2.0000 1.0000 2.0000 2.0000 1.0000 1.0000 2.0000 2.0000 1.0000 2.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 1.0000 2.0000 2.0000 1.0000 2.0000 2.0000 1.0000 2.0000 2.0000 2.0000 1.0000 2.0000 2.0000 1.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 2.0000 2.0000 2.0000 1.0000 1.0000 1.0000 2.0000 1.0000 2.0000 2.0000 2.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 2.0000 2.0000 2.0000 2.0000 1.0000 1.0000 2.0000 2.0000 2.0000 1.0000 2.0000 1.0000 1.0000 1.0000 1.0000 1.0000 2.0000 1.0000 2.0000 1.0000 1.0000 2.0000 1.0000 1.0000 2.0000

selected_col = 6;
other_cols = setdiff(1:size(M,2), selected_col);
to_compare = M(:,selected_col);
subset_to_compare_to = M(:,other_cols);
matches = false(size(M));
matches(:,other_cols) = abs(to_compare - subset_to_compare_to) < tol;

matches

matches = 7×21 logical array
   0   0   0   1   1   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0   0   0   0   0   0
   1   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   1   0   1   0
   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
   0   0   0   0   0   0   0   0   0   0   0   0   0   1   0   0   0   0   0   0   0

[HelpAStudent]

Hi. Thank you, how can I put my matrix, with my value instead of a random one?

[Walter Roberson]

Remove the assignment to M, and assign your real data to M.

[HelpAStudent]

I tried and the output has just number of 1 and 0, but in the output I want to know wich value is the same from another one. Who can I do it?

[Walter Roberson]

When you ask "which" you might be asking what the location is, or you might be asking what the value is. My code returns 0 for locations that do not match and 1 for ones that match, so it answers the question about location.

any(matches, 2)

is 1 for rows that match.

to_compare(any(matches, 2))

will tell you the values but not the location

Answer 2

[yanqi liu]

sir，may be use the follow code

clc
close all
clear all
% init matrix
M = randi([1 100], 10, 8)

M = 10×8

7 36 81 22 55 59 50 4 74 14 62 38 7 3 78 43 7 38 43 90 12 11 7 53 38 10 58 46 90 4 37 3 33 29 93 20 98 8 30 36 59 26 84 36 100 6 56 96 13 56 68 25 51 39 42 69 38 19 89 23 1 96 33 49 24 100 84 39 41 50 39 64 55 86 48 44 9 12 48 72

% choose one column 
use = 4;
% find value
Mu = M(:, use)

Mu = 10×1

22 38 90 46 20 36 25 23 39 44

M2 = M;
M2(:, use) = NaN;
% the different range 
th = 0;
for i = 1 : length(Mu)
    mi = Mu(i);
    di = abs(M2 - mi);
    [r, c] = find(di <= th);
    fprintf('\n\nthe %d data of column %d is %d\n', i, use, mi);
    if length(r) < 1
         fprintf('\n can not find other column has this value!');
    end
    for j = 1 : length(r)
        fprintf('\nM(%d,%d)=%d', r(j), c(j), M(r(j), c(j)));
    end
end

the 1 data of column 4 is 22

 can not find other column has this value!

the 2 data of column 4 is 38

M(4,1)=38
M(8,1)=38
M(3,2)=38

the 3 data of column 4 is 90

M(4,5)=90

the 4 data of column 4 is 46

 can not find other column has this value!

the 5 data of column 4 is 20

 can not find other column has this value!

the 6 data of column 4 is 36

M(1,2)=36
M(5,8)=36

the 7 data of column 4 is 25

 can not find other column has this value!

the 8 data of column 4 is 23

 can not find other column has this value!

the 9 data of column 4 is 39

M(7,6)=39
M(9,7)=39

the 10 data of column 4 is 44

 can not find other column has this value!

 fprintf('\n');

you can read the book 《计算机视觉与深度学习实战》

[HelpAStudent]

Can you make an else statemant if he finds the same value like "I find the same value in here...."?

[HelpAStudent]

This are the question about your code:

for i = 1 : length(Mu)
    mi = Mu(i);
    di = abs(M2 - mi);
    [r, c] = find(di <= th);
    fprintf('\n\nthe %d data of column %d is %d\n', i, use, mi);
    if length(r) < 1
        fprintf('\n can not find other column has this value!');
    else 
         fprintf('\n I find the same value in here!'); %I want to know where is it the same value and wich is the value 
    end
    for j = 1 : length(r)
        fprintf('\nM(%d,%d)=%d', r(j), c(j), M(r(j), c(j))); %What did you print here?
    end
end

Answer 3

[yanqi liu]

clc
close all
clear all
% init matrix
M = randi([1 100], 10, 8)

M = 10×8

85 51 44 55 49 2 1 73 5 99 30 99 77 99 6 50 6 92 97 49 35 57 24 44 27 96 96 45 69 8 89 94 64 29 82 78 13 89 77 9 1 41 53 91 33 87 12 30 52 64 85 17 54 62 63 58 74 64 21 94 28 87 80 16 56 58 77 16 59 28 94 91 96 83 2 84 54 64 39 1

% choose one column
use = 4;
% find value
Mu = M(:, use)

Mu = 10×1

55 99 49 45 78 91 17 94 16 84

M2 = M;
M2(:, use) = NaN;
% the different range
th = 0;
for i = 1 : length(Mu)
    mi = Mu(i);
    di = abs(M2 - mi);
    [r, c] = find(di <= th);
    fprintf('\n\n the %d data of column %d is %d\n', i, use, mi);
    if length(r) < 1
        fprintf('\n can not find other column has this value!');
        continue;
    else
        fprintf('\n I find the same value in here! Please view the follow %d data for detail!', length(r)); 
    end
    for j = 1 : length(r)
        fprintf('\n    M(%d,%d)=%d', r(j), c(j), M(r(j), c(j)));
    end
end

 the 1 data of column 4 is 55

 can not find other column has this value!

 the 2 data of column 4 is 99

 I find the same value in here! Please view the follow 2 data for detail!

    M(2,2)=99
    M(2,6)=99

 the 3 data of column 4 is 49

 I find the same value in here! Please view the follow 1 data for detail!

    M(1,5)=49

 the 4 data of column 4 is 45

 can not find other column has this value!

 the 5 data of column 4 is 78

 can not find other column has this value!

 the 6 data of column 4 is 91

 I find the same value in here! Please view the follow 1 data for detail!

    M(9,8)=91

 the 7 data of column 4 is 17

 can not find other column has this value!

 the 8 data of column 4 is 94

 I find the same value in here! Please view the follow 2 data for detail!

    M(9,7)=94
    M(4,8)=94

 the 9 data of column 4 is 16

 I find the same value in here! Please view the follow 1 data for detail!

    M(8,8)=16

 the 10 data of column 4 is 84

 can not find other column has this value!

fprintf('\n');