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I want to find the maximum value using the second derivative of the the expression when x is between 0 and 1. In other words I am taking the derivative of cox(x^2) twice to get the second derivative resulting in - 2*sin(x^2) - 4*x^2*cos(x^2), then I want to evaluate this second derivative at x = 0 to x = 1, and display the maximum value of the populated values.
I have:
syms x
f = cos(x^2);
secondD = diff(diff(f));
for i = 0:1
y = max(secondD(i))
end
Can someone help?
You can do it easily by subs and double:
syms x
f = cos(x^2);
secondD = diff(diff(f));
% instead of the for loop
epsilon = 0.01;
specified_range = 0:epsilon:1;
[max_val, max_ind] = max(double(subs(secondD, specified_range)));
Please note that it is a numerical approach to find the maximum and the returned answer is not completely correct all the time. However, by increasing the epsilon, you can expect a better result in general (again in some cases it is not completely correct).
I'm trying to use GPflow for a multidimensional regression. But I'm confused by the shapes of the mean and variance.
For example: A 2-dimensional input space X of shape (20,20) is supposed to be predicted. My training samples are of shape (8,2) which means 8 training samples overall for the two dimensions. The y-values are of shape (8,1) which of course means one value of the ground truth per combination of the 2 input dimensions.
If I now use model.predict_y(X) I would expect to receive a mean of shape (20,20) but obtain a shape of (20,1). Same goes for the variance. I think that this problem comes from the shape of the y-values but I have have no idea how to fix it.
bound = 3
num = 20
X = np.random.uniform(-bound, bound, (num,num))
print(X_sample.shape) # (8,2)
print(Y_sample.shape) # (8,1)
k = gpflow.kernels.RBF(input_dim=2)
m = gpflow.models.GPR(X_sample, Y_sample, kern=k)
m.likelihood.variance = sigma_n
m.compile()
gpflow.train.ScipyOptimizer().minimize(m)
mean, var = m.predict_y(X)
print(mean.shape) # (20, 1)
print(var.shape) # (20, 1)
It sounds like you may be confused between the shape of a grid of input positions and the shape of the numpy arrays: if you want to predict on a 20 x 20 grid in two dimensions, you have 400 points in total, each with 2 values. So X (the one that you pass to m.predict_y()) should have shape (400, 2). (Note that the second dimension needs to have the same shape as X_sample!)
To construct this array of shape (400,2) you can use np.meshgrid (e.g., see What is the purpose of meshgrid in Python / NumPy?).
m.predict_y(X) only predicts the marginal variance at each test point, so the returned mean and var both have shape (400,1) (same length as X). You can of course reshape them to the 20 x 20 values on your grid.
(It is also possible to compute the full covariance, for the latent f this is implemented as m.predict_f_full_cov, which for X of shape (400,2) would return a 400x400 matrix. This is relevant if you want consistent samples from the GP, but I suspect that goes well beyond this question.)
I was indeed making the mistake to not flatten the arrays which in return produced the mistake. Thank you for the fast response STJ!
Here is an example of the working code:
# Generate data
bound = 3.
x1 = np.linspace(-bound, bound, num)
x2 = np.linspace(-bound, bound, num)
x1_mesh,x2_mesh = np.meshgrid(x1, x2)
X = np.dstack([x1_mesh, x2_mesh]).reshape(-1, 2)
z = f(x1_mesh, x2_mesh) # evaluation of the function on the grid
# Draw samples from feature vectors and function by a given index
size = 2
np.random.seed(1991)
index = np.random.choice(range(len(x1)), size=(size,X.ndim), replace=False)
samples = utils.sampleFeature([x1,x2], index)
X1_sample = samples[0]
X2_sample = samples[1]
X_sample = np.column_stack((X1_sample, X2_sample))
Y_sample = utils.samplefromFunc(f=z, ind=index)
# Change noise parameter
sigma_n = 0.0
# Construct models with initial guess
k = gpflow.kernels.RBF(2,active_dims=[0,1], lengthscales=1.0,ARD=True)
m = gpflow.models.GPR(X_sample, Y_sample, kern=k)
m.likelihood.variance = sigma_n
m.compile()
#print(X.shape)
mean, var = m.predict_y(X)
mean_square = mean.reshape(x1_mesh.shape) # Shape: (num,num)
var_square = var.reshape(x1_mesh.shape) # Shape: (num,num)
# Plot mean
fig = plt.figure(figsize=(16, 12))
ax = plt.axes(projection='3d')
ax.plot_surface(x1_mesh, x2_mesh, mean_square, cmap=cm.viridis, linewidth=0.5, antialiased=True, alpha=0.8)
cbar = ax.contourf(x1_mesh, x2_mesh, mean_square, zdir='z', offset=offset, cmap=cm.viridis, antialiased=True)
ax.scatter3D(X1_sample, X2_sample, offset, marker='o',edgecolors='k', color='r', s=150)
fig.colorbar(cbar)
for t in ax.zaxis.get_major_ticks(): t.label.set_fontsize(fontsize_ticks)
ax.set_title("$\mu(x_1,x_2)$", fontsize=fontsize_title)
ax.set_xlabel("\n$x_1$", fontsize=fontsize_label)
ax.set_ylabel("\n$x_2$", fontsize=fontsize_label)
ax.set_zlabel('\n\n$\mu(x_1,x_2)$', fontsize=fontsize_label)
plt.xticks(fontsize=fontsize_ticks)
plt.yticks(fontsize=fontsize_ticks)
plt.xlim(left=-bound, right=bound)
plt.ylim(bottom=-bound, top=bound)
ax.set_zlim3d(offset,np.max(z))
which leads to (red dots are the sample points drawn from the function). Note: Code not refactored what so ever :)
I found that calculating an index to specify outliers of a dataset according to how the boxplot works does not give the same results. Please find below an example where I create some data, extract the values from the boxplot (as seen in datatips in the figure window) and compare them to the values I calculated.
