将值分配给NumPy数组

Question

有人可以向我解释为什么尝试#1不起作用？

import numpy as np    
x = np.zeros(1, dtype=np.dtype([('field', '<f8', (1,2))]))

尝试#1:

x[0]['field'] = np.array([3.,4.], dtype=np.double)
print x, '\n'

[([[ 3. 0.]])](为什么只'3'复制了？)

尝试#2:

x['field'][0] = np.array([3.,4.], dtype=np.double)
print x

[([[ 3. 4.]])] (这有效)

Answer 1

老实说...我也不确定我是否得到了结果。看起来不一致/损坏。部分原因是形状不一致，但并非全部。有些数据似乎正在消失。

例如（注意形状）：

In [1]: import numpy as np

In [2]: x = np.zeros(1, dtype=np.dtype([('field', '<f8', (1, 2))]))

In [3]: y = x[0]['field'].copy()

In [4]: y[0] = 3

In [5]: y[1] = 4
---------------------------------------------------------------------------
IndexError                                Traceback (most recent call last)
<ipython-input-5-cba72439f97c> in <module>()
----> 1 y[1] = 4

IndexError: index 1 is out of bounds for axis 0 with size 1

In [6]: y[0][1] = 4

In [7]: x
Out[7]:
array([([[0.0, 0.0]],)],
      dtype=[('field', '<f8', (1, 2))])

In [8]: y
Out[8]: array([[ 3.,  4.]])

In [9]: x[0]['field'] = y

In [10]: x
Out[10]:
array([([[3.0, 0.0]],)],
      dtype=[('field', '<f8', (1, 2))])

所以……为了更容易掌握，让我们把形状变得更简单。

In [1]: import numpy as np

In [2]: x = np.zeros(1, dtype=np.dtype([('field', '<f8', 2)]))

In [3]: y = x[0]['field'].copy()

In [4]: y[0] = 3

In [5]: y[1] = 4

In [6]: x[0]['field'] = y

In [7]: x
Out[7]:
array([([3.0, 0.0],)],
      dtype=[('field', '<f8', (2,))])

In [8]: y
Out[8]: array([ 3.,  4.])

在这种情况下数据去向......没有线索。不过，以存储数据的方式进行分配似乎很容易。

几种选择：

In [9]: x['field'][0] = y

In [10]: x
Out[10]:
array([([3.0, 4.0],)],
      dtype=[('field', '<f8', (2,))])

In [11]: x['field'] = y * 2

In [12]: x
Out[12]:
array([([6.0, 8.0],)],
      dtype=[('field', '<f8', (2,))])

In [13]: x['field'][:] = y

In [14]: x
Out[14]:
array([([3.0, 4.0],)],
      dtype=[('field', '<f8', (2,))])

In [15]: x[0]['field'][:] = y * 2

In [16]: x
Out[16]:
array([([6.0, 8.0],)],
      dtype=[('field', '<f8', (2,))])