Create 3D Binary Image

Question

I have a 2D array, acontaining a set of 100 x, y, z coordinates:

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

I would like to create a three-dimensional binary image bwith 101 pixels in each of the sizes x, y, z with coordinates from 0.00 to 1.00. The pixels in the places defined amust be set to 1, all other pixels must be set to 0.

I can create an array of zeros of the correct form with b = np.zeros((101,101,101)), but how do I assign a coordinate and a slice to it to create the ones that are used with a?

Answer 1

ints. . .

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

b 1. , list , . , ( @Divakar!: -)

b[list(a_indices.T)] = 1

10 100 2 3, :

>>> 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

- -

# 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)

---

-

.

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