numpy:要加入结构化数组吗?

问题描述

我在类似这样的列表中有很多 numpy结构化数组这个例子:

I have many numpy structured arrays in a list like this example:

import numpy

a1 = numpy.array([(1, 2), (3, 4), (5, 6)], dtype=[('x', int), ('y', int)])

a2 = numpy.array([(7,10), (8,11), (9,12)], dtype=[('z', int), ('w', float)])

arrays = [a1, a2]

所需的输出

将它们连接在一起以创建一个统一的结构化数组的正确方法是什么?

Desired Output

What is the correct way to join them all together to create a unified structured array like the following?

desired_result = numpy.array([(1, 2, 7, 10), (3, 4, 8, 11), (5, 6, 9, 12)],
                             dtype=[('x', int), ('y', int), ('z', int), ('w', float)])

当前方法

这是我目前正在使用的方法，但是它很慢，因此我怀疑必须有一种更有效的方法.

Current Approach

This is what I'm currently using, but it is very slow, so I suspect there must be a more efficent way.

from numpy.lib.recfunctions import append_fields

def join_struct_arrays(arrays):
    for array in arrays:
        try:
            result = append_fields(result, array.dtype.names, [array[name] for name in array.dtype.names], usemask=False)
        except NameError:
            result = array

    return result

推荐答案

这是一个应该更快的实现.它将所有内容转换为numpy.uint8的数组，并且不使用任何临时对象.

Here is an implementation that should be faster. It converts everything to arrays of numpy.uint8 and does not use any temporaries.

def join_struct_arrays(arrays):
    sizes = numpy.array([a.itemsize for a in arrays])
    offsets = numpy.r_[0, sizes.cumsum()]
    n = len(arrays[0])
    joint = numpy.empty((n, offsets[-1]), dtype=numpy.uint8)
    for a, size, offset in zip(arrays, sizes, offsets):
        joint[:,offset:offset+size] = a.view(numpy.uint8).reshape(n,size)
    dtype = sum((a.dtype.descr for a in arrays), [])
    return joint.ravel().view(dtype)

编辑:简化了代码，避免了不必要的as_strided().

Edit: Simplified the code and avoided the unnecessary as_strided().

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