Identify, for each column of a matrix, numbers that are smaller than the previous ones, creating a 0 1 output having the same size of the original matrix

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Simone A.
Simone A. 2023-1-13
编辑： Simone A. 2023-1-14
Dear all,
I've got a matrix, let's call it M.
M = 1000x1200
The first 3 columns of M look like:
1.05 21.22 17.2 ...
2.01 22.35 18.2 ...
3.52 23.63 19.2 ...
4.32 23.54 20.2 ...
5.11 23.56 18.2 ...
6.89 23.64 19.2 ...
7.01 23.65 20.3 ...
8.87 23.99 21.3 ...
6.52 24.52 22.3 ...
7.91 24.51 23.3 ...
8.42 24.515 24.3 ...
9.03 24.525 22.3 ...
10.88 25.23 24.4 ...
11.01 45.12 25.4 ...
11.32 48.13 24.4 ...
11.33 46.14 26.4 ...
12.12 46.15 31.1 ...
12.09 48.142 32.2 ...
12.11 48.143 30.02 ...
12.13 48.151 31.6 ...
12.52 48.36 33.1 ...
Now, what I am trying to do is to loop through each column of M, from column 1 to column 1200, and identify all of the numbers that are smaller than the last, greater, number of a sequence. From this loop I'm trying to obtain a matrix, let's call it Mx of the same size of M, made of 0 and 1, where 1s denote the numbers that are smaller than the preavious ones.
For the example above, Mx would be:
0 0 0 ...
0 0 0 ...
0 0 0 ...
0 1 0 ...
0 1 1 ...
0 0 1 ...
0 0 1 ...
0 0 0 ...
1 0 0 ...
1 1 0 ...
1 1 0 ...
0 0 1 ...
0 0 0 ...
0 0 0 ...
0 0 1 ...
0 1 0 ...
0 1 0 ...
1 1 0 ...
1 0 1 ...
0 0 1 ...
0 0 1 ...
Any help would be massively appreciated!
  2 个评论
Walter Roberson
Walter Roberson 2023-1-13
To confirm that I understand correctly:
You go down each column, keeping a running maximum-seen-so-far. For each entry that is less than the maximum seen so far in the column, a 1 should be emitted, and 0 for values that are new maxima ?
To confirm, if you encounter a value that is equal to the maximum seen so far, then because equal is not "less than", 0 should be emitted for the duplicate ?
Simone A.
Simone A. 2023-1-14
编辑：Simone A. 2023-1-14
* EDIT: Attached a sample of my dataset.*
Hi @Walter Roberson, it's me again. Altought the code runs brilliantly, I encountered a situation which i did not considered in my original question.
The previous code correctly identifies all of the values less than the maximum seen so far. The figure below shows colum 1 of my matrix with red circles indicating the values more than the minimum seen so far (namely the values identified with 1s):
However, I encountered a situation where values are not more than the minimum seen so far, but deviate from a linear trend, or better say "make some steps":
These are not considered in the script you kindly suggested.
I wonder if these deviations from the linear trend could also be identified as 1s?
I have included a sample of my actual dataset.
Thank a lot!

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回答（1 个）

the cyclist
the cyclist 2023-1-13
Ran in:
My output is a bit different from yours, but I think I followed the algorithm you wanted.
M = [ ...
1.05 21.22 17.2; ...
2.01 22.35 18.2; ...
3.52 23.63 19.2; ...
4.32 23.54 20.2; ...
5.11 23.56 18.2; ...
6.89 23.64 19.2; ...
7.01 23.65 20.3; ...
8.87 23.99 21.3; ...
6.52 24.52 22.3; ...
7.91 24.51 23.3; ...
8.42 24.515 24.3; ...
9.03 24.525 22.3; ...
10.88 25.23 24.4; ...
11.01 45.12 25.4; ...
11.32 48.13 24.4; ...
11.33 46.14 26.4; ...
12.12 46.15 31.1; ...
12.09 48.142 32.2; ...
12.11 48.143 30.02; ...
12.13 48.151 31.6; ...
12.52 48.36 33.1]; ...
nrows = height(M);
Mx = zeros(size(M));
for nr = 2:nrows
Mx(nr,:) = M(nr,:) < max(M(1:nr,:));
end
Mx
Mx = 21×3
0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 0 1 0 0 0 0 0 0 1 0 0 1 1 0
  8 个评论
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Walter Roberson
Walter Roberson 2023-1-13
Ran in:
"however, in that scenario, the second number should be considered as "1"."
That takes more code
M = [ ...
1.05 21.22 17.2; ...
2.01 22.35 18.2; ...
3.52 23.63 19.2; ...
4.32 23.54 20.2; ...
4.32 23.56 18.2; ... %identical
6.89 23.56 19.2; ... %identical
7.01 23.65 20.3; ...
8.87 23.99 21.3; ...
6.52 24.52 22.3; ...
7.91 24.51 23.3; ...
8.42 24.515 24.3; ...
9.03 24.525 22.3; ...
10.88 25.23 24.4; ...
11.01 45.12 25.4; ...
11.32 48.13 24.4; ...
11.33 46.14 26.4; ...
12.12 46.15 31.1; ...
12.09 48.142 32.2; ...
12.11 48.143 30.02; ...
12.13 48.151 31.6; ...
12.52 48.36 33.1]; ...
nrows = height(M);
Mx = zeros(size(M));
runningmax = M(1,:);
for nr = 2:nrows
oldrunningmax = runningmax;
runningmax = max(runningmax, M(nr, :));
Mx(nr,:) = M(nr,:) < runningmax | M(nr, :) == oldrunningmax;
end
Mx
Mx = 21×3
0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 1 1 0 0 0 0 0 0 1 0 0 1 1 0

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