i wanted the slope with respect to time frame

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CalebJones 2019 年 9 月 4 日
コメント済み: Star Strider 2019 年 9 月 20 日
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????
Something similar to the url i have posted above.
Thank you
  2 件のコメント
CalebJones 2019 年 9 月 4 日
Jan HbO is Heomoglobin values. Highlighted 1st column is HbO amplitude for channel 1 which is outputed from the machine.
Time frame is row index which starts at 1 and ends at 1510.
So 15 columns shows HbO amplitude of 15 channels.
So i wanted to perform polyfit func on curve from channel 1.
So one by one i wanted to calculate the slope of each channel so.



Star Strider
Star Strider 2019 年 9 月 4 日
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)
plot(T, HbO)
legend(lgdc, 'Location','eastoutside')
for k = 1:size(HbO,2)
cfs(k,:) = polyfit(T(:), HbO(:,k), 3); % Coefficient Vectors: ‘polyfit’
hold all
for k = 1:size(HbO,2)
pf(:,k) = polyval(cfs(k,:), T(:)); % Evaluate Fitted Polynomials
plot(T, pf(:,k))
hold off
ylabel('Regression Fit')
Experiment to get the resultl you want.
  20 件のコメント
Star Strider
Star Strider 2019 年 9 月 20 日
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.


その他の回答 (1 件)

Jan 2019 年 9 月 4 日
編集済み: Jan 2019 年 9 月 4 日
Maybe all you need is to call the gradient(X.') function, where X is the complete matrix?

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