无法将列表转换为集合,引发“unhashable type: 'list'”错误

家宁张*_*家宁张 6 python list set python-3.x

所以我试图找到一个列表的所有子列表,这就是我现在所拥有的。我是 Python 新手,我不明白为什么“Q3_ans=set(ans)”会引发错误。我已经尝试将列表转换为之前设置的并且它有效。

def f2(seq):
    '''
    This is the base case of the recursion from function all_sublists
    '''
    assert len(seq)==2
    assert isinstance(x,list)
    a,b=seq
    return [[a],[b],[a,b]]


def all_sublists(x):
    '''
    This function will generate all of the sublists of a list, not including the empty one, using recursion
    '''
    assert isinstance(x,list)
    ans=[]
    for i in range(0,len(x)-1):
        for j in range(1,len(x)):
            temp=[x[i],x[j]]
            temp=[f2(temp)]
            ans.extend(temp)
    Q3_ans=set(ans) 
    return Q3_ans
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这是我运行代码时的错误 y=[1,2,3,4,5]

all_sublists(y)
---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-108-f8b1bb0a7001> in <module>
----> 1 all_sublists(y)

<ipython-input-106-84f4f752e98e> in all_sublists(x)
     10             temp=[f2(temp)]
     11             ans.extend(temp)
---> 12     Q3_ans=set(ans)
     13     return Q3_ans

TypeError: unhashable type: 'list'
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ozg*_*gur 1

正如您可以弄清楚为什么,像列表这样的可变类型不能是可散列的,因此不能转换为set. 您可以尝试返回tuple;的不可变对应项list

def f2(seq):
    assert len(seq)==2
    assert isinstance(x, tuple) # what's `x` actually?
    a, b = seq
    return ((a), (b), (a,b))

def all_sublists(x):
    assert isinstance(x, list)
    ans = []
    for i in range(0, len(x) - 1):
        for j in range(1, len(x)):
            temp = (x[i], x[j])
            temp = [f2(temp)]
            ans.extend(temp)
    Q3_ans = set(tuple(ans))
    return Q3_ans
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然后

all_sublists([1, 2, 3])
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tuple您可以在文档中阅读有关类型的更多信息。