I need to compute (and plot) a
histogram2d but my binning grid is rotated and also non-orthogonal.
A way of doing this could be to apply a transformation to my data so I get it into a cartesian system, compute my
histogram2d and then apply the inverse transformation.
Can this be done directly without this overhead transformation ?
I guess my question is: how do I define the
bins for my
histogram2d in this case ? (AFAIK,
histogram2d will only accept x and y aligned
My data is 2 huge lists of points (10k~100k each), the coordinates of which are given in a cartesian coordinate system (actually a projected CRS because these are real-world locations) but they are organized in a regular grid that is not aligned to X and Y axis (rotated) and that may or may not be orthogonal. The binning grid will be derived from it so it will be a (rotated) regular quadrilaterals grid.
I have seen that
matplotlib has a
QuadMesh object (see here) so I’m being hopeful but I’m not sure how to handle this in
Basically this is what I want to achieve:
After some testing, I came to the conclusion that the overhead of transforming the coordinates into a Cartesian grid to compute the histogram and back for plotting is acceptable. Matrix operations in
NumPy are fairly efficient and I can handle 115+ million points in less than 7 sec.
However, the “back” part can be handled by
Matplotlib directly with
imshow all accept a
transform keyword which can be used to plot the Cartesian data into the desired coordinates like so:
# set I, J, bins (in the Cartesian system) and cmap
# a, b, c, d, e, f are values of the transformation matrix
transform = matplotlib.transforms.Affine2D.from_values(a, b, c, f, d, e, f)
fig, ax = plt.subplots(figsize=figsize)
_, _, _, im = ax.hist2d(I, J, bins=bins, cmap=cmap, transform=transform + ax.transData)
It is not really much faster than handling the “back” conversion with
NumPy but it can make the code lighter as it only requires 1 additional line and 1 additional keyword.
imshow can be a little bit of a pain as it won’t update the display extent after using
ax.autoscale() and it handles coordinates as images or matrix so the
transform has to be adjusted accordingly. For these reasons, I prefer
sorry for posting on your thread but I have the exact same problem. In my case rotating the data instead of rotating the binning grid takes a lot of time because I need to do so for multiple angles.
If you find something on the topic rotating the binning grid / creating a non-orthogonal binning grid, could you post it here?