我有一个服务器,它以多维数组的格式访问和获取数据,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,仅访问 s1 值并将它们存储到将作为结果返回的新数组中。
选项 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
选项 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
有区别吗?如果有区别,哪种方式会更高效、更快速?考虑到第二维的大小最多为 20,第三维的大小最多为 500
最佳答案
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)
示例输出:
[['n1s1:2', 'n1s2:8'], ['n2s1:3', 'n2s2:1'], ['n3s1:9', 'n3s2:1']]
而且它比 map
方法快 2 倍以上:
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)
关于python - 几个大嵌套循环的小循环 vs 小嵌套循环的大循环性能?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/56859047/