调用列表值的变量-python
我需要我的游戏来运行这个: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)
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。