I strongly suspect that your data, while on a grid, is not ordered so as to allow a simple reshape of the values. You have two solutions available, both involving reordering the data in different ways.
Solution 1
Since you’re already using np.unique
to extract the grid, you can get the correct ordering of vs
using the return_inverse
parameter:
px, ix = np.unique(xs, return_inverse=True)
py, iy = np.unique(ys, return_inverse=True)
pz, iz = np.unique(zs, return_inverse=True)
points = (px, py, pz)
values = np.empty_like(vs, shape=(px.size, py.size, pz.size))
values[ix, iy, iz] = vs
return_inverse
is sort of magical, largely because it’s so counterintuitive. In this case, for each element of values, it tells you which unique, sorted gross location it corresponds to.
By the way, if you are missing grid elements, you may want to replace np.empty_like(vs, shape=(px.size, py.size, pz.size))
with either np.zeros_like(vs, shape=(px.size, py.size, pz.size))
or np.empty_like(vs, np.nan, shape=(px.size, py.size, pz.size))
. In the latter case, you could interpolate the nan
s in the grid first.
Solution 2
The more obvious solution would be to rearrange the indices so you can reshape vs
as you tried to do. That only works if you’re sure that there are no missing grid elements. The easiest way would be to sort the whole dataframe, since the pandas methods are less annoying than np.lexsort
(IMO):
df.sort_values(['x', 'y', 'z'], inplace=True, ignore_index=True)
When you extract, do it efficiently:
xs, ys, zs, vs = df.to_numpy().T
Since everything is sorted, you don’t need np.unique
to identify the grid any more. The number of unique x
values is:
nx = np.count_nonzero(np.diff(xs)) + 1
And the unique values are:
bx = xs.size // nx
ux = xs[::bx]
y
values go through a full cycle every bx
elements, so
ny = np.count_nonzero(np.diff(ys[:bx])) + 1
by = bx // ny
uy = ys[:bx:by]
And for z
(bz == 1
):
nz = by
uz = zs[:nz]
Now you can construct your original arrays:
points = (ux, uy, uz)
values = vs.reshape(nx, ny, nz)
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