我想这可能行得通：

import numpy as np

nTimes = 100
nParticles = 10
A_ij = np.full((nTimes, nParticles, nParticles), False)
E_ij = np.random.randint(0, 9, (100, 10))

np.put_along_axis(A_ij, E_ij[..., None], True, axis=2)

另一种方法，接近你在最后实际尝试的方法，是这样的：

i, j = np.mgrid[:nTimes, :nParticles]
A_ij[i, j, E_ij] = True

但公认的答案肯定是解决问题的更好方法，不需要构建指数。

为了防止可能对其他人有帮助，我还找到了一种方法在这个问题上做花哨的索引，但@ Chrysophylax的答案更快更简单（我猜我对索引感到困惑，无法思考它）。我还添加了@Mercury的答案以供比较。

代码：

import numpy as np
import matplotlib.pyplot as plt
import time

nTimes = 1000000
nParticles = 10
A_ij1 = np.full((nTimes, nParticles, nParticles), False)
A_ij2 = np.full((nTimes, nParticles, nParticles), False)
A_ij3 = np.full((nTimes, nParticles, nParticles), False)
A_ij4 = np.full((nTimes, nParticles, nParticles), False)

E_ij = np.random.randint(0, 9, (nTimes, 10))

start_time = time.time()
for t in range(nTimes):
    for i in range(nParticles):
        A_ij1[t, i, E_ij[t,i]] = True
print("Loop: %s s" % (time.time() - start_time))

        
start_time = time.time()
A_ij2[np.arange(nTimes)[:,None],np.arange(nParticles)[None,:], E_ij[np.arange(nTimes)[:,None],np.arange(nParticles)[None,:]]] = True
print("Fancy indexing: %s s" % (time.time() - start_time))

start_time = time.time()
np.put_along_axis(A_ij3, E_ij[..., None], True, axis=2)
print("Put along axis: %s s" % (time.time() - start_time))

start_time = time.time()
i, j = np.mgrid[:nTimes, :nParticles]
A_ij4[i, j, E_ij] = True
print("mgrid: %s s" % (time.time() - start_time))

print(np.allclose(A_ij1, A_ij2))
print(np.allclose(A_ij1, A_ij3))
print(np.allclose(A_ij1, A_ij4))

输出：

Loop: 2.5006823539733887 s
Fancy indexing: 0.11996173858642578 s
Put along axis: 0.0814671516418457 s
mgrid: 0.19223332405090332 s
True
True
True