Filter löschen
Filter löschen

closest value to zero excluding the first one?

29 Ansichten (letzte 30 Tage)
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
Kommentiert: Bruno Luong am 12 Jan. 2023
If given a long vector of values. How can I find the index of the closest value to zero excluding the first value of the vector out of our list of stuff to search for
  2 Kommentare
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
I am attaching an example variable and the answer your code should result in should be 175 or the 174 index.
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
I = find(diff(dis_y>=0),1);

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
"dis_y" being the variable:
I = find(diff(dis_y>=0),1);
  1 Kommentar
Bruno Luong
Bruno Luong am 12 Jan. 2023
That is NOT your question asks for, that is first consecutive points where 0 is cross from positive to negative.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (2)

Torsten
Torsten am 12 Jan. 2023
Bearbeitet: Torsten am 12 Jan. 2023
out of our list of stuff to search for
?
v = [-0.1 2 4 -0.5 8];
[~,i] = min(abs(v(2:end)))
i = 3
v(i+1)
ans = -0.5000
  3 Kommentare
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
Bearbeitet: Ali Almakhmari am 12 Jan. 2023
hmm I dont think this works. For example, I attached this variable here. The answer to my question should be 175 or the 174 index. But with this code you have I do not get that.
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
Actually I found the solution: I = find(diff(dis_y>=0),1);

Melden Sie sich an, um zu kommentieren.


Bora Eryilmaz
Bora Eryilmaz am 12 Jan. 2023
Bearbeitet: Bora Eryilmaz am 12 Jan. 2023
load('var.mat')
[~,I] = sort(abs(dis_y), 'ascend');
% Closest value to 0:
dis_y(I(1))
ans = 2.3761e-05
% Second closest value to 0:
dis_y(I(2))
ans = -0.0014
  2 Kommentare
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
This is still not giving me 175 or 174.
Ali Almakhmari
Ali Almakhmari am 12 Jan. 2023
I = find(diff(dis_y>=0),1);

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Produkte


Version

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by