Question

我有这个递归函数

def recursive_search(x):
    y = []
    for i in x:
        if (i == tuple(i)) or (i == list(i)) or i == set(i):
            recursive_search(i)
        else:
            y.append(i)
    print(y)
print(recursive_search(("re",("cur",("sion",(" ",("foo",["bar",{"baz"}])))))))

打印出来

['baz']
['bar']
['foo']
[' ']
['sion']
['cur']
['re']
None

当我将“print（y）”设置为“return y”时，它只打印出“['re']”。如果使用全局y是不安全的，那么我可以采用另一种方式。

Answer 1

您需要捕获返回值并使用它来构建答案：

def recursive_search(x):
    y = []
    for i in x:
        if type(i) in (tuple, list, set):
            y.append(recursive_search(i))
        else:
            y.append(i)
    return y

print(recursive_search(("re",("cur",("sion",(" ",("foo",["bar",{"baz"}])))))))
# ['re', ['cur', ['sion', [' ', ['foo', ['bar', ['baz']]]]]]]

如果您需要嵌套以外的其他东西，您可以使用返回来执行其他操作。

Answer 2

声明第二个参数，每次调用递归函数时都会这样：

def recursive_search(x,y=None):
    if y is None:
        y = []
    for i in x:
        if (i == tuple(i)) or (i == list(i)) or i == set(i):
            recursive_search(i, y)
        else:
            y +=[i]
    return y[:]

y = recursive_search(("re",("cur",("sion",(" ",("foo",["bar",{"baz"}]))))))
print(y)
z = recursive_search(("a",("brown",("fox",("jumps ",("over",["foo",{"bar"}]))))))
print(z)

['re', 'cur', 'sion', ' ', 'foo', 'bar', 'baz']
['a', 'brown', 'fox', 'jumps ', 'over', 'foo', 'bar']

有没有办法可以存储这个变量而不会重置

2 个答案: