1

我有以下 numpy 结构化数组：

In [250]: x
Out[250]: 
array([(22, 2, -1000000000, 2000), (22, 2, 400, 2000),
       (22, 2, 804846, 2000), (44, 2, 800, 4000), (55, 5, 900, 5000),
       (55, 5, 1000, 5000), (55, 5, 8900, 5000), (55, 5, 11400, 5000),
       (33, 3, 14500, 3000), (33, 3, 40550, 3000), (33, 3, 40990, 3000),
       (33, 3, 44400, 3000)], 
      dtype=[('f1', '<i4'), ('f2', '<i4'), ('f3', '<i4'), ('f4', '<i4')])

下面的数组是上面数组的一个子集（也是一个视图）：

In [251]: fields=['f1','f3']

In [252]: y=x.getfield(np.dtype(
     ...:                    {name: x.dtype.fields[name] for name in fields}
     ...:                    ))

In [253]: y
Out[253]: 
array([(22, -1000000000), (22, 400), (22, 804846), (44, 800), (55, 900),
       (55, 1000), (55, 8900), (55, 11400), (33, 14500), (33, 40550),
       (33, 40990), (33, 44400)], 
      dtype={'names':['f1','f3'], 'formats':['<i4','<i4'], 'offsets':[0,8], 'itemsize':12})

我正在尝试将 y 转换为常规的 numpy 数组。我希望数组是一个视图。问题是以下给了我一个错误：

In [254]: y.view(('<i4',2))
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-254-88440f106a89> in <module>()
----> 1 y.view(('<i4',2))

C:\numpy\core\_internal.pyc in _view_is_safe(oldtype, newtype)
    499 
    500     # raises if there is a problem
--> 501     _check_field_overlap(new_fieldtile, old_fieldtile)
    502 
    503 # Given a string containing a PEP 3118 format specifier,

C:\numpy\core\_internal.pyc in _check_field_overlap(new_fields, old_fields)
    402         old_bytes.update(set(range(off, off+tp.itemsize)))
    403     if new_bytes.difference(old_bytes):
--> 404         raise TypeError("view would access data parent array doesn't own")
    405 
    406     #next check that we do not interpret non-Objects as Objects, and vv

TypeError: view would access data parent array doesn't own

但是，如果我选择连续的字段，它会起作用：

In [255]: fields=['f1','f2']
     ...: 
     ...: y=x.getfield(np.dtype(
     ...:                    {name: x.dtype.fields[name] for name in fields}
     ...:                    ))
     ...: 

In [256]: y
Out[256]: 
array([(22, 2), (22, 2), (22, 2), (44, 2), (55, 5), (55, 5), (55, 5),
       (55, 5), (33, 3), (33, 3), (33, 3), (33, 3)], 
      dtype=[('f1', '<i4'), ('f2', '<i4')])

In [257]: y.view(('<i4',2))
Out[257]: 
array([[22,  2],
       [22,  2],
       [22,  2],
       [44,  2],
       [55,  5],
       [55,  5],
       [55,  5],
       [55,  5],
       [33,  3],
       [33,  3],
       [33,  3],
       [33,  3]])

当字段不连续时，视图转换似乎不起作用，是否有替代方法？

4

2 回答 2

3

是的，ndarray直接使用构造函数：

x = np.array([(22, 2, -1000000000, 2000), 
              (22, 2,         400, 2000),
              (22, 2,      804846, 2000), 
              (44, 2,         800, 4000), 
              (55, 5,         900, 5000), 
              (55, 5,        1000, 5000)], 
             dtype=[('f1','i'),('f2','i'),('f3','i'),('f4','i')])

fields = ['f4', 'f1']
shape = x.shape + (len(fields),)
offsets = [x.dtype.fields[name][1] for name in fields]
assert not any(np.diff(offsets, n=2))
strides = x.strides + (offsets[1] - offsets[0],)
y = np.ndarray(shape=shape, dtype='i', buffer=x,
               offset=offsets[0], strides=strides)
print repr(y)

给出：

array([[2000,   22],
       [2000,   22],
       [2000,   22],
       [4000,   44],
       [5000,   55],
       [5000,   55]])

