How do I get the maximum numbers of every "increasing pattern" in each row in a matrix?

7 Ansichten (letzte 30 Tage)
Here is a matrix A consisting of "increasing patterns".
A=[1,2,3,1,2,3,4,1,2,1,2,3,4,1,1;1,2,1,1,2,3,1,2,3,4,1,2,3,4,5]
In each row of matrix A, all "increasing patterns" begin with 1.
Visually, all "patterns" are underlined separately below. The largest numbers in all "patterns" are in bold.
A =
1 2 3 1 2 3 4 1 2 1 2 3 4 1 1
1 2 1 1 2 3 1 2 3 4 1 2 3 4 5
How do I get B? This is, get the maximum numbers of all "patterns" in each row.
B=3 4 2 4 1 1
2 1 3 4 5 0
% use zero to fuilfill matrix B to make dimensions of each row consistent

Akzeptierte Antwort

Stephen23
Stephen23 am 27 Dez. 2018
>> A = [1,2,3,1,2,3,4,1,2,1,2,3,4,1,1;1,2,1,1,2,3,1,2,3,4,1,2,3,4,5]
A =
1 2 3 1 2 3 4 1 2 1 2 3 4 1 1
1 2 1 1 2 3 1 2 3 4 1 2 3 4 5
>> X = diff(A,1,2)~=1;
>> X(:,end+1) = true;
>> B = A.*X
B =
0 0 3 0 0 0 4 0 2 0 0 0 4 1 1
0 2 1 0 0 3 0 0 0 4 0 0 0 0 5
  3 Kommentare
Stephen23
Stephen23 am 27 Dez. 2018
Bearbeitet: Stephen23 am 27 Dez. 2018
>> Y = B.';
>> Z = +~sort(Y==0,1);
>> Z(Z==1) = Y(Y~=0);
>> B = Z(any(Z,2),:).'
B =
3 4 2 4 1 1
2 1 3 4 5 0

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Weitere Antworten (1)

Linjun He
Linjun He am 27 Dez. 2018
Bearbeitet: Linjun He am 27 Dez. 2018
A2=[A,ones(size(A,1),1)]-[ones(size(A,1),1),A]<=0
B=A2(:,2:end).*A
Here is what I get.
B =
0 0 3 0 0 0 4 0 2 0 0 0 4 1 1
0 2 1 0 0 3 0 0 0 4 0 0 0 0 5
It's not a good answer but it's ok for me.
I am still looking forward to a better answer.
  2 Kommentare
madhan ravi
madhan ravi am 27 Dez. 2018
but in the orginal question your desired output is different right I mean 5 was padded with 0 at the end??
Linjun He
Linjun He am 27 Dez. 2018
It's not a good answer but it's ok for me.
I am still looking forward to a better answer.

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