Question

我想'剪切'一个numpy阵列。我不确定我是否正确使用“剪切”一词;通过剪切，我的意思是：

将第一列移动0位置
将第二列移动1个位置将第三列移动2个位置
等...

所以这个数组：

array([[11, 12, 13],
       [17, 18, 19],
       [35, 36, 37]])

会变成这个数组：

array([[11, 36, 19],
       [17, 12, 37],
       [35, 18, 13]])

或类似这样的数组：

array([[11,  0,  0],
       [17, 12,  0],
       [35, 18, 13]])

取决于我们如何处理边缘。我对边缘行为并不太了解。

这是我尝试执行此功能的函数：

import numpy

def shear(a, strength=1, shift_axis=0, increase_axis=1, edges='clip'):
    strength = int(strength)
    shift_axis = int(shift_axis)
    increase_axis = int(increase_axis)
    if shift_axis == increase_axis:
        raise UserWarning("Shear can't shift in the direction it increases")
    temp = numpy.zeros(a.shape, dtype=int)
    indices = []
    for d, num in enumerate(a.shape):
        coords = numpy.arange(num)
        shape = [1] * len(a.shape)
        shape[d] = num
        coords = coords.reshape(shape) + temp
        indices.append(coords)
    indices[shift_axis] -= strength * indices[increase_axis]
    if edges == 'clip':
        indices[shift_axis][indices[shift_axis] < 0] = -1
        indices[shift_axis][indices[shift_axis] >= a.shape[shift_axis]] = -1
        res = a[indices]
        res[indices[shift_axis] == -1] = 0
    elif edges == 'roll':
        indices[shift_axis] %= a.shape[shift_axis]
        res = a[indices]
    return res

if __name__ == '__main__':
    a = numpy.random.random((3,4))
    print a
    print shear(a)

似乎有效。如果没有，请告诉我！

它看起来也很笨重而且不优雅。我是否会忽略内置的numpy / scipy功能呢？在numpy中有更清洁/更好/更有效的方法吗？我重新发明了轮子吗？

编辑：
如果这适用于N维数组，而不仅仅是2D情况，则可获得奖励积分。

这个函数将位于循环的中心，我将在数据处理中重复多次，所以我怀疑它实际上值得优化。

第二次编辑：我终于做了一些基准测试。尽管有循环，看起来像numpy.roll是要走的路。谢谢，tom10和Sven Marnach！

基准测试代码:(在Windows上运行，我认为不要在Linux上使用time.clock）

import time, numpy

def shear_1(a, strength=1, shift_axis=0, increase_axis=1, edges='roll'):
    strength = int(strength)
    shift_axis = int(shift_axis)
    increase_axis = int(increase_axis)
    if shift_axis == increase_axis:
        raise UserWarning("Shear can't shift in the direction it increases")
    temp = numpy.zeros(a.shape, dtype=int)
    indices = []
    for d, num in enumerate(a.shape):
        coords = numpy.arange(num)
        shape = [1] * len(a.shape)
        shape[d] = num
        coords = coords.reshape(shape) + temp
        indices.append(coords)
    indices[shift_axis] -= strength * indices[increase_axis]
    if edges == 'clip':
        indices[shift_axis][indices[shift_axis] < 0] = -1
        indices[shift_axis][indices[shift_axis] >= a.shape[shift_axis]] = -1
        res = a[indices]
        res[indices[shift_axis] == -1] = 0
    elif edges == 'roll':
        indices[shift_axis] %= a.shape[shift_axis]
        res = a[indices]
    return res

def shear_2(a, strength=1, shift_axis=0, increase_axis=1, edges='roll'):
    indices = numpy.indices(a.shape)
    indices[shift_axis] -= strength * indices[increase_axis]
    indices[shift_axis] %= a.shape[shift_axis]
    res = a[tuple(indices)]
    if edges == 'clip':
        res[indices[shift_axis] < 0] = 0
        res[indices[shift_axis] >= a.shape[shift_axis]] = 0
    return res

