How do I extract only the rows from a matrix that meet a specific requirement?
3 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Jialin HE
le 19 Fév 2022
Commenté : Star Strider
le 20 Fév 2022
I have some matrixes containing the movement of certain subjects doing the same task, all of them of different length (some people take more time and some less). However, I need to compare them therefore the matrixes have to be same length. Since the matrixes have 4 columns, where first three are x y z and forth is time, I converted time in % of the cycle (since i know first and last row of each matrix correspond to start and end of a task). Now I have something like:
0.473487217592520 0.472650945192777 0.0963353018423272 0
0.451556907216931 0.493711477180010 0.104843749333787 0.116660590594240
0.428662781457986 0.515192918720937 0.114246578015789 0.233321181188480
0.403546122766580 0.537143904136897 0.126402289179415 0.360918702150930
My idea is to get a new matrix where I extract the row of data where time(% cicle) is close to 0.5, 1, 1.5, 2 and so on, so that from every initial matrix of data I get a new matrix with same length and correspond to same % of the cycle not depending on time. How can I extract only these data?
I know this isn't for sure the smartest way, I appreciate any suggestion or solving ways you have! Thank you!
0 commentaires
Réponse acceptée
Star Strider
le 19 Fév 2022
‘My idea is to get a new matrix where I extract the row of data where time(% cicle) is close to 0.5, 1, 1.5, 2 and so on ...’
Assuming that ‘0.5’ is ‘0.5%’ (and so for the rest) —
M = [0.473487217592520 0.472650945192777 0.0963353018423272 0
0.451556907216931 0.493711477180010 0.104843749333787 0.116660590594240
0.428662781457986 0.515192918720937 0.114246578015789 0.233321181188480
0.403546122766580 0.537143904136897 0.126402289179415 0.360918702150930];
M4q = 0:0.005:1;
Mi = [interp1(M(:,4), M(:,1:3), M4q(:)) M4q(:)]
If I interpreted your request correctly, that should be the desired result.
.
2 commentaires
Star Strider
le 20 Fév 2022
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Plus de réponses (0)
Voir également
Catégories
En savoir plus sur Multirate Signal Processing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!