While the median and quartiles match up the upper and lower adjacent values do not. According to the Matlab help under 'Whisker', the adjacent values are calculated as q3 + w*(q3-q1) where q3 and q1 are the quantiles and w is the specified whisker length.
Am I calculating this wrong or is there any other mistake? I would like to be able to explain the error.
Screenshot of results table (please note the results vary due to random data)
%Boxplot test
% create random, normally distributed dataset
data = round(randn(1000,1)*10,2);
figure(10)
clf
boxplot(data,'Whisker',1.5)
clear stats tmp
% read data from boxplot, same values as can be seen in datatips in the figure window
h = findobj(gcf,'tag','Median');
tmp = get(h,'YData');
stats(1,1) = tmp(1);
h = findobj(gcf,'tag','Box');
tmp = get(h,'YData');
stats(1,2) = tmp(1);
stats(1,3) = tmp(2);
h = findobj(gcf,'tag','Upper Adjacent Value');
tmp = get(h,'YData');
stats(1,4) = tmp(1);
h = findobj(gcf,'tag','Lower Adjacent Value');
tmp = get(h,'YData');
stats(1,5) = tmp(1);
% calculated data
stats(2,1) = median(data);
stats(2,2) = quantile(data,0.25);
stats(2,3) = quantile(data,0.75);
range = stats(2,3) - stats(2,2);
stats(2,4) = stats(2,3) + 1.5*range;
stats(2,5) = stats(2,2) - 1.5*range;
% error calculation
for k=1:size(stats,2)
stats(3,k) = stats(2,k)-stats(1,k);
end %for k
% convert results to table with labels
T = array2table(stats,'VariableNames',{'Median','P25','P75','Upper','Lower'}, ...
'RowNames',{'Boxplot','Calculation','Error'});
While the calculation of the boundaries, e.g. q3 = q3 + w*(q3-q1), is correct, it is not displayed in the boxplot. What is actually displayed and marked as upper/lower adjacent value is the minimum and maximum of the values within the aforementioned boundaries.
Regarding the initial task leading to the question: For applying the same filtering of outliers as in the boxplot the calculated boundaries can be used.
Is there some simple way of calculating of p-value of t-Test in MATLAB.
I found something like it however I think that it does not return correct values:
Pval=2*(1-tcdf(abs(t),n-2))
I want to calculate the p-value for the test that the slope of regression is equal to 0. Therefore I calculate the Standard Error
$SE= \sqrt{\frac{\sum_{s = i-w }^{i+w}{(y_{s}-\widehat{y}s})^2}{(w-2)\sum{s=i-w}^{i+w}{(x_{s}-\bar{x}})^2}}$
where $y_s$ is the value of analyzed parameter in time period $s$,
$\widehat{y}_s$ is the estimated value of the analyzed parameter in time period $s$,
$x_i$ is the time point of the observed value of the analysed parameter,
$\bar{x}$ is the mean of time points from analysed period and then
$t_{score} = (a - a_{0})/SE$ where $a_{0}$ where $a_{0} = 0$.
I checked that p values from ttest function and the one calculated using this formula:
% Let n be your sample size
% Let v be your degrees of freedom
% Then:
pvalues = 2*(1-tcdf(abs(t),n-v))
and they are the same!
Example with Matlab demo dataset:
load accidents
x = hwydata(:,2:3);
y = hwydata(:,4);
stats = regstats(y,x,eye(size(x,2)));
fprintf('T stat using built-in function: \t %.4f\n', stats.tstat.t);
fprintf('P value using built-in function: \t %.4f\n', stats.tstat.pval);
fprintf('\n\n');
n = size(x,1);
v = size(x,2);
b = x\y;
se = diag(sqrt(sumsqr(y-x*b)/(n-v)*inv(x'*x)));
t = b./se;
p = 2*(1-tcdf(abs(t),n-v));
fprintf('T stat using own calculation: \t\t %.4f\n', t);
fprintf('P value using own calculation: \t\t %.4f\n', p);
I am trying to write my own function for scaling up an input image by using the Nearest-neighbor interpolation algorithm. The bad part is I am able to see how it works but cannot find the algorithm itself. I will be grateful for any help.