def shear_3(a, strength=1, shift_axis=0, increase_axis=1):
    if shift_axis > increase_axis:
        shift_axis -= 1
    res = numpy.empty_like(a)
    index = numpy.index_exp[:] * increase_axis
    roll = numpy.roll
    for i in range(0, a.shape[increase_axis]):
        index_i = index + (i,)
        res[index_i] = roll(a[index_i], i * strength, shift_axis)
    return res

numpy.random.seed(0)
for a in (
    numpy.random.random((3, 3, 3, 3)),
    numpy.random.random((50, 50, 50, 50)),
    numpy.random.random((300, 300, 10, 10)),
    ):
    print 'Array dimensions:', a.shape
    for sa, ia in ((0, 1), (1, 0), (2, 3), (0, 3)):
        print 'Shift axis:', sa
        print 'Increase axis:', ia
        ref = shear_1(a, shift_axis=sa, increase_axis=ia)
        for shear, label in ((shear_1, '1'), (shear_2, '2'), (shear_3, '3')):
            start = time.clock()
            b = shear(a, shift_axis=sa, increase_axis=ia)
            end = time.clock()
            print label + ': %0.6f seconds'%(end-start)
            if (b - ref).max() > 1e-9:
                print "Something's wrong."
        print

Answer 1

这可以使用this answer by Joe Kington中描述的技巧来完成：

from numpy.lib.stride_tricks import as_strided
a = numpy.array([[11, 12, 13],
                 [17, 18, 19],
                 [35, 36, 37]])
shift_axis = 0
increase_axis = 1
b = numpy.vstack((a, a))
strides = list(b.strides)
strides[increase_axis] -= strides[shift_axis]
strides = (b.strides[0], b.strides[1] - b.strides[0])
as_strided(b, shape=b.shape, strides=strides)[a.shape[0]:]
# array([[11, 36, 19],
#        [17, 12, 37],
#        [35, 18, 13]])

要获取“剪辑”而不是“滚动”，请使用

b = numpy.vstack((numpy.zeros(a.shape, int), a))

这可能是最有效的方法，因为它根本不使用任何Python循环。

Answer 2

numpy roll这样做。例如，如果原始数组是x，那么

for i in range(x.shape[1]):
    x[:,i] = np.roll(x[:,i], i)

产生

[[11 36 19]
 [17 12 37]
 [35 18 13]]

Answer 3

tom10's answer中的方法可以扩展到任意维度：

def shear3(a, strength=1, shift_axis=0, increase_axis=1):
    if shift_axis > increase_axis:
        shift_axis -= 1
    res = numpy.empty_like(a)
    index = numpy.index_exp[:] * increase_axis
    roll = numpy.roll
    for i in range(0, a.shape[increase_axis]):
        index_i = index + (i,)
        res[index_i] = roll(a[index_i], -i * strength, shift_axis)
    return res

Answer 4

以下是您自己方法的清理版本：

def shear2(a, strength=1, shift_axis=0, increase_axis=1, edges='clip'):
    indices = numpy.indices(a.shape)
    indices[shift_axis] -= strength * indices[increase_axis]
    indices[shift_axis] %= a.shape[shift_axis]
    res = a[tuple(indices)]
    if edges == 'clip':
        res[indices[shift_axis] < 0] = 0
        res[indices[shift_axis] >= a.shape[shift_axis]] = 0
    return res

主要区别在于它使用numpy.indices()而不是滚动您自己的版本。

Answer 5

r = lambda l, n: l[n:]+l[:n]

transpose(map(r, transpose(a), range(0, len(a)))

我想。您应该考虑这个伪代码而不是实际的Python。基本上转置数组，在其上映射一般旋转函数进行旋转，然后将其转置回来。

剪一个numpy阵列

5 个答案: