nda.shape为(2,2),将其转换为(2,2,2)
dtypes = [('a', np.float64), ('b', object)]
nda = np.zeros((2,2), dtype = dtypes)
nda['b'][0,0] = [1,2]
nda['b'][1,0] = [2,3]
nda['b'][0,1] = [3,4]
nda['b'][1,1] = [9,5]
解决方案应提供:nda['b'][0,0,1]==2
,nda['b'][1,1,0]==9
等
尝试以下操作
nda = np.resize(nda, (2,2,2))
nda.shape
结果
(2,2,2)
您创建了一个奇怪的结构;您不能简单地重塑它:
In [1]: dtypes = [('a', np.float64), ('b', object)]
...: nda = np.zeros((2,2), dtype = dtypes)
...:
...: nda['b'][0,0] = [1,2]
...: nda['b'][1,0] = [2,3]
...: nda['b'][0,1] = [3,4]
...: nda['b'][1,1] = [9,5]
它有2个字段,一个带有数字,另一个带有列表:
In [2]: nda
Out[2]:
array([[(0., list([1, 2])), (0., list([3, 4]))],
[(0., list([2, 3])), (0., list([9, 5]))]],
dtype=[('a', '<f8'), ('b', 'O')])
列表字段:
In [3]: nda['b']
Out[3]:
array([[list([1, 2]), list([3, 4])],
[list([2, 3]), list([9, 5])]], dtype=object)
In [4]: _.shape
Out[4]: (2, 2)
如果转换为1d,我们可以stack
(或与concatenate
结合使用:]
In [5]: nda['b'].ravel()
Out[5]:
array([list([1, 2]), list([3, 4]), list([2, 3]), list([9, 5])],
dtype=object)
In [6]: np.stack(nda['b'].ravel())
Out[6]:
array([[1, 2],
[3, 4],
[2, 3],
[9, 5]])
In [7]: np.stack(nda['b'].ravel()).reshape(2,2,2)
Out[7]:
array([[[1, 2],
[3, 4]],
[[2, 3],
[9, 5]]])
[通常,如果您具有列表或数组的对象dtype数组,则可以将其合并为一个具有concatenate
排序版本的(数字)数组,但必须为1d,即数组/列表的“可迭代”。] >
而且,是的,将字段解压缩到嵌套列表中会产生一些可以转换回(2,2,2)数组的东西:
In [14]: _2['b'].tolist() Out[14]: [[[1, 2], [3, 4]], [[2, 3], [9, 5]]]
((您不能简单地将这些数组(或列表)放回
nda
数组。dtype错误。)
具有不同的dtype
(`b字段是2个整数,不是更通用的对象):
In [10]: dtypes = [('a', np.float64), ('b', int, (2,))]
...: nda = np.zeros((2,2), dtype = dtypes)
...:
...: nda['b'][0,0] = [1,2]
...: nda['b'][1,0] = [2,3]
...: nda['b'][0,1] = [3,4]
...: nda['b'][1,1] = [9,5]
In [11]: nda
Out[11]:
array([[(0., [1, 2]), (0., [3, 4])],
[(0., [2, 3]), (0., [9, 5])]],
dtype=[('a', '<f8'), ('b', '<i8', (2,))])
In [12]: nda['b']
Out[12]:
array([[[1, 2],
[3, 4]],
[[2, 3],
[9, 5]]])