写下2个六面骰子的滚动模拟。你的程序应该有一个Roll()函数,它返回掷骰子的总和。你可以假设六个方面中的每一个都同样可能被滚动(也就是说,骰子是“公平的”)。运行模拟1000次并报告每次总和发生的频率。
到目前为止我有这个,但我的程序似乎不会加总。我可能完全错了。请帮忙。我认为我的主要问题在于我的印刷声明。我需要输出打印2次显示总和的次数,3的总和,4的总和,直到12。
def Roll():
for i in range(1000):
one = 0
two = 0
three = 0
four = 0
five = 0
six = 0
dice1= float(0)
dice2= float(0)
dice1 = random.randint(1,6)
if dice1 == 1:
one = one + 1
count= 1
return count
elif dice1 == 2:
two = two + 1
count= 1
return count
elif dice1 == 3:
three = three + 1
count= 1
return count
elif dice1 == 4:
four = four + 1
count= 1
return count
elif dice1 == 5:
five = five + 1
count= 1
return count
else:
six = six + 1
count= 1
return count
dice2 = random.randint(1,6)
if dice2 == 1:
one = one + 1
elif dice2 == 2:
two = two + 1
elif dice2 == 3:
three = three + 1
elif dice2 == 4:
four = four + 1
elif dice2 == 5:
five = five + 1
else:
six = six + 1
total = one + two + three + four + five + six
print("2", dice1 + dice2)
print("3", dice1 + dice2)
print("4", dice1 + dice2)
print("5", dice1 + dice2)
print("6", dice1 + dice2)
print("7", dice1 + dice2)
print("8", dice1 + dice2)
print("9", dice1 + dice2)
print("10", dice1 + dice2)
print("11", dice1 + dice2)
print("12", dice1 + dice2)
答案 0 :(得分:2)
我已经回答了你的一位朋友,他的任务相同:
import random
from collections import defaultdict
def main():
dice = int(input("Enter the number of dice: "))
sides = int(input("Enter the number of sides: "))
rolls = int(input("Enter the number of rolls to simulate: "))
result = roll(dice, sides, rolls)
maxH = 0
for i in range(dice, dice * sides + 1):
if result[i] / rolls > maxH: maxH = result[i] / rolls
for i in range(dice, dice * sides + 1):
print('{:2d}{:10d}{:8.2%} {}'.format(i, result[i], result[i] / rolls, '#' * int(result[i] / rolls / maxH * 40)))
def roll(dice, sides, rolls):
d = defaultdict(int)
for _ in range(rolls):
d[sum(random.randint(1, sides) for _ in range(dice))] += 1
return d
main()
您可以取出对您有用的部分。这还应该包括后续问题,例如:“我如何整齐地打印”和“如何绘制直方图”。
示例:
Enter the number of dice: 2
Enter the number of sides: 6
Enter the number of rolls to simulate: 1000
2 28 2.80% ######
3 59 5.90% #############
4 84 8.40% ###################
5 96 9.60% ######################
6 155 15.50% ####################################
7 170 17.00% ########################################
8 147 14.70% ##################################
9 102 10.20% #######################
10 80 8.00% ##################
11 50 5.00% ###########
12 29 2.90% ######
答案 1 :(得分:1)
这是一个快速而肮脏的方法
from collections import Counter
import random
def roll():
return random.randint(1,6) + random.randint(1,6)
counter = Counter( roll() for _ in range(1000) )
for num, cnt in counter.iteritems():
print '%2d: %.3f' % (num, cnt / 1000.0)
导致
2: 0.022
3: 0.063
4: 0.072
5: 0.104
6: 0.154
7: 0.174
8: 0.141
9: 0.112
10: 0.077
11: 0.057
12: 0.024