将参数传递给我的函数时,它不会识别列表并输出字符串。带有列表的PYTHON参数
该游戏被称为通过猪,并需要输出一个猪的状态。
我知道该代码是效率低下的地方,虽然这是由于这样的事实,我一直试图有事先没有成功:(
感谢您的帮助不同的方法!
这里是代码:
norolls = int(input("Enter the number of rolls: "))
counter = 0
def roll(nothrows,counter):
rollList = []
while counter < nothrows:
rollrand = randint(0,100)
rollList.append(rollrand)
counter = (counter + 1)
return rollList
rollList = roll(norolls,counter)
rollList = list(map(int, rollList))
listlen = len(rollList)
def rollout(List, listpos, ListLen):
listpos = 0
for x in range(ListLen):
if List[listpos] == 1< 35:
print("Pink")
elif List[listpos] == 35 < 65:
print("Dot")
elif List[listpos] == 65< 85:
print("Razorback")
elif List[listpos] == 85 < 95:
print("Trotter")
elif List[listpos] == 95 < 99:
print("Snouter")
else:
List[listpos] == 99 < 100
print("Leaning Jewler")
listpos = (listpos + 1)
rollout(rollList, counter, listlen)
我不明白* if ...== x
guidot
¯| _(ツ)_ /¯umm不知道 –
@guidot:不完全。 'a == b
DSM