This won't work, since this contains all events, even those that are not in any bins in x. The most common case where this occurs is after slicing. You should be able to write a unit test that still shows the initial error, even after this fix: da['y', 0].bin(x=4). Sorry I misread. But you can just use x.bins.size().sum() to achieve correctly what you did by hand.

Instead of counting events, can you just use the min and max and check those? Do we get a similar error when there is just a single event?

Good, I'll use x.bins.size().sum() instead.

Instead of counting events, can you just use the min and max and check those?

I don't understand, do you mean min and max of the coordinate?

With a single event we get a binning where every edge has the same value.

import scipp as sc
import numpy as np
rng = np.random.default_rng(123)

lati = rng.random(10)
# modify a single entry to be in the 'lati' bin with index 1
lati[-1] = 1.5

da = sc.DataArray(
    data=sc.ones(sizes={"row": 10}, unit="counts"),
    coords={
        'longi': sc.array(dims=['row'], values=rng.random(10)),
        'lati': sc.array(dims=['row'], values=lati),
    }
)

empty_b = da.bin(lati=sc.array(dims=['lati'], values=[0.0, 1.2, 2.0]))['lati', 1]
empty_b.bin(longi=5)
# <scipp.DataArray>Dimensions: Sizes[longi:5, ] Coordinates: ... longi float64  [dimensionless]  (longi [bin-edge])  [0.518165, 0.518165, ..., 0.518165, 0.518165] Data: ...

Yes, I mean use what is computed below anyway, and check if the range is valid?

Aha okay, I'll try to do that instead.

or something like that?

Not sure this belongs in the same test?

It feels like handling the start and stop below may be simpler?

Now I wonder: If you give bad manual bounds (similar to the ones computed automatically with no events, e.g., not increasing), do we also get a weird error?

If we manually give bad edges like

table = sc.data.table_xyz(0)
table.bin(x=sc.array(dims=['x'], values=[3.0, 1.2, 2.0]))
# Raises: BinEdgeError: Bin edges in dim x must be sorted.

the error message makes a lot of sense.
Was that what you meant?

Yes, good 👍