# NumPy histogram2d with a rotated, non-orthogonal binning grid

## Question:

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 `bins`

)

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 `NumPy`

.

Basically this is what I want to achieve:

## Answers:

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 `matplotlib.transforms`

.

`pcolormesh`

, `hist2d`

and `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)
fig.colorbar(im)
ax.autoscale()
```

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 `hist2d`

.

References:

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?