Issue indexing by xarray's own time values + offset

[leifdenby]

I'm struggling to work out how to index by a xarray time value + an offset (either created using np.timedelta64 or datetime.timedelta). I read through #1240 and #1240 because they appear related, but I'm not sure how to correctly achieve this.

MCVE Code Sample

import xarray as xr
import numpy as np
import datetime as dt

now = dt.datetime.now()
dt_array = xr.DataArray(
   range(10), dims=('time', ),
   coords=dict(time=[now + dt.timedelta(seconds=i) for i in range(10)])
)

# this works
dt_array.loc[dt_array.time.min():dt_array.time.max()].count() == 10

# this fails, only the first value is returned (adding 
# the time delta appears to have no effect)
dt_array.loc[dt_array.time.min():dt_array.time.min() + np.timedelta64(seconds=4)].count() == 4

# this fails, an exception is raised when trying to add 
# a datetime.timedelta to the xarray value
dt_array.loc[dt_array.time.min():dt_array.time.max() + dt.timedelta(seconds=4)].count() == 4

# also fails, I got the impression from issue #1240 
# that `.loc[...]` should work for indexing too, but just to double-check
dt_array.sel(time=slice(dt_array.time.min(), dt_array.time.min() + np.timedelta64(seconds=4))).count() == 4

# fails, showing that adding a time increment has no effect
dt_array.time.min() + np.timedelta64(seconds=10) != dt_array.time.min()

Expected Output

Where I am indexing by the minimum time plus a np.timedelta64 offset of 4 seconds I would expect a DataArray of length 4 to be return. It would be nice if it was possible to add an increment with a native datetime.timedelta object.

Problem Description

I can't work out how to correctly add an increment to a time value in an xarray DataArray. It would be nice if one of the above approaches worked. Or maybe if I'm missing something obvious I could add an example to the documentation on datetime-indexing?

Versions

Output of xr.show_versions()

INSTALLED VERSIONS ------------------ commit: None python: 3.6.7 |Anaconda, Inc.| (default, Oct 23 2018, 19:16:44) [GCC 7.3.0] python-bits: 64 OS: Linux OS-release: 3.10.0-957.27.2.el7.x86_64 machine: x86_64 processor: x86_64 byteorder: little LC_ALL: None LANG: en_GB.UTF-8 LOCALE: en_GB.UTF-8 libhdf5: 1.10.1 libnetcdf: 4.5.0

xarray: 0.15.1
pandas: 0.25.3
numpy: 1.15.4
scipy: 1.1.0
netCDF4: 1.4.0
pydap: None
h5netcdf: 0.7.4
h5py: 2.10.0
Nio: None
zarr: None
cftime: 1.0.2.1
nc_time_axis: None
PseudoNetCDF: None
rasterio: None
cfgrib: None
iris: 2.2.0
bottleneck: None
dask: 0.20.0
distributed: 1.24.0
matplotlib: 2.2.3
cartopy: 0.16.0
seaborn: 0.9.0
numbagg: None
setuptools: 46.1.3
pip: 10.0.1
conda: None
pytest: 5.3.2
IPython: 7.1.1
sphinx: None

[keewis]

take a look at the result of np.timedelta64(seconds=4):

In [2]: np.timedelta64(seconds=4)
Out[2]: numpy.timedelta64(0)

In [3]: np.timedelta64(dt.timedelta(seconds=4))
Out[3]: numpy.timedelta64(4000000,'us')

In [4]: np.timedelta64(4, 's').astype("timedelta64[us]")
Out[4]: numpy.timedelta64(4000000,'us')

so you need to also specify the desired unit or cast a dt.timedelta object. I'm not sure why timedelta64 ignores the seconds kwarg, though.

[leifdenby]

Ah! It's an issue with how I am using np.timedelta64 then. I (stupidly) assumed that np.timedelta64 has the same call signature as datetime.timedelta. It appears that np.timedelta64 silently ignores any kwargs 😕

Anyway, I won't be making that mistake again. Thank you @keewis !