can someone explain this code?
显示 更早的评论
x1= (Subject5_leftfall_1(:,44)-Subject5_leftfall_1(10,44));
x1_Y=find(x1 >= mean(x1(10:300))+3*std(x1(10:300)));
plot(x1(x1_Y(1):length(x1)),'g')
S5_Peak_left_1= max(x1(x1_Y(1):x1_Y(1)+70));
S5_index_peak_left_1=find(x1(x1_Y(1):x1_Y(1)+70)==S5_Peak_left_1)
I don't know what x1_Y is trying to do. Also, I don't know what the last two lines of the code are trying to do.
采纳的回答
Geoff
2012-6-28
x1_Y is an array of indices. You're doing a boolean test on x1(10:300) to determine if a point is an outlier. That is, it is larger than (or equal to) 3 standard deviations over the mean. Actually, this is only testing outlying peaks - it's not testing standard deviations below the mean. Anyway, that's what find does. It takes a boolean expression and returns the index of every true value. So, if I had an array:
x = [ 0 1 1 0 0 0 1];
Then find(x) would return:
ans = [2 3 7]
Now, the last two lines are looking for the largest x1 value between the first detected outlier and 70 positions ahead. It then asks for the indices of any value equal to that in the same range. However, this is a little bogus. If there are two identical maximum peaks there, then find will return an array. I think this code intends to get a single index for the largest value. This is more correct:
[S5_Peak_left_1, S5_index_peak_left_1] = max( x1( x1_Y(1):x1_Y(1)+70 ) );
This code might still be wrong, depending on the reason for searching that 71-element range. Remember the index x1_Y is indexing into a slice of x1: (10:300), so if x1_Y(1) is 10, that means the actual position of that outlier in x1 is 20.
Hope that helps a little.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Logical 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
