Cant understand how the values are stored
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Rehman Tanim
am 27 Nov. 2020
Kommentiert: Rehman Tanim
am 27 Nov. 2020
A = load('r08_edfm.mat');
B = load('r08_edfm.mat');
d = A.val(1,:);
d_T = d';
input_channel = 3;
noisy_sig = B.val(input_channel,:);
noisy_sig_T = noisy_sig'
at workspace im getting these values
6 Kommentare
KSSV
am 27 Nov. 2020
What you are looking into are completely two different data sets. There is no conversion of numbers taking place.
Akzeptierte Antwort
Image Analyst
am 27 Nov. 2020
You forgot to attach the .mat files so we can't do much. The first row of the val matrix stored in the mat file is extracted and put into a variable you cryptically call "d". We don't know what that first row means and so we can't "explain" it. All we can tell is that the first row of val from A is stored in a row vector "d" and then you're transposing it and putting it into column vector d_T, and you're then getting the third row of the val matrix from a different mat file and calling that noise, and also transposing that row vector into a column vector called "noisy_sig_T".
Can't you ask the people who created the mat file? If you don't even know what it means, then why do you want to do anything with it? Where did you get that script, or did you write it?
4 Kommentare
Image Analyst
am 27 Nov. 2020
There is no "Direct_1" column in the mat file:
A =
struct with fields:
val: [6×10000 double]
B =
struct with fields:
val: [6×10000 double]
so I'm not sure what you're talking about. And all the values in val are around 4-500 in the first part of the array, not around 42.
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Variables finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!