多个大型嵌套循环的小型循环与小型嵌套循环的大型循环性能比较？

Question

我有一个服务器，可以访问和获取多维数组格式的数据，因此最终结果是：

 [
    [
        [n1t1:1, n1s1:2, n1o1:5],
        [n1t2:3, n1s2:8, n1o2:9]
    ],
    [
        [n2t1:9, n2s1:3, n2o1:2],
        [n2t2:5, n2s2:1, n2o2:7]
    ],
    [
        [n3t1:4, n3s1:9, n3o1:2],
        [n3t2:7, n3s2:1, n3o2:5]
    ]
 ]

我需要遍历该数组，只访问s1值，并将它们存储到一个新数组中，该数组将作为结果返回。你知道吗

方案1：

result = []
parent_enum = 0
while len(array) > parent_enum:
    child_enum = 0
    result.append([])
    while len(child_enum) > array_num:
        result[parent_enum].append(array[parent_enum][child_enum][1])
        child_enum += 1
    parent_enum += 1

方案2：

result = [[] for i in range(len(array))]
parent_enum = 0
while len(array[0]) > parent_enum:
    child_enum = 0
    while len(array) > child_enum:
         result[child_enum].append(array[child_enum][parent_enum][1])
         child_enum += 1
    parent_enum += 1

有什么区别吗？如果有，哪种方法更有效、更快速？考虑到第二维度的大小是20，第三维度的大小是500

Answer 1

通过使用内置函数，以下代码应该更具可读性并具有良好的性能。你知道吗

data = [ ...your data... ]
result = map(lambda first:  # for each first-level entry
                 map(lambda second:  # for each second-level entry within first
                         second[1],  # return the second value
                     first
                 ),
             data
         )

[
    [
        2,
        8
    ],
    [
        3,
        1
    ],
    [
        9,
        1
    ]
 ]

Answer 2

为什么不使用简单的列表理解：

arr = [
    [
        ["n1t1:1", "n1s1:2", "n1o1:5"],
        ["n1t2:3", "n1s2:8", "n1o2:9"]
    ],
    [
        ["n2t1:9", "n2s1:3", "n2o1:2"],
        ["n2t2:5", "n2s2:1", "n2o2:7"]
    ],
    [
        ["n3t1:4", "n3s1:9", "n3o1:2"],
        ["n3t2:7", "n3s2:1", "n3o2:5"]
    ]
 ]


result = [[arr_lev3[1] for arr_lev3 in arr_lev2] for arr_lev2 in arr]

print(result)

样本输出：

[['n1s1:2', 'n1s2:8'], ['n2s1:3', 'n2s2:1'], ['n3s1:9', 'n3s2:1']]

---

它比map方法快2倍多：

In [38]: %timeit result = [[arr_lev3[1] for arr_lev3 in arr_lev2] for arr_lev2 in arr]
753 ns ± 2.24 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

In [39]: %timeit result2 = list(map(lambda first: list(map(lambda second: second[1], first)), arr))
1.63 µs ± 20.4 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)