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python - 如何在一种方法中递归地使参数传递的数据类型不可变？

问题描述

我正在尝试解决一个需要递归地使数据类型不变的练习练习，但是递归案例存在问题：

该函数返回一个不可变的等效数据结构（可用于集合中的值或字典中的键）。int、str 和frozenset 类型已经是不可变的。将集合转换为frozenset；以相同的顺序将元组中的所有值转换为它们的不可变等价物）；将列表转换为元组（其值的不可变等价物，以相同的顺序）；将 dict 转换为 2 元组的元组

我得到了基本情况，但没有太多其他...

def immutify(a : 'an int, str, list, tuple, set, or dict') -> 'an int, str, tuple, or forzenset':
    if (isinstance(a, int)):
        return a
    if (isinstance(a, str)):
        return a
    if (isinstance(a, tuple)):
        # immutify each element in it?
    if (isinstance(a, frozenset)):
        # immutify each element in it?
    if (isinstance(a, set)):
        # return an immutified frozenset of a?
    if (isinstance(a, list)):
         # return an immutified tuple of a?
    if (isinstance(a, dict)):
        #return an immutified tuple of a?

预期的输出是：

immutify(1) # -->1
immutify('a') # -->'a'
immutify( (1, 2, 3)) # -->(1, 2, 3)
immutify( frozenset([1, 2, 3])) # -->frozenset({1, 2, 3})
immutify( [1, 2, 3, 4, 5, 6]) # -->(1, 2, 3, 4, 5, 6)
immutify( [1, 2, [3, [4], 5], 6])# -->(1, 2, (3, (4,), 5), 6)
immutify( [1, 2, (3, [4], 5), 6]) # -->(1, 2, (3, (4,), 5), 6)
immutify( [{1,2}, {3,frozenset([4,5])}, {6,7}]) #-->(frozenset({1, 2}), frozenset({3, frozenset({4, 5})}), frozenset({6, 7}))
immutify( [{1,2}, {3,frozenset([4,5])}, [{5,6}]]) #-->(frozenset({1, 2}), frozenset({3, frozenset({4, 5})}), (frozenset({5, 6}),))
immutify( {'b' : [1,2], 'a' : {'ab': {1,2}, 'aa' : (1,2)}}) #-->(('a', (('aa', (1, 2)), ('ab', frozenset({1, 2})))), ('b', (1, 2)))

编辑：我几乎到了那里：

def immutify(a : 'an int, str, list, tuple, set, or dict') -> 'an int, str, tuple, or forzenset':
    if (isinstance(a, int)):
        return a
    if (isinstance(a, str)):
        return a
    if (isinstance(a, tuple)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)
    if (isinstance(a, frozenset)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return frozenset(out)
    if (isinstance(a, set)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)

    if (isinstance(a, list)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)
    if (isinstance(a, dict)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)

标签： pythonpython-3.xrecursionmethodstypes

解决方案

我知道了。

def immutify(a : 'an int, str, list, tuple, set, or dict') -> 'an int, str, tuple, or forzenset':
    if (isinstance(a, int)):
        return a
    if (isinstance(a, str)):
        return a
    if (isinstance(a, tuple)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)
    if (isinstance(a, frozenset)):
        #out = []
        #for elem in a:
            #new = immutify(elem)
            #out.append(new)
        #return frozenset(out)
        return a
    if (isinstance(a, set)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return frozenset(out)

    if (isinstance(a, list)):
        out = []
        for elem in a:
            new = immutify(elem)
            out.append(new)
        return tuple(out)
    if (isinstance(a, dict)):
        tup = []
        pair = []
        for elem in a:
            new1 = immutify(elem)
            new2 = immutify(a[elem])
            pair.append(new1)
            pair.append(new2)
            tup.append(tuple(pair))
            pair = []
        return tuple(tup)

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