问题描述
我有一个混合类型的结构化数组:
I have a structured array with mixed types:
dt = np.dtype([('x', np.float64), ('y', np.float64), ('n', np.uint32)])
arr = np.empty(10, dtype=dt)
从numpy 1.16开始,如果我查看x
和y
,则会得到一个视图:
As of numpy 1.16 or so, if I view x
and y
, I get a view:
>>> sub = arr[['x', 'y']]
>>> sub
array([(6.23042070e-307, 4.67296746e-307),
(1.15710088e-306, 1.60221615e-306),
(1.95821574e-306, 6.23062102e-307),
(1.78019082e-306, 1.37959740e-306),
(1.37959129e-306, 1.33511562e-306),
(1.33511018e-306, 1.33511969e-306),
(1.11261027e-306, 1.11261502e-306),
(8.45593934e-307, 9.34600963e-307),
(6.23038336e-307, 1.29061142e-306),
(2.22522596e-306, 2.22522596e-306)],
dtype={'names':['x','y'], 'formats':['<f8','<f8'], 'offsets':[0,8], 'itemsize':20})
这是一个问题,因为我希望能够将子集sub
转换为x
和y
字段的(10, 2)
视图.
This is a problem because I would like to be able to convert the subset sub
into a (10, 2)
view of the x
and y
fields.
我不能只使用sub.view(dtype=np.float64)
.这引发了错误
I can not just use sub.view(dtype=np.float64)
. That raises the error
ValueError: When changing to a smaller dtype, its size must be a divisor of the size of original dtype
我可以使用np.lib.stride_tricks.as_strided
,但这很麻烦而且有问题,因为它仅在我需要两个字段(或任意数量的均匀间隔的字段)时才起作用:
I can use np.lib.stride_tricks.as_strided
, but that is hacky and problematic because it only works when I want two fields (or alternatively any number of evenly-spaced fields):
>>> shape = sub.shape + (2,)
>>> strides = (sub.dtype.itemsize,
np.diff([x[1] for x in sub.dtype.fields.values()]).item())
>>> np.lib.stride_tricks.as_strided(sub, shape=shape, strides=strides)['x']
array([[6.23042070e-307, 4.67296746e-307],
[1.15710088e-306, 1.60221615e-306],
[1.95821574e-306, 6.23062102e-307],
[1.78019082e-306, 1.37959740e-306],
[1.37959129e-306, 1.33511562e-306],
[1.33511018e-306, 1.33511969e-306],
[1.11261027e-306, 1.11261502e-306],
[8.45593934e-307, 9.34600963e-307],
[6.23038336e-307, 1.29061142e-306],
[2.22522596e-306, 2.22522596e-306]])
如果sub
是副本,那么我可以简单地将其视为浮点数的(10, 2)
数组.如何通过复制所选内容或任何其他方式将所选字段视为这样的数组?
If sub
were a copy, then I could simply view it as a (10, 2)
array of floats. How can I view the selected fields as such an array, either by copying the selection or any other means?
推荐答案
repack_fields
以及多字段视图中的更改:
repack_fields
goes along with the change in multifield view:
In [135]: dt = np.dtype([('x', np.float64), ('y', np.float64), ('n', np.uint32)])
...: arr = np.empty(3, dtype=dt)
In [136]: sub = arr[['x','y']]
In [137]: import numpy.lib.recfunctions as rf
In [138]: rf.repack_fields(sub)
Out[138]:
array([(4.04359530e-316, 4.04349886e-316),
(0.00000000e+000, 0.00000000e+000),
(4.04355735e-316, 0.00000000e+000)],
dtype=[('x', '<f8'), ('y', '<f8')])
In [139]: sub
Out[139]:
array([(4.04359530e-316, 4.04349886e-316),
(0.00000000e+000, 0.00000000e+000),
(4.04355735e-316, 0.00000000e+000)],
dtype={'names':['x','y'], 'formats':['<f8','<f8'], 'offsets':[0,8], 'itemsize':20})
它是副本,而不是视图.
It is a copy, not a view.
对于(n,2)个副本:
And for a (n,2) copy:
In [140]: rf.structured_to_unstructured(sub)
Out[140]:
array([[4.04359530e-316, 4.04349886e-316],
[0.00000000e+000, 0.00000000e+000],
[4.04355735e-316, 0.00000000e+000]])
In [141]: rf.structured_to_unstructured(rf.repack_fields(sub))
Out[141]:
array([[4.04359530e-316, 4.04349886e-316],
[0.00000000e+000, 0.00000000e+000],
[4.04355735e-316, 0.00000000e+000]])
这篇关于如何获取结构化数组选择的副本的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!