使用NaN值反向排序Numpy数组

Question

I have a numpy array with some NaN values:

>>> a
array([  1.,  -1.,   nan,  0.,  nan], dtype=float32)

I can sort it in ascending or 'descending' order:

>>> numpy.sort(a)
array([ -1.,   0.,   1.,  nan,  nan], dtype=float32)
>>> numpy.sort(a)[::-1]
array([ nan,  nan,   1.,   0.,  -1.], dtype=float32)

However, what I want is descending order with NaN values at the end, like this:

>>> numpy.genuine_reverse_sort(a)
array([ 1.,   0.,   -1.,  nan,  nan], dtype=float32)

How could this be accomplished? I suspect that there is no special method for this.

Answer 1

I think you're probably right -- there is no built in special method to do this. But you could do it in two steps as follows by rolling your NaNs into the place you want them:

a = np.array([  1.,  -1.,   np.nan,  0.,  np.nan], dtype=np.float32)
sa = np.sort(a)[::-1]
np.roll(sa,-np.count_nonzero(np.isnan(a)))

array([  1.,   0.,  -1.,  nan,  nan], dtype=float32)

Answer 2

You can do the following:

>>> np.concatenate((np.sort(a[~np.isnan(a)])[::-1], [np.nan] * np.isnan(a).sum()))
array([ 1.,   0.,   -1.,  nan,  nan])

With this fragment you reverse sort the numerical entries of the input array, and then you concatenate with the appropriate number of nan values.

Answer 3

You can use np.argpartition to sort only the non-NaNs , like so -

a[np.argpartition(-a, np.arange((~np.isnan(a)).sum()) )]

Sample run -

In [253]: a
Out[253]: array([  1.,  -1.,  nan,   0.,  nan,   2.,   4.,  -2., -10.,  nan])

In [254]: a[np.argpartition(-a, np.arange((~np.isnan(a)).sum()) )]
Out[254]: array([  4.,   2.,   1.,   0.,  -1.,  -2., -10.,  nan,  nan,  nan])

Answer 4

What about negating the values twice:

>>> a = np.array([2., -1., nan,  0., nan])
>>> np.sort(a)
array([ -1.,   0.,   2.,  nan,  nan])
>>> -np.sort(-a)
array([  2.,   0.,  -1.,  nan,  nan])