Question

我对Python还是比较陌生，并且正在努力寻找一种执行以下操作的pythonic方法：

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）。从每个1000 x 1000矩阵（保持3和4不变）中，我想提取每列的前n个元素。但是n在列之间的变化类似于逐步函数。

因此，例如，我想要一个布尔掩码，看起来像下面的矩阵：

0 1 1 1 1
0 0 1 1 1
0 0 0 0 1
0 0 0 0 0

尽管如此，我读到索引应该比布尔掩码更快。因此，我目前的方法是使用许多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

是否有一种有效且pythonic的方式来解决此问题？

非常感谢！

编辑：这是一个（希望是最少的）可复制示例：

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]

Python：逐步布尔索引/掩码

0 个答案: