我如何在 python 中模拟有偏差的模具？
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How do I simulate biased die in python?
提问by Pratik Deoghare
I want to simulate N-sided biased die?
我想模拟 N 面偏置模具？
def roll(N,bias):
'''this function rolls N dimensional die with biasing provided'''
# do something
return result
>> N=6
>> bias=( 0.20,0.20,0.15,0.15,0.14,0.16,)
>> roll(N,bias)
2
回答by Yuval Adam
A little bit of math here.
这里有一点数学。
A regular die will give each number 1-6 with equal probability, namely 1/6. This is referred to as uniform distribution(the discrete version of it, as opposed to the continuous version). Meaning that if Xis a random variable describing the result of a single role then X~U[1,6]- meaning Xis distributed equally against all possible results of the die roll, 1 through 6.
一个普通的骰子会给每个数字 1-6 的概率相等，即1/6。这被称为均匀分布（它的离散版本，与连续版本相反）。这意味着 ifX是一个随机变量，描述单个角色的结果，则X~U[1,6]- 意义X针对骰子的所有可能结果平均分配，1 到 6。
This is equal to choosing a number in [0,1)while dividing it into 6 sections: [0,1/6), [1/6,2/6), [2/6,3/6), [3/6,4/6), [4/6,5/6), [5/6,1).
这相当于在选择数目[0,1)而将其分成6个部分：[0,1/6)，[1/6,2/6)，[2/6,3/6)，[3/6,4/6)，[4/6,5/6)，[5/6,1)。
You are requesting a different distribution, which is biased.
The easiest way to achieve this is to divide the section [0,1)to 6 parts depending on the bias you want. So in your case you would want to divide it into the following:
[0,0.2), [0.2,0.4), [0.4,0.55), 0.55,0.7), [0.7,0.84), [0.84,1).
您正在请求不同的分布，这是有偏见的。实现这一目标的最简单方法是[0,1)根据您想要的偏差将部分分成6 个部分。因此，在您的情况下，您希望将其分为以下几部分：
[0,0.2), [0.2,0.4), [0.4,0.55), 0.55,0.7), [0.7,0.84), [0.84,1)。
If you take a look at the wikipedia entry, you will see that in this case, the cumulative probability function will not be composed of 6 equal-length parts but rather of 6 parts which differ in length according to the biasyou gave them. Same goes for the mass distribution.
如果您查看维基百科条目，您会发现在这种情况下，累积概率函数不会由 6 个等长的部分组成，而是由 6 个长度不同的部分组成，这些部分的长度根据您给它们的偏差而定。质量分布也是如此。
Back to the question, depending on the language you are using, translate this back to your die roll. In Python, here is a very sketchy, albeit working, example:
回到问题，根据您使用的语言，将其翻译回您的掷骰子。在 Python 中，这是一个非常粗略但有效的示例：
import random
sampleMassDist = (0.2, 0.1, 0.15, 0.15, 0.25, 0.15)
# assume sum of bias is 1
def roll(massDist):
randRoll = random.random() # in [0,1]
sum = 0
result = 1
for mass in massDist:
sum += mass
if randRoll < sum:
return result
result+=1
print(roll(sampleMassDist))
回答by Toon Krijthe
More language agnostic, but you could use a lookup table.
更多的语言不可知，但你可以使用查找表。
Use a random number in the range 0-1 and lookup the value in a table:
使用 0-1 范围内的随机数并在表中查找值：
0.00 - 0.20 1
0.20 - 0.40 2
0.40 - 0.55 3
0.55 - 0.70 4
0.70 - 0.84 5
0.84 - 1.00 6
回答by nosklo
import random
def roll(sides, bias_list):
assert len(bias_list) == sides
number = random.uniform(0, sum(bias_list))
current = 0
for i, bias in enumerate(bias_list):
current += bias
if number <= current:
return i + 1
The bias will be proportional.
