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Let me exhaust a few other ideas first. I'll definitely share my thoughts here first before making any commit. Thanks.

I'm fine with this approach for now. It would be great if we could convince bottleneck to help us out with a keyword argument of some kind.

@chunweiyuan I agree, this seems worth doing, and I think you have a pretty sensible approach here. For large arrays (especially with ndim > 1), this should add only minimal performance overhead. If you can fit this into the existing framework for rolling that would be awesome!

My opinion is that the nan has got to go. If we want to (1) maintain pandas-consistency and (2) use bottleneck without mucking it up, then I think we need to add some logic in either rolling.reduce() or rolling._center_result().

So here's my failed attempt:

``` def reverse_and_roll_1d(data, window_size, min_periods=1): """ Implements a concept to fix the end-of-array problem with xarray.core.rolling._center_shift(), by 1.) take slice of the back-end of the array 2.) flip it 3.) compute centered-window arithmetic 4.) flip it again 5.) replace back-end of default result with (4)

:param DataArray data: 1-D data array, with dim name 'x'.
:param int window_size: size of window.
"""
# first the default way to computing centered window
r = data.rolling(x=window_size, center=True, min_periods=min_periods)
avg = r.mean()
# now we need to fix the back-end of the array
rev_start = len(data.x) # an index
rev_end = len(data.x) - window_size - 1 \
                 if len(data.data) > window_size \
                else None  # another index
tail_slice = slice(rev_start, rev_end, -1) # back end of array, flipped
r2 = data[dict(x=tail_slice)].\
    rolling(x=window_size, center=True, min_periods=min_periods)
avg[dict(x=slice(-window_size+1, None))] = \
    r2.mean()[dict(x=slice(window_size-2, None, -1))] # replacement

return avg

```

This algorithm is consistently 8 times slower than pd.DataFrame.rolling(), for various 1d array sizes.

I'm open to ideas as well :)

We do try to stay consistent with pandas except for the last position. Here's the unit test where we verify that behavior.

Using x=0 from your example in Pandas:

``` Python In [1]: import pandas as pd s In [2]: data = pd.Series([0, 3, 6])

In [3]: data.rolling(3, center=True, min_periods=1).mean() Out[3]: 0 1.5 1 3.0 2 4.5 ```

If I remember correctly, and my brain is a bit like mush right now so I could be wrong, bottleneck and pandas handle this case differently so we had to make a decision. We choose to use bottleneck (for speed) but to do our best to stay consistent with pandas. Back to your example, this time just with bottleneck:

Python In [4]: bn.move_mean(data, 3, min_count=1) Out[4]: array([ 0. , 1.5, 3. ])

So, as you can see, bottleneck does something totally different that wouldn't otherwise work with center=True unless we did our little shift trick. I'm not really sure the best way to correct for this difference in the last position except to either a) try to push a center=True option into bottleneck (may not be possible), or b) write a bunch of logic on our end bridge the gap between these two (may be laborious). Of course, I'm open to ideas.

I think we mostly tried to make this consistent with pandas. To be honest I don't entirely understand the logic myself.

Cc @jhamman