contour | Contour of a Union of Iso-Oriented Rectangles

You can do this efficiently with a dictionary:

• Extract the index, idx, of the maximum value from each row of array XA
• Use idx on T and XA to extract the x-axis and y-axis values.
  • Indexing the array is slightly faster than using y = XA.max(axis=1) to get the max XA values.
• The shape of XA is (8, 120000), so there are 8 maximums. I'm not certain why only 7 contour lines are showing.
  • Use x[:-1] and y[:-1] to not plot the last point.

Restating the problem - My understanding given the different comments and your updates is the following:

• someone other than you is in possession of data, which as it happens is 2D data, i.e. an Nx2 matrix;
• using the covariance matrix, they are effectively saying something about the joint distribution of these two dimensions, specifically about the variance;
• if they assume a Gaussian distribution, as is implied by your comment regarding 68%, 95% and 99.7% for 1sigma, 2sigma and 3sigma, they can draw ellipses which represent the 2D-normal distribution: these are in fact some of the contour lines associated with the 3D "bell" surface;
• you have obtained the contour lines in a graph and are trying to obtain the covariance matrix (not the original data...);
• you are concerned about the complexity of having to extract the information from each ellipsis.

Partial answer:

• It is impossible to recover the original data, I hope you are already aware of that, but in case you are not let's just note that the covariance matrix is a summary statistic of the data, much like the average, and although it says something about the data many different datasets could happen to have the same summary statistic (the same way many different sets of numbers can give you an average of 10).
• It is possible to somewhat recover the covariance matrix, i.e. the 3 numbers a, b and c in the matrix [a,b;b,c], though the error in doing so will likely be large because of how imprecise the pixel representation is. Essentially, you will be looking for the dimensions of the two axes, for the variances, as well as the angle of one of the axes, for the covariance.
• Unless I am mistaken, under the Gaussian assumption above, you only need to measure this for one of the three ellipses, and then factor by whatever number of sigmas that contour represents. Here you might want to either use the best-defined ellipse, or attempt to use the largest one, which will provide the maximum precision for your measurements (cf. pixelization).
• Also, the problem of finding the axes and angle for the ellipse need not be as complex as what it seems like in your first trials: instead of trying to find the contour of the ellipses, find the bounding rectangle.
• In order to further simplify this process, if your images are color-coded the way you show, then a filter on blue pixels might be enough in terms of image processing. Then simply take the minimum and maximum (x,y) coordinates in order to obtain the bounding rectangle.
• Once the bounding rectangle is obtained, find the equation to your ellipse (that's a question for a math group, but you could start here for example).

Happy filtering!

It seems that you're looking for a tile plot. Here is an approach with ggplot2:

You need to set self.theta to be an array, not a scalar (at least in this specific problem).

In your case, (intercepted-augmented) X is a '3 by n' array, so try self.theta = [0, 0, 0] for example. This will correct the specific error 'bool' object has no attribute 'mean'. Still, this will just produce preds as a zero vector; you haven't fit the model yet.

To let you know how I approached the error, I first went to the exact line the error message was pointing to, and put print(preds == y) before the line, and it printed out False. I guess what you expected was a vector of True and Falses. Your y seemed okay; it was a vector (a list to be specific). So I tried print(pred), which showed me a '3 by n' array, which is weird. Now going up from that line, I found out that pred comes from predict_prob(), especially np.dot(X, self.theta). Here, when X is a '3 by n' array and self.theta is a scalar, numpy seems to multiply the scalar to each item in the array and return the array (having the same dimension as the original array), instead of doing matrix multiplication! So you need to explicitly provide self.theta as an array (conforming to the dimension of X).

Hope the answer and the reasoning behind it helped.

As for the red line you mentioned in the comment, I guess it is also because you are not fitting the model. (To see the problem, put print(probs) before plt.countour(...). You'll see an array with 0.5 only.)

So try putting model.fit(X, y) before preds = model.predict(X). (You'll also need to put self.verbose = verbose in the __init__().)

After that, I get the following:

The short answer is "yes": it is a good idea to have as many neurons in output as you have in input, i.e. to output an image with the same resolution of the input images.

The network architecture will have an input layer with a neuron for each pixel, then typically the hidden layers will shrink to less neurons, probably with convolutional layers, and then some more layers will re-expand the number of neurons, up to the output layer, which in principle may have the same number of neurons as the input layer.

The most common architecture in this type of problem is the U-net architecture, described in the article "U-Net: Convolutional Networks for Biomedical Image Segmentation", by Ronneberger, Fischer, and Brox, published on the open arxiv: https://arxiv.org/abs/1505.04597.

[

The cv2 pre-processing is unecessary here, tesseract is able to find the text on its own. See the example below, commented inline:

I found solution for it I changed the format of messages into required format of gifted chat it will not work fine until we modify our response into the gifted chat required format Here it is what i did

There are auxilliary keywords that can modify binary filetype=png w rgbimage. The ones you want are dx and dy (see "help binary keywords"). In your case the image is 2000x2000 pixels and each pixel represents an area 57599/2000 = 28.8, so the plot command becomes