对在python中执行两个代码感到困惑？

Question

代码1：

%%timeit
students = [['Zack',38],['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]
second_highest = sorted(list(set([x[1] for x in students])))[1]
([a for a,b in sorted(students) if b == second_highest])

代码2：

%%timeit
students = [['Zack',38],['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]
s = sorted(set([x[1] for x in students]))
for name in sorted(x[0] for x in students if x[1] == s[1]):
    name

现在我对两个程序的执行感到困惑，尽管在code2中使用嵌套for循环，但code2如何比代码code1快。下面的图片来自Jupyter笔记本，它显示了100000个循环的代码所需的平均时间。虽然差异很小但我感到很困惑，因为嵌套for循环的工作方式比单循环方式更快。<​​/ p>

我应该打印输出，所以可以在最后一行代码之前放置

Answer 1

这不是嵌套循环。

看来你错了这条线

for name in sorted(x[0] for x in students if x[1] == s[1]):
    name

在这里，您要创建一个生成器并将其传递给sorted函数

sorted(x[0] for x in students if x[1] == s[1])

首先计算一个列表。然后，在评估该表达式并获得列表之后，它由for循环迭代。所以它是两个循环，一个接一个，而不是嵌套。

Answer 2

这是因为第二种情况下，生成器被传递到sorted()循环中的for。 sorted()函数首先复制序列，这个副本对于生成器列表要比列表列表慢，因为列表需要首先从生成器生成。请参阅this问题。因此，如果您通过列表替换生成器，它会变得更快。

In [11]: %%timeit
students = [['Zack',38],['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]
s = sorted(set([x[1] for x in students]))
for name in sorted([x[0] for x in students if x[1] == s[1]]):
    name
   ....: 
The slowest run took 5.71 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 4.72 µs per loop

In [6]: %%timeit                         
students = [['Zack',38],['Harry', 37.21], ['Berry', 37.21], ['Tina', 37.2], ['Akriti', 41], ['Harsh', 39]]
second_highest = sorted(list(set([x[1] for x in students])))[1]
([a for a,b in sorted(students) if b == second_highest])
   ...: 
The slowest run took 11.50 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.31 µs per loop