沿着numpy数组的任意轴迭代连续的1-D切片

Question

我正在编写一个python包，它沿着任意形状的numpy数组的任意轴执行各种复杂的统计分析任务。

目前，为了使阵列形状和轴可以是任意的，我只是置换数组，使感兴趣的轴放在远RHS上，并将LHS轴压缩成一个。例如，如果数组形状是(3,4,5)，并且我们想要沿着轴1执行一些操作，它将被转换为形状(15,4)，操作沿着轴-1执行，然后它被转换回形状(3,4,5)并由功能。

由于所有这些阵列操作，我觉得这种方法可能会不必要地慢。有没有一种方法可以干净地迭代除阵列的一个维度？也就是说，在上面的例子中，这将是[0,:,0]，[0,:,1]，...，[2,:,3]，[2,:,4]，但是这应该适用于任意数组形状和轴位置。

也许np.ndenumerate，np.ndindex和np.take可以用于某种方式吗？

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编辑：有没有办法用np.nditer做到这一点？也许这可以匹配置换/重塑的速度。

Answer 1

结果只是转置和重塑确实更快。所以我猜答案是......不要这样做，最好像我一直在做的那样进行置换和重塑。

这是我项目的代码。

# Benchmark
f = lambda x: x # can change this to any arbitrary function
def test1(data, axis=-1):
    # Test the lead flatten approach
    data, shape = lead_flatten(permute(data, axis))
    output = np.empty(data.shape)
    for i in range(data.shape[0]): # iterate along first dimension; each row is an autocor
        output[i,:] = f(data[i,:]) # arbitrary complex equation
    return unpermute(lead_unflatten(output, shape), axis)
def test2(data, axis=-1):
    # Test the new approach
    output = np.empty(data.shape)
    for d,o in zip(iter_1d(data, axis), iter_1d(output, axis)):
        o[...] = f(d)
    return output

# Iterator class
class iter_1d(object):
    def __init__(self, data, axis=-1):
        axis = (axis % data.ndim) # e.g. for 3D array, -1 becomes 2
        self.data = data
        self.axis = axis
    def __iter__(self):
        shape = (s for i,s in enumerate(self.data.shape) if i!=self.axis)
        self.iter = np.ndindex(*shape)
        return self
    def __next__(self):
        idx = self.iter.next()
        idx = [*idx]
        idx.insert(self.axis, slice(None))
        return self.data[idx]

# Permute and reshape functions
def lead_flatten(data, nflat=None):
    shape = list(data.shape)
    if nflat is None:
        nflat = data.ndim-1 # all but last dimension
    if nflat<=0: # just apply singleton dimension
        return data[None,...], shape
    return np.reshape(data, (np.prod(data.shape[:nflat]).astype(int), *data.shape[nflat:]), order='C'), shape # make column major

def lead_unflatten(data, shape, nflat=None):
    if nflat is None:
        nflat = len(shape) - 1 # all but last dimension
    if nflat<=0: # we artificially added a singleton dimension; remove it
        return data[0,...]
    if data.shape[0] != np.prod(shape[:nflat]):
        raise ValueError(f'Number of leading elements {data.shape[0]} does not match leading shape {shape[nflat:]}.')
    if not all(s1==s2 for s1,s2 in zip(data.shape[1:], shape[nflat:])):
        raise ValueError(f'Trailing dimensions on data, {data.shape[1:]}, do not match trailing dimensions on new shape, {shape[nflat:]}.')
    return np.reshape(data, shape, order='C')

def permute(data, source=-1, destination=-1):
    data = np.moveaxis(data, source, destination)
    return data

def unpermute(data, source=-1, destination=-1):
    data = np.moveaxis(data, destination, source)
    return data

以下是一些%timeit操作的结果。

import numpy as np
a = np.random.rand(10,20,30,40)
%timeit -r10 -n10 test1(a, axis=2) # around 12ms
%timeit -r10 -n10 test2(a, axis=2) # around 22ms