Question

我正在上课。我重新编辑了我的帖子，以显示我想要的结果。到目前为止，我被困在频率部分。我想达到这些结果： Results

谢谢！

导入随机

num = int(input("How many times would you like to roll the dice? "))
sides = int(input("How many sides does the dice have? "))

def throwDice(num=1, sides=6):
    return [random.randint(1,sides) for i in range(num)]


#-- main ----------------------------------------------------------------------   

numberOfTrials = int(input('How many trials? Enter:'))



# perform simulation, record and print frequencies
frequency= 13*[0]  # same as [0,0,0,0,0,0,0,0,0,0,0,0,0]

for i in range(numberOfTrials):
   t = throwDice()
   frequency[t[0]+t[0]] += 1;

# end for
print()
print("Frequencies:")
print(frequency)


# calculate relative frequencies, probabilities and errors
relativeFrequency = [0, 0]
probability = [0,0]
error = [0,0]
for i in range(2, len(frequency)):
   relativeFrequency.append(frequency[i]/numberOfTrials)
   probability.append(min(i-1,13-i)/36)
   error.append(abs(probability[i]-relativeFrequency[i]))
# end for


#print(relativeFrequency)
#print(probability)
#print(error)
print()


# print results
f1 = "{0:<10}{1:<22}{2:<22}{3:<22}"
f2 = 71*"-"
f3 = "{0:>3}       {1:<22.15f}{2:<22.15f}{3:<.15f}"
print(f1.format("Sum","Relative Frequency","Probability","Error"))
print(f2)
for i in range(2, len(frequency)):
   print(f3.format(i, relativeFrequency[i], probability[i], error[i]))
#end for
print()

Answer 1

右。看看你的日常工作：

def throwDice(rolls=1, dice_sides=6):
    results = 0
    throwDice = 0
    for i in range(0, rolls):
            results = random.randint(1, dice_sides)
    return throwDice

...

   t = throwDice()
   frequency[t[0]+t[1]] += 1;

你制作给定数量的卷，将最后一卷保持在results。然后忽略该值并返回最初分配给局部变量0的{{1}}（顺便说一下，在本地禁用函数定义）。然后将0返回给调用程序。

您不能取整数的第一个和第二个元素。

如果不知道你的功能是不正常的，你是如何累积这么多代码的？在继续之前尝试测试小黑人。事实上，尝试测试：几分钟，有良好位置的打印语句会指出问题所在。查看这个可爱的debug博客以获取帮助。

Answer 2

t是一个int，因此t[0]没有任何有效含义。

完成作业：骰子概率

2 个答案: