遍历 xarray。 DataArray 第一维及其坐标

Question

Suppose I have the following DataArray

arr = xarray.DataArray(np.arange(6).reshape(2,3),
                        dims=['A', 'B'],
                        coords=dict(A=['a0', 'a1'], 
                                    B=['b0', 'b1', 'b2']))

I want to iterate over the first dimension and do the following (of course I want to do something more complex than printing)

for coor in arr.A.values:
    print(coor, arr.sel(A=coor).values)

and get

a0 [0 1 2]
a1 [3 4 5]

I am new to xarray , so I was wondering whether there was some more natural way to achieve this, something like

for coor, sub_arr in arr.some_method():
    print(coor, sub_arr)

Answer 1

You can simply iterate over the DataArray - each element of the iterator will itself be a DataArray with a single value for the first coordinate:

for a in arr:
    print(a.A.values, a.values)

prints

a0 [0 1 2]
a1 [3 4 5]

Note that here aAvalues is a zero-dimensional numpy array, use the .item() method to access the underlying scalar.

To iterate over the second dimension, you can just transpose the data:

for b in arr.transpose():  # or arr.T
    print(b.B.values, b.values)

prints

b0 [0 3]
b1 [1 4]
b2 [2 5]

For multidimensional data, you can move the dimension you want to iterate over to the first place using ellipsis:

for x in arr.transpose("B", ...):
    ...

The documentation on reshaping and reorganizing data has further details.

Answer 2

It's an old question, but I find that using groupby is cleaner and makes more intuitive sense to me than using transpose when you want to iterate some dimension other than the first:

for coor, sub_arr in arr.groupby('A'):
    print(coor)
    print(sub_arr)

a0
<xarray.DataArray (B: 3)>
array([0, 1, 2])
Coordinates:
  * B        (B) <U2 'b0' 'b1' 'b2'
    A        <U2 'a0'
a1
<xarray.DataArray (B: 3)>
array([3, 4, 5])
Coordinates:
  * B        (B) <U2 'b0' 'b1' 'b2'
    A        <U2 'a1'

Also it seems that older versions of xarray don't handle the ellipsis correctly (see mgunyho's answer), but groupby still works correctly.