Here's what I tried for scaling up the input image by a factor of 2:
function output = nearest(input)
[x,y]=size(input);
output = repmat(uint8(0),x*2,y*2);
[newwidth,newheight]=size(output);
for i=1:y
for j=1:x
xloc = round ((j * (newwidth+1)) / (x+1));
yloc = round ((i * (newheight+1)) / (y+1));
output(xloc,yloc) = input(j,i);
end
end
Here is the output after Mark's suggestion
This answer is more explanatory than trying to be concise and efficient. I think gnovice's solution is best in that regard. In case you are trying to understand how it works, keep reading...
Now the problem with your code is that you are mapping locations from the input image to the output image, which is why you are getting the spotty output. Consider an example where input image is all white and output initialized to black, we get the following:
What you should be doing is the opposite (from output to input). To illustrate, consider the following notation:
1 c 1 scaleC*c
+-----------+ 1 +----------------------+ 1
| | | | | |
|----o | <=== | | |
| (ii,jj) | |--------o |
+-----------+ r | (i,j) |
inputImage | |
| |
+----------------------+ scaleR*r
ouputImage
Note: I am using matrix notation (row/col), so:
i ranges on [1,scaleR*r] , and j on [1,scaleC*c]
and ii on [1,r], jj on [1,c]
The idea is that for each location (i,j) in the output image, we want to map it to the "nearest" location in the input image coordinates. Since this is a simple mapping we use the formula that maps a given x to y (given all the other params):
x-minX y-minY
--------- = ---------
maxX-minX maxY-minY
in our case, x is the i/j coordinate and y is the ii/jj coordinate. Therefore substituting for each gives us:
jj = (j-1)*(c-1)/(scaleC*c-1) + 1
ii = (i-1)*(r-1)/(scaleR*r-1) + 1
Putting pieces together, we get the following code:
% read a sample image
inputI = imread('coins.png');
[r,c] = size(inputI);
scale = [2 2]; % you could scale each dimension differently
outputI = zeros(scale(1)*r,scale(2)*c, class(inputI));
for i=1:scale(1)*r
for j=1:scale(2)*c
% map from output image location to input image location
ii = round( (i-1)*(r-1)/(scale(1)*r-1)+1 );
jj = round( (j-1)*(c-1)/(scale(2)*c-1)+1 );
% assign value
outputI(i,j) = inputI(ii,jj);
end
end
figure(1), imshow(inputI)
figure(2), imshow(outputI)
A while back I went through the code of the imresize function in the MATLAB Image Processing Toolbox to create a simplified version for just nearest neighbor interpolation of images. Here's how it would be applied to your problem:
%# Initializations:
scale = [2 2]; %# The resolution scale factors: [rows columns]
oldSize = size(inputImage); %# Get the size of your image
newSize = max(floor(scale.*oldSize(1:2)),1); %# Compute the new image size
%# Compute an upsampled set of indices:
rowIndex = min(round(((1:newSize(1))-0.5)./scale(1)+0.5),oldSize(1));
colIndex = min(round(((1:newSize(2))-0.5)./scale(2)+0.5),oldSize(2));
%# Index old image to get new image:
outputImage = inputImage(rowIndex,colIndex,:);
Another option would be to use the built-in interp2 function, although you mentioned not wanting to use built-in functions in one of your comments.
EDIT: EXPLANATION
In case anyone is interested, I thought I'd explain how the solution above works...
newSize = max(floor(scale.*oldSize(1:2)),1);
First, to get the new row and column sizes the old row and column sizes are multiplied by the scale factor. This result is rounded down to the nearest integer with floor. If the scale factor is less than 1 you could end up with a weird case of one of the size values being 0, which is why the call to max is there to replace anything less than 1 with 1.
rowIndex = min(round(((1:newSize(1))-0.5)./scale(1)+0.5),oldSize(1));
colIndex = min(round(((1:newSize(2))-0.5)./scale(2)+0.5),oldSize(2));
Next, a new set of indices is computed for both the rows and columns. First, a set of indices for the upsampled image is computed: 1:newSize(...). Each image pixel is considered as having a given width, such that pixel 1 spans from 0 to 1, pixel 2 spans from 1 to 2, etc.. The "coordinate" of the pixel is thus treated as the center, which is why 0.5 is subtracted from the indices. These coordinates are then divided by the scale factor to give a set of pixel-center coordinates for the original image, which then have 0.5 added to them and are rounded off to get a set of integer indices for the original image. The call to min ensures that none of these indices are larger than the original image size oldSize(...).
outputImage = inputImage(rowIndex,colIndex,:);
Finally, the new upsampled image is created by simply indexing into the original image.
MATLAB has already done it for you. Use imresize:
output = imresize(input,size(input)*2,'nearest');
or if you want to scale both x & y equally,
output = imresize(input,2,'nearest');
You just need a more generalized formula for calculating xloc and yloc.
xloc = (j * (newwidth+1)) / (x+1);
yloc = (i * (newheight+1)) / (y+1);
This assumes your variables have enough range for the multiplication results.