Python矢量化嵌套for循环

Question

我很感激在寻找和理解pythonic方法的一些帮助,以优化嵌套for循环中的以下数组操作:

def _func(a, b, radius):
    "Return 0 if a>b, otherwise return 1"
    if distance.euclidean(a, b) < radius:
        return 1
    else:
        return 0

def _make_mask(volume, roi, radius):
    mask = numpy.zeros(volume.shape)
    for x in range(volume.shape[0]):
        for y in range(volume.shape[1]):
            for z in range(volume.shape[2]):
                mask[x, y, z] = _func((x, y, z), roi, radius)
    return mask

其中volume.shape(182,218,200)和roi.shape(3,)都是ndarray类型; 而且radius是一个int

Answer 1

方法#1

这是一个矢量化的方法 -

m,n,r = volume.shape
x,y,z = np.mgrid[0:m,0:n,0:r]
X = x - roi[0]
Y = y - roi[1]
Z = z - roi[2]
mask = X**2 + Y**2 + Z**2 < radius**2

可能的改进:我们可以用numexpr模块加速最后一步-

import numexpr as ne

mask = ne.evaluate('X**2 + Y**2 + Z**2 < radius**2')

方法#2

我们还可以逐渐构建与形状参数相对应的三个范围,并针对动态的三个元素执行减法,roi而无需实际创建网格,如前所述np.mgrid.broadcasting出于效率目的,这将有益于使用.实现看起来像这样 -

m,n,r = volume.shape
vals = ((np.arange(m)-roi[0])**2)[:,None,None] + \
       ((np.arange(n)-roi[1])**2)[:,None] + ((np.arange(r)-roi[2])**2)
mask = vals < radius**2

简化版本:感谢@Bi Rico建议改进,因为我们可以用np.ogrid更简洁的方式执行这些操作,就像这样 -

m,n,r = volume.shape    
x,y,z = np.ogrid[0:m,0:n,0:r]-roi
mask = (x**2+y**2+z**2) < radius**2

运行时测试

功能定义 -

def vectorized_app1(volume, roi, radius):
    m,n,r = volume.shape
    x,y,z = np.mgrid[0:m,0:n,0:r]
    X = x - roi[0]
    Y = y - roi[1]
    Z = z - roi[2]
    return X**2 + Y**2 + Z**2 < radius**2

def vectorized_app1_improved(volume, roi, radius):
    m,n,r = volume.shape
    x,y,z = np.mgrid[0:m,0:n,0:r]
    X = x - roi[0]
    Y = y - roi[1]
    Z = z - roi[2]
    return ne.evaluate('X**2 + Y**2 + Z**2 < radius**2')

def vectorized_app2(volume, roi, radius):
    m,n,r = volume.shape
    vals = ((np.arange(m)-roi[0])**2)[:,None,None] + \
           ((np.arange(n)-roi[1])**2)[:,None] + ((np.arange(r)-roi[2])**2)
    return vals < radius**2

def vectorized_app2_simplified(volume, roi, radius):
    m,n,r = volume.shape    
    x,y,z = np.ogrid[0:m,0:n,0:r]-roi
    return (x**2+y**2+z**2) < radius**2

计时 -

In [106]: # Setup input arrays  
     ...: volume = np.random.rand(90,110,100) # Half of original input sizes 
     ...: roi = np.random.rand(3)
     ...: radius = 3.4
     ...: 

In [107]: %timeit _make_mask(volume, roi, radius)
1 loops, best of 3: 41.4 s per loop

In [108]: %timeit vectorized_app1(volume, roi, radius)
10 loops, best of 3: 62.3 ms per loop

In [109]: %timeit vectorized_app1_improved(volume, roi, radius)
10 loops, best of 3: 47 ms per loop

In [110]: %timeit vectorized_app2(volume, roi, radius)
100 loops, best of 3: 4.26 ms per loop

In [139]: %timeit vectorized_app2_simplified(volume, roi, radius)
100 loops, best of 3: 4.36 ms per loop

因此,一如既往地broadcasting展示其10,000x对原始代码的疯狂几乎加速的魔力,并且比10x通过使用即时广播操作创建网格更好!

Answer 2

假设您首先构建一个xyzy数组:

import itertools

xyz = [np.array(p) for p in itertools.product(range(volume.shape[0]), range(volume.shape[1]), range(volume.shape[2]))]

现在,使用numpy.linalg.norm,

np.linalg.norm(xyz - roi, axis=1) < radius

检查每个元组的距离roi是否小于半径.

最后,只reshape需要您需要的尺寸的结果.