Question

尝试像在地下城和龙中掷骰子但显示每个掷骰子。我不太清楚我做错了什么，并感谢所有的帮助。

from random import randint
def d(y): #basic die roll
    return randint(1, y)
def die(x, y): #multiple die roll 2d20 could roll 13 and 7 being 20 
    for [x*d(y)]:
        print (sum(int(y)))

print (die(3, 20))

好的，所以我接受了建议并改变了它，但仍然在我的回程线上收到错误

#
#trying to roll dice like in dungeons and dragon but display each roll
from random import randint 
def d(sides): 
    return randint(1, sides) 
def roll(n, sides): 
    return tuple(d(sides) for _ in range(n))
def dice(n, sides):
    print (roll(n, sides))
    return sum(dice)
print(dice(3,20))

Answer 1

您只能将d()一次通话的结果复用多次，您需要n对d()进行不同的调用：

from random import randint

def d(sides):
    return randint(1, sides)

def roll(n, sides):
    return tuple(d(sides) for _ in range(n))

dice = roll(3, 20)
print(dice, sum(dice))
# (20, 18, 1) 39

Answer 2

注意：这是使用Python 3.6或更高版本

from random import randint

def roll(roll)

    rolling = []

    try:
        for x in range(int(roll.split('d')[0])):
            rolling.append(randint(1,int(roll.split('d')[1])))
    except Exception as err:
        print(f'I got bungled @_@ \n Error: {err}')

    print(f'You rolled {" ".join(str(x) for x in rolling)} which has a total'
          f' of {sum(rolling)}')

此功能的作用是将任何掷骰子组合（例如3d10、10d10等）接收，并使用fstring打印每个掷骰子，然后打印组合的总数。

编辑：f字符串可用于在必要时将其压缩得更多

D＆D Dice掷蟒蛇

2 个答案: