matlab | matlab for python

Since both columns are pandas Timestamp, you can do this:

So, if your file looks anything like this:

You're probably looking for a lookup table:

Yes. The colon operator is a "select all, from this index" operator. Note that in the first line, its unnecesary as long as errors was already 2D, as "select the entire variable" is not something you need to specify, its by default. So it could have been

errors = (errors*trials + 0.5)./(trials + 1);.

If errors had more dimensions than 2, then the colon operator in the right hand side of the equal sign is doing something, in particular "A(:,:) reshapes all elements of A into a two-dimensional matrix. This has no effect if A is already a matrix or vector.". The one in the left hand side is useless anyway, as that line overwrites the variable.

The use of the colon operator in the second line however is well justified.

There are few other things that the colon operator means in MATLAB, read them all here: https://uk.mathworks.com/help/matlab/ref/colon.html

In MATLAB, manipulating arrays and matrices can be done much easier than for-loop solutions, in your task, can do find and delete negative value in the array, simply, as follows:

I added a loop within the "hold" portion of your code.

Given the current comments, you are trying to understand how two series of random draws from two distributions can have zero correlation. Specifically, exercise 4.6.9 to which you refer mentions draws from two normal distributions.

An issue with your approach is that you are hoping to derive a link between a theoretical property and experimentation, in this case using Matlab. And, as you seem to have noticed, unless you are looking at specific degenerate cases, your experimentation will fail. That is because although the true correlation parameter rho in the exercise might be zero, a sample of random draws will always have some level of correlation. Here is an illustration, and as you'll notice if you run it the actual correlations span the whole spectrum between -1 and 1 despite their average being zero (as it should be since both generators are pseudo-uncorrelated):

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!

This works without the sprintf which is returning an error: