finding the increase in values

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johnson saldanha
johnson saldanha el 6 de Dic. de 2018
Comentada: johnson saldanha el 6 de Dic. de 2018
suppose i have a column matrix a=( 2 2 2 3 3 3 5 5 5 4 4 3 3 3 2 2 3 3 3 4 4 4 5 5 5 4 4 3 3 )
i want to find the constant and increasing elements and put them separately in different columns of a matrix. eliminating the decreasing elements.
output(:,1) = ( 2 2 2 3 3 3 5 5 5 )
output(:,2) = (2 2 3 3 3 4 4 4 5 5 5 )

Respuesta aceptada

Stephen23
Stephen23 el 6 de Dic. de 2018
Editada: Stephen23 el 6 de Dic. de 2018
a = [2;2;2;3;3;3;5;5;5;4;4;3;3;3;2;2;3;3;3;4;4;4;5;5;5;4;4;3;3]
idx = [true;diff(a)~=0];
idy = diff(a(idx))>0;
idd = diff([0;idy;0]);
idb = find(idd>0);
ide = find(idd<0);
ids = cumsum(idx);
fun = @(b,e)a(ismember(ids,b:e));
C = arrayfun(fun,idb,ide,'uni',0);
And checking the output:
>> C{:}
ans =
2
2
2
3
3
3
5
5
5
ans =
2
2
3
3
3
4
4
4
5
5
5
  2 comentarios
johnson saldanha
johnson saldanha el 6 de Dic. de 2018
Error using horzcat
Dimensions of matrices being concatenated are not consistent.
Error in accelerationrate (line 21)
idx = [true,diff(a)~=0];
im getting this error
Guillaume
Guillaume el 6 de Dic. de 2018
You're certainly not getting this error with the exact code that stephen gave. Of course, if your actual vector is a column vector instead of a row vector you either need to use vertical concatenation (i.e. use ; instead of , resulting in [true; diff(a)~=0]) or transpose your input into a row vector (i.e. [true, diff(a.')~=0]). First option is more efficient.

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johnson saldanha
johnson saldanha el 6 de Dic. de 2018
i did that but im not getting the desired output. im getting the decreasing values too
(5 5 5 5 3.75 3.75 3.75 3.75) im getting this in the output
  10 comentarios
Stephen23
Stephen23 el 6 de Dic. de 2018
Editada: Stephen23 el 6 de Dic. de 2018
See my edited answer, it should work now. For your .xlsx data it gives this:
>> C{:}
ans =
11.250
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
13.750
13.750
13.750
13.750
13.750
13.750
13.750
13.750
13.750
ans =
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
12.500
13.750
13.750
13.750
13.750
13.750
13.750
15.000
15.000
15.000
15.000
15.000
15.000
15.000
15.000
15.000
15.000
And for your .csv data it gives this:
>> C{:}
ans =
1.2500
1.2500
2.5000
3.7500
3.7500
3.7500
5.0000
5.0000
6.2500
6.2500
7.5000
7.5000
7.5000
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
8.7500
ans =
2.5000
2.5000
2.5000
2.5000
3.7500
3.7500
3.7500
3.7500
3.7500
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
ans =
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
3.7500
5.0000
5.0000
5.0000
5.0000
5.0000
5.0000
6.2500
6.2500
6.2500
6.2500
7.5000
7.5000
7.5000
7.5000
7.5000
8.7500
8.7500
8.7500
8.7500
8.7500
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
10.0000
ans =
8.7500
10.0000
10.0000
10.0000
10.0000
10.0000
11.2500
11.2500
11.2500
12.5000
12.5000
13.7500
13.7500
13.7500
15.0000
15.0000
15.0000
15.0000
16.2500
16.2500
16.2500
16.2500
16.2500
16.2500
17.5000
17.5000
17.5000
17.5000
17.5000
17.5000
18.7500
18.7500
18.7500
johnson saldanha
johnson saldanha el 6 de Dic. de 2018
thanks a lot. it works. and suppose i have to do the decrease they i have to change the > with < right

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