偏差将成正比。
>>> print roll(6, (0.20, 0.20, 0.15, 0.15, 0.14, 0.16))
6
>>> print roll(6, (0.20, 0.20, 0.15, 0.15, 0.14, 0.16))
2
Could use integers too (better):
也可以使用整数（更好）：
>>> print roll(6, (10, 1, 1, 1, 1, 1))
5
>>> print roll(6, (10, 1, 1, 1, 1, 1))
1
>>> print roll(6, (10, 1, 1, 1, 1, 1))
1
>>> print roll(6, (10, 5, 5, 10, 4, 8))
2
>>> print roll(6, (1,) * 6)
4
回答by shrokmel
It is a little surprising that the np.random.choiceanswer hasn't been given here.
np.random.choice这里没有给出答案，这有点令人惊讶。
from numpy import random
def roll(N,bias):
'''this function rolls N dimensional die with biasing provided'''
return random.choice(np.range(N),p=bias)
The p option gives "the probabilities associated with each entry in a", where ais np.range(N)for us. "If not given the sample assumes a uniform distribution over all entries in a".
该p选项使“与每一项目相关的概率一个，其中”一个是np.range(N)我们。“如果没有给出，样本假设在一个”中的所有条目上均匀分布。
回答by denis
See the recipe for Walker's alias methodfor random objects with different probablities.
An example, strings A B C or D with probabilities .1 .2 .3 .4 --
请参阅Walker对具有不同概率的随机对象的别名方法的配方。
例如，字符串 ABC 或 D 的概率为 .1 .2 .3 .4 --
abcd = dict( A=1, D=4, C=3, B=2 )
# keys can be any immutables: 2d points, colors, atoms ...
wrand = Walkerrandom( abcd.values(), abcd.keys() )
wrand.random() # each call -> "A" "B" "C" or "D"
# fast: 1 randint(), 1 uniform(), table lookup
cheers
-- denis
干杯
——丹尼斯
回答by Mapio
Just to suggest a more efficient (and pythonic3) solution, one can use bisectto search in the vector of accumulated values — that can moreover be precomputed and stored in the hope that subsequent calls to the function will refer to the same "bias" (to follow the question parlance).
只是为了提出一个更有效的（和 pythonic3）解决方案，可以使用bisect来搜索累积值的向量——而且可以预先计算和存储，希望随后对该函数的调用将引用相同的“偏差”（遵循问题的说法）。
from bisect import bisect
from itertools import accumulate
from random import uniform
def pick( amplitudes ):
if pick.amplitudes != amplitudes:
pick.dist = list( accumulate( amplitudes ) )
pick.amplitudes = amplitudes
return bisect( pick.dist, uniform( 0, pick.dist[ -1 ] ) )
pick.amplitudes = None
In absence of Python 3 accumulate, one can just write a simple loop to compute the cumulative sum.
在没有 Python 3 累加的情况下，我们可以编写一个简单的循环来计算累加和。
回答by edo liberty
from random import random
biases = [0.0,0.3,0.5,0.99]
coins = [1 if random()<bias else 0 for bias in biases]
回答by niksy
i have created a code for a dictionary giving a event and corresponding probability, it gives back the corresponding key ie the event of that probability.
我为字典创建了一个代码，给出了一个事件和相应的概率，它返回了相应的键，即该概率的事件。
import random
def WeightedDie(Probabilities):
high_p = 0
rand = random.uniform(0,1)
for j,i in Probabilities.items():
high_p = high_p + i
if rand< high_p:
return j
回答by Sandipan Dey
We could use numpy's multinomialdistribution too
我们也可以使用numpy'smultinomial分布
import numpy as np
bias = [0.10,0.10,0.15,0.15,0.14,0.16,0.05,0.06,0.04,0.05] # a 10-sided biased die
np.where(np.random.multinomial(1, bias, size=1)[0]==1)[0][0]+1 # just 1 roll
# 4
If you want to roll the biased die (with the given biasprobabilities) for ntimes, use the following function
如果您想多次掷有偏差的骰子（具有给定的bias概率）n，请使用以下函数
def roll(probs, ntimes): # roll a len(probs) sided biased die with bias probs for ntimes
return np.apply_along_axis(lambda x: x.tolist().index(1)+1, 1,
np.random.multinomial(1, bias, size=10)).tolist()
roll(probs=bias, ntimes=10) # 10 rolls
# [5, 6, 8, 4, 8, 3, 1, 5, 8, 6]
