issue_comments

Oh, I see the release of xgcm of 0.7.0. It is really a great update! I also find the boundary condition and grid_ufunc examples on the docs (still 0.6.0), which indeed may solve many of my problems. The grid-ufunc provides flexible building blocks for complicated cases. I'll spend some times trying the new version, re-think my cases in this great architecture, and report soon if I have problems with that. Thanks to you guys' great work!

A quite question is that has the xgcm been refactored using grid_ufunc? (I hope I could catch up with you guys).

Thanks to you guys here @Illviljan @TomNicholas @dcherian. I've been a user of xgcm for quite a time. So you can see my proposal just follows the style of xgcm.

I am working on my xinvert package, in which I may need some partial differential calculations. This can be done by xgcm quite well, but I am still worried about the metrics concept introduced by xgcm. I think this should be discussed over xgcm's repo.

For most of the cases, lat/lon-type grids are uniform and on the Arakawa A grid. So xarray's differentiate() is good enough with pad() (although it is experimental) for BCs, as suggested by @dcherian. We don't need stagged grid point and metrics, as in xgcm, but centered difference (a[i+1]-a[i-1]) will be good enough for A grid. This is simpler and do not make heavy dependence of the third-party package like xgcm.

I'll give a try with differentiate() and pad() to implement grad/div/vor... But some designs in xgcm also inspire me to make things much natural.

OK, thanks again. This is clear to me now. So this should be a matplotlib problem. I am closing this now.

Thanks @mathause , that works well. After searching the doc on infer_intervals, I still don't know why this is happening.

Thanks @dcherian. I use the get_group function and solve my problem. But a warning arises at the line if name not in self.groups: as: DeprecationWarning: elementwise comparison failed; this will raise an error in the future.

When the group contains nothing, I may assign nan or 0 to the corresponding bin. If it is dropped silently, then I need to find out which bin is missing its group. I like this PR #1027 by Ryan with a kwarg drop_empty_bins but not sure this option is eventually merged into master or not.

When I use counts = grouped.count(), I can get all the bins and counts in the correct order, and the empty bin just returns nan. But I don't want aggregation operations here. I need the samples in each group to index another array. Maybe this is not the best solution. But my case do need all the groups in the same order as the given bins, including possible empty ones.

Minor comment / nit: you don't really need the astype(np.float), the result should already be of dtype float since there are missing values after the rolling sum.

You're right. That is used for debuging the intermediate results.

Thanks again @keewis @max-sixty.

Hi @keewis, this is really a smart way, using rolling twice. I've refactored the code slightly as: ```python def continuous_meet(cond, count, dim): """ Continuously meet a given condition along a dimension. """ _found = cond.rolling(dim={dim:count}, center=True).sum().fillna(0).astype(np.float)

detected = (
    _found.rolling(dim={dim:count}, center=True) 
    .reduce(lambda a, axis: (a == count).any(axis=axis)) 
    .fillna(False) 
    .astype(bool) 
)

if count % 2 == 0:
    return detected.shift({dim:-1}).fillna(False)

return detected

sst = xr.DataArray( np.array( [0., 0., 0., 0., 0., 1., 1., 1., 1., 1., 0., 0., 1., 0., 0., 1., 1., 1., 1., 1., 1., 0., 0., 0.] ), dims="time", coords={"time": np.arange(24)}, name="sst", )

ElNino = continuous_meet(sst > 0.5, count=5, dim='time')

sst.plot.step(linewidth=3) sst.where(ElNino).plot.step(linewidth=2) `` Note that whencountis a even number, truly centeredrollingcannot be obtained. So we need toshift` the result by -1. Is this perfect? I didn't check the performance.

Hi @max-sixty, thanks for your kind help. But I found it works not as I expected. If the SST has the values [..., 0, 0, 1, 1, 1, 1, 1, 0, 0, ...], then the method you suggested will give [..., F, F, F, F, T, F, F, F, F, ...]. That is only one True for a 5 consecutive 1s. But I would expect five True like [..., F, F, T, T, T, T, T, F, F, ...]. Any suggestion?