order 参数仅适用于结构化数组:
In [383]: arr=np.zeros((10,),dtype='i,i')
In [385]: for i in range(10):
...: arr[i] = (i,10-i)
In [386]: arr
Out[386]:
array([(0, 10), (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)],
dtype=[('f0', '<i4'), ('f1', '<i4')])
In [387]: np.sort(arr, order=['f0','f1'])
Out[387]:
array([(0, 10), (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3), (8, 2), (9, 1)],
dtype=[('f0', '<i4'), ('f1', '<i4')])
In [388]: np.sort(arr, order=['f1','f0'])
Out[388]:
array([(9, 1), (8, 2), (7, 3), (6, 4), (5, 5), (4, 6), (3, 7), (2, 8),
(1, 9), (0, 10)],
dtype=[('f0', '<i4'), ('f1', '<i4')])
对于二维数组,lexsort 提供类似的“有序”排序
In [402]: arr=np.column_stack((np.arange(10),10-np.arange(10)))
In [403]: np.lexsort((arr[:,1],arr[:,0]))
Out[403]: array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9], dtype=int32)
In [404]: np.lexsort((arr[:,0],arr[:,1]))
Out[404]: array([9, 8, 7, 6, 5, 4, 3, 2, 1, 0], dtype=int32)
使用您的对象数组,我可以将属性提取到以下任一结构中:
In [407]: np.array([(a.a, a.b) for a in arr])
Out[407]:
array([[ 0, 10],
[ 1, 9],
[ 2, 8],
....
[ 7, 3],
[ 8, 2],
[ 9, 1]])
In [408]: np.array([(a.a, a.b) for a in arr],dtype='i,i')
Out[408]:
array([(0, 10), (1, 9), (2, 8), (3, 7), (4, 6), (5, 5), (6, 4), (7, 3),
(8, 2), (9, 1)],
dtype=[('f0', '<i4'), ('f1', '<i4')])
Python sorted 函数将适用于 arr(或其等效列表)
In [421]: arr
Out[421]:
array([<__main__.Obj object at 0xb0f2d24c>,
<__main__.Obj object at 0xb0f2dc0c>,
....
<__main__.Obj object at 0xb0f35ecc>], dtype=object)
In [422]: sorted(arr, key=lambda a: (a.b,a.a))
Out[422]:
[<__main__.Obj at 0xb0f35ecc>,
<__main__.Obj at 0xb0f3570c>,
...
<__main__.Obj at 0xb0f2dc0c>,
<__main__.Obj at 0xb0f2d24c>]
您的 Obj 类缺少一个不错的 __str__ 方法。我必须使用 [(i.a, i.b) for i in arr] 之类的东西来查看 arr 元素的值。
正如我在评论中所说,对于这个例子,列表比对象数组好得多。
In [423]: alist=[]
In [424]: for i in range(10):
...: alist.append(Obj(i,10-i))
list append 比重复的数组追加更快。与列表相比,对象数组并没有增加太多功能,尤其是在 1d 时,并且对象是这样的自定义类。您无法对 arr 进行任何数学运算,而且正如您所见,排序并不容易。