Python 向量化嵌套 for 循环

我希望能够帮助您找到和理解一种 Pythonic 方法来优化嵌套 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

Where volume.shape（182、218、200）和roi.shape(3,) 都是ndarray类型；和radius is an int

方法#1

这是一种矢量化方法 -

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：我们可以加快最后一步numexpr模块 -

import numexpr as ne

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

方法#2

我们还可以逐步构建形状参数对应的三个范围，并对形状参数的三个元素进行减法roi即时创建网格，而不像之前那样实际创建网格np.mgrid。这将受益于使用broadcasting出于效率目的。实施看起来像这样 -

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 提出改进建议，我们可以使用np.ogrid以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

函数定义 -

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

时间安排 -

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

所以，一如既往broadcasting几乎疯狂地展现出它的魔力10,000x比原始代码加速并超过10x比使用即时广播操作创建网格更好！