根据Stef的观察，我们只需要跟踪前1%，但我用NumPy数组和分区来做到这一点：

import numpy as np

# Simulated files (3x5 grid instead of 361x576)
def files():
    np.random.seed(123)
    for year in range(43):
        yield np.random.random((8760, 3, 5))

# Just for testing; Compute expectation with np.percentile
def reference(files):
    whole = np.vstack(files)
    return np.percentile(whole, 99, 0, method='closest_observation')

# The actual solution
def solution(files):

    # Preparation for getting top 1% = top k
    k = round(43 * 8760 / 100)
    def top_k(a):
        return np.partition(a, ~k, 0)[~k:]

    # Compute the result
    result = top_k(next(files))
    for other in files:
        other = top_k(other)
        both = np.vstack((result, other))
        result = top_k(both)
    return result[0]

expect = reference(files())
result = solution(files())

print('expect:')
print(expect)

print('\nresult:')
print(result)

print('\ndifference:')
print(result - expect)

输出（Attempt This Online!）：

expect:
[[0.98989099 0.98997275 0.99009619 0.98969514 0.99034828]
 [0.9898165  0.99010689 0.98995486 0.99006558 0.98968308]
 [0.98996179 0.98979965 0.9896849  0.98996829 0.99012452]]

result:
[[0.98989099 0.98997275 0.99009619 0.98969514 0.99034828]
 [0.9898165  0.99010689 0.98995486 0.99006558 0.98968308]
 [0.98996179 0.98979965 0.9896849  0.98996829 0.99012452]]

difference:
[[0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0.]]

使用模拟的36x58网格，在ATO网站上，“计算结果”部分花费了不到30秒。因此，对于您自己的计算机上的真实的数据，它应该花费不到一个小时。