unique values inside a matrix
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Hello,
Assuming i have a matrix like this m=
1 30
1 20
1 20
1 20
2 30
2 28
2 28
... ...
i want it to be like this:
1 30
1 20
2 30
2 28
PS: THE MATRIX IS HUGE CONSISTING OF 11000 ROWS AND 9 COLUMNS
unique is not working since it has it's taking a unique value of all, for example if i apply it for the second column it will take 28 for 1 and not for 2
thank you in advance
Accepted Answer
Star Strider
on 21 Sep 2019
Edited: Star Strider
on 21 Sep 2019
Try this:
A = [1 30
1 20
1 20
1 20
2 30
2 28
2 28];
Out = unique(A, 'rows', 'stable')
producing:
Out =
1 30
1 20
2 30
2 28
Experiment to get different results.
9 Comments
Michel tawil
on 21 Sep 2019
Star Strider
on 21 Sep 2019
It changes when I run it:
A = [1 74
1 73
1 72
1 70
1 69
1 67
1 65
1 63
1 61
1 58
1 57
1 55
1 53
1 51
1 50
1 49
1 48
1 47
1 46
1 45
1 44
1 43
1 42
1 42
1 41
1 41
1 40
1 32
1 31
1 30
1 30
1 29
1 28
1 28
1 27
1 26
1 26
1 25
1 24
1 24
1 23
1 22
1 21
1 21
1 20
1 20
1 20
1 20
1 20];
Out = unique(A, 'rows', 'stable')
produces:
Out =
1 74
1 73
1 72
1 70
1 69
1 67
1 65
1 63
1 61
1 58
1 57
1 55
1 53
1 51
1 50
1 49
1 48
1 47
1 46
1 45
1 44
1 43
1 42
1 41
1 40
1 32
1 31
1 30
1 29
1 28
1 27
1 26
1 25
1 24
1 23
1 22
1 21
1 20
With ‘A’ being (49x2) and ‘Out’ being (38x2).
If nothing truly changes, we may not be getting all the information we need to have.
Michel tawil
on 21 Sep 2019
Star Strider
on 21 Sep 2019
You may need to use uniquetol instead. See specifically the section on Perpare Vectors for Exact Comparison, since the 'stable' argument is not supported, and 'rows' becomes 'ByRows',1.
Star Strider
on 21 Sep 2019
Using uniquetol with your .mat file:
D = load('C.mat');
A = D.C(:,1:2)
[Out,ia,ic] = uniquetol(A, 0.1, 'ByRows',1)
OutStable = A(ia,:)
produces:
A =
1 74
1 73
1 72
1 70
1 69
1 67
1 65
1 63
1 61
1 58
1 57
1 55
1 53
1 51
1 50
1 49
1 48
1 47
1 46
1 45
1 44
1 43
1 42
1 42
1 41
1 41
1 40
1 40
1 32
1 31
1 30
1 30
1 29
1 28
1 28
1 27
1 26
1 26
1 25
1 24
1 24
1 23
1 22
1 21
1 21
1 20
1 20
1 20
1 20
1 20
2 85
2 84
2 82
2 80
2 78
2 77
2 75
2 73
2 72
2 71
2 69
2 68
2 67
2 66
2 65
2 63
2 62
2 61
2 60
2 58
2 57
2 56
2 54
2 53
2 52
2 51
2 50
2 49
2 48
2 47
2 46
2 45
2 44
2 44
2 43
2 42
2 42
2 41
2 33
2 33
2 33
2 32
3 91
3 89
3 87
3 85
3 83
3 80
3 78
3 75
OutStable =
1 20
1 30
1 40
1 50
1 61
1 72
2 32
2 42
2 52
2 62
2 72
2 82
3 75
3 85
Experiment to get the result you want.
Michel tawil
on 21 Sep 2019
Star Strider
on 22 Sep 2019
Sure!
The code changes to:
D = load('C.mat');
C = D.C;
[Out,ia,ic] = uniquetol(C(:,[1 2]), 0.1, 'ByRows',1);
OutStable = C(ia,:)
producing:
OutStable =
1 20 568 28.4 1 27.719 63.087 27.719 63.087
1 30 755 37.75 1 27.719 63.087 27.719 63.087
1 40 810 40.5 1 27.719 63.087 27.719 63.087
1 50 890 44.5 1 27.719 63.087 27.719 63.087
1 61 1005 50.25 1 27.719 63.087 27.719 63.087
1 72 1108 55.4 1 27.719 63.087 27.719 63.087
2 32 784 39.2 1 27.719 63.087 27.719 63.087
2 42 831 41.55 1 27.719 63.087 27.719 63.087
2 52 904 45.2 1 27.719 63.087 27.719 63.087
2 62 983 49.15 1 27.719 63.087 27.719 63.087
2 72 1020 51 1 27.719 63.087 27.719 63.087
2 82 1068 53.4 1 27.719 63.087 27.719 63.087
3 75 1040 52 1 27.719 63.087 27.719 63.088
3 85 1106 55.3 1 27.719 63.087 27.719 63.088
This is the shortg format option, since it’s easier to read (and post). The full precision remains internally.
Michel tawil
on 22 Sep 2019
Edited: Michel tawil
on 22 Sep 2019
Star Strider
on 22 Sep 2019
As always, my pleasure!
More Answers (1)
Guillaume
on 21 Sep 2019
I'm not sure what you mean by unique is not working. You haven't explained the reasoning for going from your input to your output, but it does look like you want unique rows:
result = unique(m, 'rows')
With your demo data:
>> m = [
1 30
1 20
1 20
1 20
2 30
2 28
2 28];
>> unique(m, 'rows')
ans =
1 20
1 30
2 28
2 30
>> unique(m, 'rows', 'stable') %with stable to get the same order as your example
ans =
1 30
1 20
2 30
2 28
5 Comments
Michel tawil
on 21 Sep 2019
Guillaume
on 21 Sep 2019
The simplest way for us to understand what you're talking about is to attach a mat file with your data to your question.
>> unique([20; round(19.8)])
ans =
20
As you can see it's not a problem unique returns just one value. Whatever your data is, it's not exactly how you describe it.
Possibly, it's something like:
>> v = [20; 20+1e-14]
v =
20
20
>> unique(v) %return 2 elements because the 2nd one is not exactly 20.
ans =
20
20
>> ans - 20
ans =
0
1.06581410364015e-14
Michel tawil
on 21 Sep 2019
Michel tawil
on 21 Sep 2019
Michel tawil
on 22 Sep 2019
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