Question

有人可以向我解释为什么尝试＃1不起作用吗？

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.]])]（为什么只复制了'3'？）

尝试＃2：

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

[([[ 3. 4.]])]（这有效）

Answer 1

说实话......我不确定我是否也得到了结果。它似乎不一致/破碎。部分原因是形状不一致，但不是全部。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...为了让它更容易掌握，让我们的形状更简单。

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.])

在这种情况下数据的来源......不是线索。尽管如此，分配数据的方式似乎很容易。

有几个选择：

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,))])

Answer 2

它似乎是Numpy中公认的bug。有可能修复的讨论，但错误仍然是开放的。

将值分配给NumPy数组

2 个答案: