i wanted the slope with respect to time frame
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CalebJones
il 4 Set 2019
Commentato: Star Strider
il 20 Set 2019
I wanted to calculate slope of channel 1 to 15 with respect to the time frame. The values in the tables are HbO values which should be Y axis and X axis should be time time frame which in this case is 1510.
I have attached my data file as well.
How do i calculate the slope of channels 1 to 15 indivijually and place the values in a different table and perhaps even plot to visually see it????
- Slope: the value that fits a regression line to the given data set.
- https://journals.plos.org/plosone/article/figure/image?size=medium&id=info:doi/10.1371/journal.pone.0208843.g007
Something similar to the url i have posted above.
Thank you
2 Commenti
Jan
il 4 Set 2019
The question is not clear. What are "channels 1 to 15"? Why did you highlight the first cell? What are "HbO" values? Where do we find the "time frame"?
Risposta accettata
Star Strider
il 4 Set 2019
First, negative values for haemoglobin or oxyhaemoglobin do not make sense physiologically.
I have no idea what you want to do, so start with:
D = load('HbO_Good_channels.mat');
HbO = D.HbO_good_channel;
Ts = 35/size(HbO,1); % Create A Sampling Interval, Since None Are Provided
T = linspace(0, size(HbO,1), size(HbO,1))*Ts; % Time Vector
lgdc = sprintfc('Ch %2d', 1:size(HbO,2)); % Legend String Cell Array (Channels)
figure
plot(T, HbO)
grid
xlabel('Time')
ylabel('HbO')
legend(lgdc, 'Location','eastoutside')
for k = 1:size(HbO,2)
cfs(k,:) = polyfit(T(:), HbO(:,k), 3); % Coefficient Vectors: ‘polyfit’
end
figure
hold all
for k = 1:size(HbO,2)
pf(:,k) = polyval(cfs(k,:), T(:)); % Evaluate Fitted Polynomials
plot(T, pf(:,k))
end
hold off
grid
xlabel('Time')
ylabel('Regression Fit')
legend(lgdc)
Experiment to get the resultl you want.
20 Commenti
Star Strider
il 20 Set 2019
I have no idea. As I mentioned before, I have very little recent experience with classification, and essentially no experience with SVM.
I suggest that you open a new Question on this.
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