Question

我有一个项目列表，并希望生成所有可能的子集。因此，我使用带有项目编号的递归函数和所有选定项目的列表作为参数。调用该函数时，将0作为第一个参数，并执行以下操作：

查看索引参数
选择它
它使用递增的索引参数
取消选择项目
它使用递增的索引参数

我需要可能的子集来优化某些东西，但由于列表会变得很长，我无法查看所有这些。起初我尝试使用蛮力来考虑所有子集，但这是一个天真的想法。现在新的计划是创建一个贪婪算法，它采用第一个“有用的”选择：我想查看所有子集，直到找到一个符合我需要的子集，并认为python的yield语句是正确的选择。这是一些代码：

def bruteForceLeft(selected,index):
    #left is the list of which i need subsets
    #its a gobal variable. to test the code, just make sure that you have a 
    #list called left in scope
    if index==len(left):
        #print(selected)
        yield selected
    else:
        #the algorithm stores the selection in a tuple of two lists
        #that's necessary since there's a second list called right as well
        #I think you can just ignore this. Think of selected as a list that
        #contains the current selection, not a tuple that contains the current
        #selection on the right as well as the left side.
        selected[0].append(left[index])
        bruteForceLeft(selected,index+1)
        selected[0].pop()
        bruteForceLeft(selected,index+1)

#as you can see I pass a tuple of two empty lists to the function.
#only the first one is used in this piece of code
for option in bruteForceLeft( ([],[]) ,0):
    print(option)
    #check if the option is "good"
    #break

输出是：没有

起初我以为我在生成子集时出错了，但在if条件下你可以看到我有一个注释的print语句。如果我取消注释这个print语句，而是注释掉yield语句，那么打印所有可能的选项 - 并且for循环被打破

使用yield语句，代码运行时没有错误，但它也没有做任何事情。

Answer 1

问题在于，当您递归调用bruteForceLeft时，所产生的值不会从封闭函数中神奇地产生。所以，你需要自己重新产生它们：

def bruteForceLeft(selected,index):
    #left is the list of which i need subsets
    if index==len(left):
        #print(selected)
        yield selected
    else:
        #the algorithm stores the selection in a tuple of two lists
        #that's necessary since there's a second list called right as well
        #I think you can just ignore this. Think of selected as a list that
        #contains the current selection, not a tuple that contains the current
        #selection on the right as well as the left side.
        selected[0].append(left[index])
        for s in bruteForceLeft(selected,index+1):
            yield s
        selected[0].pop()
        for s in bruteForceLeft(selected,index+1):
            yield s

（编辑：我实际上只测试了这个，你的代码有错误，但我很确定不会重新产生问题）

如何使用python的yield语句

1 个答案: