Python 创建三维二值图像

我有一个二维数组，

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]]

我想创建一个三维二值图像，

b

，在x、y、z维度上各有101个像素，坐标范围在0.00和1.00之间。 由

a

定义的位置处的像素值应为1，所有其他像素值应为0

---

我可以用

b=np.zeros（（101101101））

创建一个正确形状的零数组，但是我如何分配坐标并将其切片以使用

a

创建这些零？

您可以这样做-

# 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

---

首先，从安全地将浮点数舍入整数开始。在上下文中，请参见问题

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

列表

而不是数组，否则会触发使用。这种方法的性能似乎与其他方法的性能相当（感谢@Divakar！：-）

我创建了一个小示例，大小为10而不是100，尺寸为2而不是3，以说明：

>>> 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.]]

---

（a*100）。aType（int）

非常危险！你可能会得到错误的索引！试试看：

a=np.arange（101）/100

，

b=（100*a）.astype（int）

，

（b==np.arange（101，dtype=int））.all（）

，你会得到

False

！例如，由于浮点精度问题，数字28在

b

中出现两次。@Praveen Damn，你说得对！四舍五入是唯一的办法。更新了帖子。现在，它和你开始时的一样。你想让我删除这篇文章吗？我不会要求你删除它。你的

np.put

方法很有趣。我对这两者之间的时间分析很感兴趣，只要使用

list（a_index.T）

@Praveen当然，让我补充一下。@Praveen补充道。性能数据似乎是可比的。

>>> 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.]]