numpy sort a structured array by two fields, ascending and descending orders

Question

I have this line of code:

TableArr = numpy.sort(TableArr, order=['destID','ATTRACT'])

I need the 'ATTRACT' order to be descending while the destID to be ascending, which is default for both. Attempts with [::-1] failed, as it reverses the entire array.

I'll add that a reverse argument that can be set to True or False was mentioned elsewhere, but it doesn't work for me and isn't mentioned in the Numpy's documentation.

Answer 1

Here's an interactive session that tests the idea in my comment:

In [1]: import numpy as np

In [2]: dt = np.dtype([('destID',int),('ATTRACT',float),('other','S10')])
In [3]: TableArr=np.zeros((10,),dt)
In [5]: TableArr['destID']=np.random.randint(10,size=(10,))
In [6]: TableArr['ATTRACT']=np.random.randint(100,size=(10,))

In [7]: TableArr
Out[7]: 
array([(2, 39.0, b''), (7, 7.0, b''), (8, 74.0, b''), (5, 83.0, b''),
       (5, 3.0, b''), (9, 26.0, b''), (8, 9.0, b''), (3, 1.0, b''),
       (1, 67.0, b''), (7, 5.0, b'')], 
      dtype=[('destID', '<i4'), ('ATTRACT', '<f8'), ('other', 'S10')])

In [13]: Tcopy=TableArr[['destID','ATTRACT']].copy()
# use copy() to avoid a FutureWarning

In [14]: Tcopy['ATTRACT'] *= -1  # 'reverse' a field

In [16]: I=np.argsort(Tcopy,order=['destID','ATTRACT'])

In [17]: I
Out[17]: array([8, 0, 7, 3, 4, 1, 9, 2, 6, 5], dtype=int32)

In [18]: TableArr[I]
Out[18]: 
array([(1, 67.0, b''), (2, 39.0, b''), (3, 1.0, b''), (5, 83.0, b''),
       (5, 3.0, b''), (7, 7.0, b''), (7, 5.0, b''), (8, 74.0, b''),
       (8, 9.0, b''), (9, 26.0, b'')], 
      dtype=[('destID', '<i4'), ('ATTRACT', '<f8'), ('other', 'S10')])

The integers are increasing, and for the 3 cases where they tie, the floats are decreasing. So it works.