顺便说一句，当原始数组中的所有字段都具有相同的 dtype 时，首先在该数组上创建一个视图然后进行切片操作要容易得多。对于与上述相同的结果：

y = x.view('i').reshape(x.shape + (-1,))[:,-1::-3]
于 2016-11-18T18:13:11.217 回答
1

以下内容有点令人困惑 - 但要点是，要使这种view工作能够正常工作，它必须能够访问具有规则数组步幅和形状的字段。np.ones((12,4))[:,[0,2]]从 ['f1','f3'] 获取视图失败的原因与生成副本的原因基本相同。

========

在您的结构化数组中，每条记录都存储为 4*'i4' 字节。该布局与 (n,4) 'i4' 数组兼容：

In [381]: x.__array_interface__['data']    # databuffer pointer
Out[381]: (160925352, False)
In [382]: x.view(('i',4)).__array_interface__['data']
Out[382]: (160925352, False)          # same buffer
In [387]: x.view(('i',4)).shape
Out[387]: (12, 4)

但是当我取这个数组的各个部分时

In [383]: x.view(('i',4))[:,[0,1]].__array_interface__['data']
Out[383]: (169894184, False)       # advance indexing - a copy

In [384]: x.view(('i',4))[:,:2].__array_interface__['data']
Out[384]: (160925352, False)       # same buffer

但是选择 ['f1','f3'] 相当于: x.view(('i',4))[:,[0,2]]，另一个副本。

或者看看步伐。使用第一个 2 个字段

In [404]: y2=x.getfield(np.dtype({name: x.dtype.fields[name] for name in ['f1','f2']}))
In [405]: y2.dtype
Out[405]: dtype([('f1', '<i4'), ('f2', '<i4')])
In [406]: y2.strides
Out[406]: (16,)
In [407]: y2.view(('i',2)).strides
Out[407]: (16, 4)

要将此数组仅视为整数，它可以将行步进 16，将列步进 4，并且只占用 2 列。

或查看 4 列和 2 列案例的完整字典

In [409]: x.view(('i',4)).__array_interface__
Out[409]: 
{'data': (160925352, False),
 'descr': [('', '<i4')],
 'shape': (12, 4),
 'strides': None,
 'typestr': '<i4',
 'version': 3}
In [410]: y2.view(('i',2)).__array_interface__
Out[410]: 
{'data': (160925352, False),
 'descr': [('', '<i4')],
 'shape': (12, 2),
 'strides': (16, 4),
 'typestr': '<i4',
 'version': 3}

相同的步幅和 dtype，只是形状不同。该y2案例有效，因为它可以通过跨步访问所需的字节，并忽略 2 列。

如果我切出 4 列案例的 2 个中间列，我会得到一个视图 - 相同的数据缓冲区，但有一个偏移量：

In [385]: x.view(('i',4))[:,2:4].__array_interface__['data']
Out[385]: (160925360, False)

但是getfield与这两个字段一起使用会产生与 ['f1','f3'] 相同的错误：

In [388]: y2=x.getfield(np.dtype({name: x.dtype.fields[name] for name in ['f2','f3']})).view(('i',2))
...
ValueError: new type not compatible with array.

view无法实现切片可以实现的数据缓冲区偏移量。

========

再次查看 2 个中间字段：

In [412]: y2=x.getfield(np.dtype({name: x.dtype.fields[name] for name in ['f2','f3']}))
     ...:  
In [413]: y2
Out[413]: 
array([(2, -1000000000), (2, 400), (2, 804846), (2, 800), (5, 900),
       (5, 1000), (5, 8900), (5, 11400), (3, 14500), (3, 40550),
       (3, 40990), (3, 44400)], 
      dtype={'names':['f2','f3'], 'formats':['<i4','<i4'], 'offsets':[4,8], 'itemsize':12})
In [414]: y2.__array_interface__['data']
Out[414]: (160925352, False)

y2指向原始数据库开始。它用dtype偏移量来实现偏移量。将其与 中的偏移量进行比较In[385]。

于 2016-11-18T17:32:41.310 回答