调用列表值的变量-python

调用列表值的变量-python,python,list,variables,Python,List,Variables,我需要我的游戏来运行这个: def w_win(l,l2,check): line = l line2 = l2 for x in range(4): if line[x] == line2[x]: if len(str(check+check)) == 1: line[x] = " "+str(check+check)+" " if len(str(check+check)

我需要我的游戏来运行这个:

def w_win(l,l2,check):
    line = l
    line2 = l2
    for x in range(4):
        if line[x] == line2[x]:
            if len(str(check+check)) == 1:
                line[x] = "  "+str(check+check)+" "
            if len(str(check+check)) == 2:
                line[x] = "  "+str(check+check)+""
            if len(str(check+check)) == 3:
                line[x] = " "+str(check+check)
            if len(str(check+check)) == 4:
                line[x] = str(check+check) 
但我明白了:

if line[x] == line2[x]:

TypeError: 'NoneType' object has no attribute '__getitem__'
我正在打电话给你

for p in range(2,1025):

    first = w_win([" ","2","16"," "],[" ","2"," ","32"],p)

有什么帮助吗?

您的w_-win函数缺少返回语句,当您将函数的值赋给变量“first”时,返回语句是必需的。在您的情况下,由于缺少return语句,首先将变量初始化为“NoneType”类

什么是line和line2?你是如何调用这个函数的?提供完整的错误消息toooIT似乎<代码>行或<代码> LIL2在调用这个函数时是非类型的,你最好考虑一下。除了上面的注释之外,for循环缺少了冒号的冒号,怀疑是否有一种方法在列表调用中使用变量,这对我来说似乎很好。不客气。我无法重现您在Python2和Python3上遇到的相同错误。