有没有一种方法可以让我这个变量存储没有它越来越重

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

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

当我设置“打印（Y）”到“回归y“只打印出”['re']“。如果使用全局y不安全，那么我可以通过另一种方式来做到这一点。

Answer 1

您需要捕获返回值和使用来构建你的答案：

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']]]]]]] 

如果你需要比嵌套你可以做别的事情与返回的其他东西。

Answer 2

声明第二个参数，这将是每一个递归函数调用时：

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']