创建三维二值图像

Question

我有一个2D数组，a，包含一组100 x，y，z坐标：

[[ 0.81  0.23  0.52]
 [ 0.63  0.45  0.13]
 ...
 [ 0.51  0.41  0.65]]

我想要创建一个三维二值图像，b，在每个x，y，z维中的101个像素，坐标在0.00到1.00之间。a定义的位置上的像素值应该是1，所有其他像素的值都应该是0。

我可以用b = np.zeros((101,101,101))创建一个形状正确的零数组，但是如何分配坐标并将其切片以创建使用a的零数组

Answer 1

首先，从安全地将您的浮点数转到ints开始。在上下文中，请参见这问题。

a_indices = np.rint(a * 100).astype(int)

接下来，将b中的索引分配给1，但是要小心使用普通的list而不是数组，否则会触发索引数组的使用。该方法的性能似乎与其他方法相当(谢谢@Divakar！:-)

b[list(a_indices.T)] = 1

我创建了一个小示例，其大小为10而不是100，二维而不是3，以举例说明：

>>> a = np.array([[0.8, 0.2], [0.6, 0.4], [0.5, 0.6]])
>>> a_indices = np.rint(a * 10).astype(int)
>>> b = np.zeros((10, 10))
>>> b[list(a_indices.T)] = 1
>>> print(b) 
[[ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  1.  0.  0.  0.]
 [ 0.  0.  0.  0.  1.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  1.  0.  0.  0.  0.  0.  0.  0.]
 [ 0.  0.  0.  0.  0.  0.  0.  0.  0.  0.]]

Answer 2

你可以这样做-

# Get the XYZ indices
idx = np.round(100 * a).astype(int)

# Initialize o/p array
b = np.zeros((101,101,101))

# Assign into o/p array based on linear index equivalents from indices array
np.put(b,np.ravel_multi_index(idx.T,b.shape),1)

分配部分的运行时-

让我们使用一个更大的网格来计时。

In [82]: # Setup input and get indices array
    ...: a = np.random.randint(0,401,(100000,3))/400.0
    ...: idx = np.round(400 * a).astype(int)
    ...: 

In [83]: b = np.zeros((401,401,401))

In [84]: %timeit b[list(idx.T)] = 1 #@Praveen soln
The slowest run took 42.16 times longer than the fastest. This could mean that an intermediate result is being cached.
1 loop, best of 3: 6.28 ms per loop

In [85]: b = np.zeros((401,401,401))

In [86]: %timeit np.put(b,np.ravel_multi_index(idx.T,b.shape),1) # From this post
The slowest run took 45.34 times longer than the fastest. This could mean that an intermediate result is being cached.
1 loop, best of 3: 5.71 ms per loop

In [87]: b = np.zeros((401,401,401))

In [88]: %timeit b[idx[:,0],idx[:,1],idx[:,2]] = 1 #Subscripted indexing
The slowest run took 40.48 times longer than the fastest. This could mean that an intermediate result is being cached.
1 loop, best of 3: 6.38 ms per loop