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 )

 Réponse acceptée

Stephen23
Stephen23 le 6 Déc 2018
Modifié(e) : Stephen23 le 6 Déc 2018

0 votes

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 commentaires

johnson saldanha
johnson saldanha le 6 Déc 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 le 6 Déc 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.

Connectez-vous pour commenter.

Plus de réponses (1)

johnson saldanha
johnson saldanha le 6 Déc 2018

0 votes

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 commentaires

johnson saldanha
johnson saldanha le 6 Déc 2018
in one or two cells im getting only the increasing but majority of them im getting the decreasing values too
Stephen23
Stephen23 le 6 Déc 2018
@johnson saldanha: please upload some data that shows this behavior.
johnson saldanha
johnson saldanha le 6 Déc 2018
i will upload the column vector.
johnson saldanha
johnson saldanha le 6 Déc 2018
this is the output of one cell
Stephen23
Stephen23 le 6 Déc 2018
Modifié(e) : Stephen23 le 6 Déc 2018
@johnson saldanha: please upload the input data in a .mat file (not in an .xlsx file) in a new comment (not in an answer).
johnson saldanha
johnson saldanha le 6 Déc 2018
Modifié(e) : johnson saldanha le 6 Déc 2018
yes u can see theres a decrease from 13.75 to 12.5. i dont want the decrease here. the next set of increasing values should fall into the next cell. not the same cell
johnson saldanha
johnson saldanha le 6 Déc 2018
csv file is okay ?
johnson saldanha
johnson saldanha le 6 Déc 2018
the input
Stephen23
Stephen23 le 6 Déc 2018
Modifié(e) : Stephen23 le 6 Déc 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 le 6 Déc 2018
thanks a lot. it works. and suppose i have to do the decrease they i have to change the > with < right

Connectez-vous pour commenter.

Catégories

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by