为什么要“产量”递归功能
当我尝试实现置换函数时，我从Web获得了此代码。
def permutations(seq):
if len(seq) == 1:
yield seq
else:
for i in range(len(seq)):
perms = permutations(seq[:i] + seq[i+1:])
for p in perms:
yield [seq[i], *p]
我试图解开代码，但失败了，因为我不了解如何递归使用屈服。我知道每次下一个（生成器）时，它都会在身体中运行代码，并在收益率上停止。但是，如果我只有一次只做一次才能达到基本状态，这很明显，我必须多次进行递归才能到达基地（len == 1）。 根据我的理解，底线的产量应返回。
When I try to implement a permutation function, I got this code from web.
def permutations(seq):
if len(seq) == 1:
yield seq
else:
for i in range(len(seq)):
perms = permutations(seq[:i] + seq[i+1:])
for p in perms:
yield [seq[i], *p]
I tried to unserstand the code but I failed, because I can't understand how to use yield recursively. I know every time I do next(generator), It run the code in body and stop at a yield statement. But how can it reach base condition if I only do next once cause it's obvious I have to do recursion multiple times before I can reach base(len == 1).
Based on my understanding, the yield from the bottom line should be return.
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此作用的原因如下：在条件的递归分支的环体中，首先要执行：
这在当前递归步骤中创建本地生成器对象。
接下来的两行：
迭代该发电机对象生成的所有值，并根据它们产生排列。
收益率语句不会“穿透”递归，但是您的代码收集了从递归调用中产生的值，然后根据它们产生结果。实际上，这里有很多屈服 - 与
seq的长度成正比，但是根据输出的大小，这是可以预期的。The reason this works is as follows: in the loops body of the recursive branch of the condition, first you perform:
This create a local generator object within the current recursion step.
The next two lines:
iterate over all the values generated by that generator object, and yields permutation based on them.
The
yieldstatement does not "penetrate" recursion, but your code collects the value yielded from a recursive call, and then yields results based on them.There is in fact a lot of yielding going on here - proportional to the factorial of the length of
seq, but that would be expected, based on the size of your output.