使用 outfor 和 while 循环索引这个列表的更快方法是什么？（Python）

Question

我试图找到一种方法来摆脱这个 while 循环，因为它的时间成本很高。我在这里做的是索引列表（数据），然后找到 [x:x+9] 之间的最高值，然后将其添加到另一个数组（结果），然后将 1 添加到 x 以索引整个列表. 这是一种愚蠢的做法吗？有没有更快更聪明的方法？任何帮助深表感谢。我希望我已经很好地解释了这一点。

def calc(data):
    result = np.zeros(len(data)) # allocating space
    x = 0
    while x < len(a):
        highest_value = max(data[x:x+9])
        print(f'{data[x:x+9]} highest value = {highest_value}')
        result[x] = b
        print(result)
        x += 1
    return result

data = [7,6,5,4,3,4,2,3,4,5,6,7,8,9,3,5,4,2,3,1]

result = calc(data)

出去：

[7, 6, 5, 4, 3, 4, 2, 3, 4] highest value = 7
[7. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[6, 5, 4, 3, 4, 2, 3, 4, 5] highest value = 6
[7. 6. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[5, 4, 3, 4, 2, 3, 4, 5, 6] highest value = 6
[7. 6. 6. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[4, 3, 4, 2, 3, 4, 5, 6, 7] highest value = 7
[7. 6. 6. 7. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[3, 4, 2, 3, 4, 5, 6, 7, 8] highest value = 8
[7. 6. 6. 7. 8. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[4, 2, 3, 4, 5, 6, 7, 8, 9] highest value = 9
[7. 6. 6. 7. 8. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[2, 3, 4, 5, 6, 7, 8, 9, 3] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[3, 4, 5, 6, 7, 8, 9, 3, 5] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[4, 5, 6, 7, 8, 9, 3, 5, 4] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[5, 6, 7, 8, 9, 3, 5, 4, 2] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[6, 7, 8, 9, 3, 5, 4, 2, 3] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7, 8, 9, 3, 5, 4, 2, 3, 1] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0.]
[8, 9, 3, 5, 4, 2, 3, 1] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0.]
[9, 3, 5, 4, 2, 3, 1] highest value = 9
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0.]
[3, 5, 4, 2, 3, 1] highest value = 5
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 0. 0. 0. 0. 0.]
[5, 4, 2, 3, 1] highest value = 5
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 0. 0. 0. 0.]
[4, 2, 3, 1] highest value = 4
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 0. 0. 0.]
[2, 3, 1] highest value = 3
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 0. 0.]
[3, 1] highest value = 3
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 3. 0.]
[1] highest value = 1
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 3. 1.]

______________________________________________________________
result: 

[7. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 0. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 0. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 0. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 0. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 0. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 3. 0.]
[7. 6. 6. 7. 8. 9. 9. 9. 9. 9. 9. 9. 9. 9. 5. 5. 4. 3. 3. 1.]

挂墙时间：7.01 毫秒

Answer 1

您可以通过为 numpy 提供每个子范围的索引列表来让 numpy 并行执行计算：

例如：

import numpy as np

data = np.array([7,6,5,4,3,4,2,3,4,5,6,7,8,9,3,5,4,2,3,1])

idx = np.arange(9)+np.arange(len(data))[:,None] # indexes of subRanges
idx = np.minimum(len(data)-1,idx)               # don't overflow indexes

rollingMax = np.max(data[idx],axis=1) # apply maximums on every subrange

print(rollingMax)
[7 6 6 7 8 9 9 9 9 9 9 9 9 9 5 5 4 3 3 1]

[编辑] 一种更快的方法是遍历值偏移而不是位置。虽然这仍然涉及一个循环，但它要快得多，并且可以在更大的数据集上保持速度改进。

def rollingMax2(data,window=9):
    result = data.copy()
    for offset in range(1,window):
        result[:-1] = np.maximum(result[:-1],result[1:])
    return result

                         speed improvement
number of values     rollingMax   rollingMax2
           20             3x          3x
          200            15x         31x  
        2,000            26x         99x
       20,000            35x        167x
      200,000            21x        220x 
    2,000,000            11x         66x 
   20,000,000            11x         35x