使用 outfor 和 while 循环对这个列表进行索引的任何更快的方法？（Python）

Question

I'm trying to find a way to get rid of this while loop because its costly in time.我正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。 what I'm doing here is indexing the list (data) to then find the highest value in between [x:x+9] and then adding that to another array (result) and then adding 1 to x to index through the entire list.我在这里做的是索引列表（数据），然后找到 [x:x+9] 之间的最大值，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. Is this a stupid way of doing it?这是一种愚蠢的做法吗？ is there a faster more clever way?.有没有更快更聪明的方法？ Any help is much appreciated.任何帮助深表感谢。 I hope that I explained this well enough.我希望我解释得足够好。

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)

out: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.]

Wall time: 7.01 ms挂墙时间：7.01 毫秒

Answer 1

You can let numpy perform the calculations in parallel by providing it with a list of indexes for each sub range:您可以让 numpy 通过为每个子范围提供索引列表来并行执行计算：

For example: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]

[EDIT] an even faster way is to loop through the value offsets rather than on the positions. [编辑] 一种更快的方法是遍历值偏移而不是位置。 While this still involves a loop, it is much faster and sustains the speed improvement on larger data sets.虽然这仍然涉及一个循环，但它要快得多，并且可以维持更大数据集的速度改进。

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

使用 outfor 和 while 循环对这个列表进行索引的任何更快的方法？（Python）

问题描述

1 个解决方案

解决方案1
5 2020-12-29 21:57:07

使用 outfor 和 while 循环对这个列表进行索引的任何更快的方法？ （Python）

问题描述

1 个解决方案

解决方案1 5 2020-12-29 21:57:07

使用 outfor 和 while 循环对这个列表进行索引的任何更快的方法？（Python）

解决方案1
5 2020-12-29 21:57:07