理论上，使用@Ed Smith 所描述的scipy.ndimage.interpolation.shift 应该可以工作，但由于存在错误（https://github.com/scipy/scipy/issues/1323），它没有给出相当于多次调用np.roll 的结果。
更新：在 numpy 版本 1.12.0 中，numpy.roll 添加了“多卷”功能。这是一个二维示例，其中第一个轴滚动一个位置，第二个轴滚动三个位置：
In [7]: x = np.arange(20).reshape(4,5)
In [8]: x
Out[8]:
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
In [9]: numpy.roll(x, [1, 3], axis=(0, 1))
Out[9]:
array([[17, 18, 19, 15, 16],
[ 2, 3, 4, 0, 1],
[ 7, 8, 9, 5, 6],
[12, 13, 14, 10, 11]])
这会使下面的代码过时。我会把它留给后代。
下面的代码定义了一个我称之为multiroll 的函数，它可以满足您的需求。这是一个将其应用于形状为 (500, 500, 500) 的数组的示例：
In [64]: x = np.random.randn(500, 500, 500)
In [65]: shift = [10, 15, 20]
使用多次调用np.roll 来生成预期结果：
In [66]: yroll3 = np.roll(np.roll(np.roll(x, shift[0], axis=0), shift[1], axis=1), shift[2], axis=2)
使用multiroll生成移位数组：
In [67]: ymulti = multiroll(x, shift)
验证我们得到了预期的结果：
In [68]: np.all(yroll3 == ymulti)
Out[68]: True
对于这种大小的数组，对np.roll 进行三次调用几乎比对multiroll 的调用慢三倍：
In [69]: %timeit yroll3 = np.roll(np.roll(np.roll(x, shift[0], axis=0), shift[1], axis=1), shift[2], axis=2)
1 loops, best of 3: 1.34 s per loop
In [70]: %timeit ymulti = multiroll(x, shift)
1 loops, best of 3: 474 ms per loop
这是multiroll的定义：
from itertools import product
import numpy as np
def multiroll(x, shift, axis=None):
"""Roll an array along each axis.
Parameters
----------
x : array_like
Array to be rolled.
shift : sequence of int
Number of indices by which to shift each axis.
axis : sequence of int, optional
The axes to be rolled. If not given, all axes is assumed, and
len(shift) must equal the number of dimensions of x.
Returns
-------
y : numpy array, with the same type and size as x
The rolled array.
Notes
-----
The length of x along each axis must be positive. The function
does not handle arrays that have axes with length 0.
See Also
--------
numpy.roll
Example
-------
Here's a two-dimensional array:
>>> x = np.arange(20).reshape(4,5)
>>> x
array([[ 0, 1, 2, 3, 4],
[ 5, 6, 7, 8, 9],
[10, 11, 12, 13, 14],
[15, 16, 17, 18, 19]])
Roll the first axis one step and the second axis three steps:
>>> multiroll(x, [1, 3])
array([[17, 18, 19, 15, 16],
[ 2, 3, 4, 0, 1],
[ 7, 8, 9, 5, 6],
[12, 13, 14, 10, 11]])
That's equivalent to:
>>> np.roll(np.roll(x, 1, axis=0), 3, axis=1)
array([[17, 18, 19, 15, 16],
[ 2, 3, 4, 0, 1],
[ 7, 8, 9, 5, 6],
[12, 13, 14, 10, 11]])
Not all the axes must be rolled. The following uses
the `axis` argument to roll just the second axis:
>>> multiroll(x, [2], axis=[1])
array([[ 3, 4, 0, 1, 2],
[ 8, 9, 5, 6, 7],
[13, 14, 10, 11, 12],
[18, 19, 15, 16, 17]])
which is equivalent to:
>>> np.roll(x, 2, axis=1)
array([[ 3, 4, 0, 1, 2],
[ 8, 9, 5, 6, 7],
[13, 14, 10, 11, 12],
[18, 19, 15, 16, 17]])
"""
x = np.asarray(x)
if axis is None:
if len(shift) != x.ndim:
raise ValueError("The array has %d axes, but len(shift) is only "
"%d. When 'axis' is not given, a shift must be "
"provided for all axes." % (x.ndim, len(shift)))
axis = range(x.ndim)
else:
# axis does not have to contain all the axes. Here we append the
# missing axes to axis, and for each missing axis, append 0 to shift.
missing_axes = set(range(x.ndim)) - set(axis)
num_missing = len(missing_axes)
axis = tuple(axis) + tuple(missing_axes)
shift = tuple(shift) + (0,)*num_missing
# Use mod to convert all shifts to be values between 0 and the length
# of the corresponding axis.
shift = [s % x.shape[ax] for s, ax in zip(shift, axis)]
# Reorder the values in shift to correspond to axes 0, 1, ..., x.ndim-1.
shift = np.take(shift, np.argsort(axis))
# Create the output array, and copy the shifted blocks from x to y.
y = np.empty_like(x)
src_slices = [(slice(n-shft, n), slice(0, n-shft))
for shft, n in zip(shift, x.shape)]
dst_slices = [(slice(0, shft), slice(shft, n))
for shft, n in zip(shift, x.shape)]
src_blks = product(*src_slices)
dst_blks = product(*dst_slices)
for src_blk, dst_blk in zip(src_blks, dst_blks):
y[dst_blk] = x[src_blk]
return y