Question

编写一个名为distribution_of_rolls的函数，该函数需要一个数字 - 滚动两个骰子的次数 - 并以下图所示的形式打印这些卷的值的分布例如：

    Distribution of dice rolls

    2:     7 ( 3.5%)  *******
    3:    14 ( 7.0%)  **************
    4:    15 ( 7.5%)  ***************
    5:    19 ( 9.5%)  *******************
    6:    24 (12.0%)  ************************
    7:    35 (17.5%)  ***********************************
    8:    24 (12.0%)  ************************
    9:    28 (14.0%)  ****************************
   10:    18 ( 9.0%)  ******************
   11:     9 ( 4.5%)  *********
   12:     7 ( 3.5%)  *******
   -----------------------------
         200 rolls

我的代码在某处错了。它无法打印出来。

def roll():
        return randrange(1,7) + randrange(1,7)

def distribution_of_rolls(n:int):
        result=({i,0} for i in range(2,13,1))
        c=''
        result=list(result)
        for i in range(n):
              a = roll()
              print(a)
              result[a]= result[a] + 1
              print(result)
        for i in range(2,13,1):
            b=(result[i]/float(n)) * 100
        for i in range(int(math.floor(n))):
            c = c + '*'
            d = "{0:0.1f}%".format(b)
        print("{0:2d}:   {1:5d} ({2:s})  {3:s}".format(i, result[i], d, c)
        print("-------------------------")
        print("{0:10d} rolls".format(n))


    distribution_of_rolls(20)

Answer 1

在概率模型中，您可以使用特征函数来计算独立定律的总和。即使有50个骰子，也可以在我的计算机上使用OpenTurns平台即时获得。

import openturns as ot
d = 2 #number of dice

#define dice distribution with possible values: 1, 2, 3, 4, 5, 6
dice_distribution = ot.UserDefined([[i] for i in range(1,7)]) 

# create same distribution d times then sum
sum_distribution = sum([dice_distribution] * d) 

# create a sample of size 200
Sample = sum_distribution.getSample(200)

如果您想计算每个实现：

import numpy as np
unique, counts = np.unique(Sample, return_counts = True)
print(unique)
print(counts/200) # frequency

>>>[ 2.  3.  4.  5.  6.  7.  8.  9. 10. 11. 12.]
   [0.05  0.05  0.09  0.115 0.13  0.175 0.125 0.115 0.07  0.04  0.04 ]

您还可以使用OpenTurns查看器绘制概率密度函数（PDF）：

import openturns.viewer as otv 
graph = sum_distribution.drawPDF()
otv.View(graph)

Rendered image of the probability distribution function

您还可以检索值：

print(sum_distribution)
>>> UserDefined(
     {x = [2], p = 0.0277778},
     {x = [3], p = 0.0555556},
     {x = [4], p = 0.0833333},
     {x = [5], p = 0.111111},
     {x = [6], p = 0.138889},
     {x = [7], p = 0.166667},
     {x = [8], p = 0.138889},
     {x = [9], p = 0.111111},
     {x = [10], p = 0.0833333},
     {x = [11], p = 0.0555556},
     {x = [12], p = 0.0277778}
   )

如果d = 50个骰子，则密度将更像是高斯分布：

Answer 2

使用更多功能，以便更容易调试。如，

def roll():
  return random.randrange(1,7) + random.randrange(1,7)

def create_data(nrolls):
  data = collections.defaultdict(int)
  for _ in range(nrolls):
    data[roll()] += 1
  return data

fmtstart = "{dsum:>2}:{count:>6} ({pct:4.1f}%)"

def format_data(data):
  result = []
  total = float( sum(data.values()) )
  tuples = sorted( data.items() )
  for k,v in tuples:
    start = fmtstart.format(dsum=k,count=v,pct=100*v/total)
    end = " " + "*"*v
    result.append(start + end)
  return "\n".join(result)

print( format_data( create_data(200) ) )

Answer 3

from random import randrange

def roll():
        return randrange(1,7) + randrange(1,7)

def distribution_of_rolls(n:int):
    results = {i:0 for i in range(2,13)}
    for i in range(n):
        results[roll()] += 1
    for i in range(2,13):
        occurrences = results[i]
        percentage = (occurrences/float(n)) * 100
        bar = occurrences * "*"
        print("{0:2d}: {1:5d} ({2:4.1f}%)  {3:s}".format(i, occurrences, percentage, bar))
    print("-------------------------")
    print("{0:9d} rolls".format(n))


distribution_of_rolls(200)

找到骰子卷的分布

3 个答案: