Python：逐步布尔索引/屏蔽

Question

我对Python比较陌生，正在努力寻找一种Python方式来完成以下工作：

我有一个相对较大的四维数组（最小情况下为1000 x 1000 x 5 x 9）。在每1000 x 1000矩阵中（保持dim 3和4不变），我想提取每列的前n个元素。但n在列之间的变化类似于一个逐步函数

因此，例如，我想要一个布尔掩码，它看起来像以下矩阵：

01
011
01
0 0 0 0

然而，我读到索引应该比布尔掩蔽更快。因此，我当前的方法是使用大量for循环：

def getElements(index,L):
"""
index - a (T x T x K x A)-array, smallest case is T = 1,000, K = 5, A = 9,
goes up to T = 50,000, K = 20, A = 10
L - a list used to compute the fraction of column elements extracted, containing 
numbers from 1 to ~10
"""

T = index.shape[0]
K = index.shape[2]
A = index.shape[3]
dimL = max(L)

elem = np.empty((T,T,K*dimL,A))
elem[:] = np.nan

m = -1

for kk in range(K):
    for ll in L:
        m +=1
        p = ll/(dimL+1)
        for aa in range(A):
            for tt in range(T):
                n = min(math.floor(tt*p),T-kk) # Floor function leads to stepwise mask/ indexing
                elem[:n, tt, m, aa] = index[:n, tt, kk, aa]
            elem[:, :, m, aa] += (kk+1)
return elem

有没有一种有效的方法来解决这个问题

非常感谢

编辑：下面是一个（希望最小）可复制的示例：

index = np.arange(100000)
index = np.reshape(index,(100,100,10))

T = index.shape[0]
K = index.shape[2]
L = 2

elem = np.full((T,T,K*L), np.nan)

m = -1

for kk in range(K):
  for ll in range(L):
    m +=1
    p = (ll+1)/(L+1)
    for tt in range(T):
        n = min(math.floor((tt+1)*p),T-kk) # tt+1 as python counts from zero
        elem[:n, tt, m] = index[:n, tt, kk]