Filter löschen
Filter löschen

Comparing elements of a vector.

2 Ansichten (letzte 30 Tage)
Madhurima Reddy
Madhurima Reddy am 10 Okt. 2019
Kommentiert: Madhurima Reddy am 10 Okt. 2019
I have two vectors named R and T, where elements in R keep on increasing and finally reaches a steady state value.
Let's say T = [1;2;3;4;5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20;21] and R = [1;2;3;4;5;6;7;8;9;10;10;10;11;12;12;12;12;12;13;13;13].
I intend to get g(:,1) = [1;2;3;4;5;6;7;8;9;10;13;14;19] and g(:,2) = [1;2;3;4;5;6;7;8;9;10;11;12;13]. But I am getting all the values in g. Could you please help me out with this.
Thanks and Regards.
L = length(R);
for h = 1:1:L-1
dffrad = R(h+1)-R(h);
dfftime= T(h+1)-T(h);
if (dffrad~=0 && dffrad>0)
g = [g; T(h+1) R(h+1)];
end
end
  3 Kommentare
Madhurima Reddy
Madhurima Reddy am 10 Okt. 2019
The values in my data(not that's given in the question) were getting rounded off to nearest decimal and that's why I thought the values were repeating. Thank you Shubham for your help.
Shubham Gupta
Shubham Gupta am 10 Okt. 2019
I am glad I could help. Also, you might wanna look at @the_cyclist 's answer for better performance.

Melden Sie sich an, um zu kommentieren.

Antworten (1)

the cyclist
the cyclist am 10 Okt. 2019
[g(:,2), idx] = unique(R);
g(:,1) = T(idx);
  3 Kommentare
the cyclist
the cyclist am 10 Okt. 2019
You are correct.
But the original question explicitly stated that R is increasing, so I assumed that that was OK.
Madhurima Reddy
Madhurima Reddy am 10 Okt. 2019
My data does decrease after reaching a max value. However, Thank you to both of you for help.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Introduction to Installation and Licensing finden Sie in Help Center und File Exchange

Produkte


Version

R2016a

Community Treasure Hunt

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

Start Hunting!

Translated by