Question

我正在尝试使用此代码来压缩我的列表。通过展平我的意思是转换像

这样的列表

[1,[2], [[[[3]]]],4]

到

[1,2,3,4]

这是我的代码

i = 0

def flatten(aList):
    ''' 
    aList: a list 
    Returns a copy of aList, which is a flattened version of aList 
    '''
    global x
    x = []

    def check(L):
        global i
        if type(L[i]) != list:
            x.append(L[i])
            i += 1
            return check(aList[i])
        else:
            return check(L[i])

    return check(aList)`

我一直收到此错误

    Traceback (most recent call last):

  File "<ipython-input-87-ee05c7b1d059>", line 1, in <module>
    flatten(l)

  File "/Users/ashwin/.spyder-py3/untitled1.py", line 20, in flatten
    return check(aList)

  File "/Users/ashwin/.spyder-py3/untitled1.py", line 18, in check
    return check(L[i])

  File "/Users/ashwin/.spyder-py3/untitled1.py", line 16, in check
    return check(aList[i])

  File "/Users/ashwin/.spyder-py3/untitled1.py", line 13, in check
    if type(L[i]) != list:

TypeError: 'int' object is not subscriptable

我需要改变什么？

Answer 1

您可以简化如下：

def flatten(a_list):
    x = []
    def check(l):
        # Note: check l, not l[i] because l can be anything in the recursive calls
        if not isinstance(l, list):  
            x.append(l)
        else:
            for sub in l:
                check(sub)
    check(a_list)
    return x

> flatten([1, 2, [3, 4, [[5], 6]]])
[1, 2, 3, 4, 5, 6]

l[i]没有硬性访问权限，因为您永远不会l。整数会引发您遇到的错误。这也消除了对全局变量的需求。

＆＃39; INT＆＃39;对象不可订阅（试图压扁列表）？

1 个答案: