英文:
How to Algorithmically Generate a List of Probabilities
问题
请原谅我对统计学术语的不熟悉。
我已经得到一个任意的值列表要进行抽样,目前如下所示:
list_to_sample = [1, 2, 3, 4, 5]
。
在这一点上,列表包含的内容并不重要,但列表的长度为5。
而且,我已经得到一个几乎任意的“帕累托式”概率列表,目前如下所示:
probability_list = [0.5, 0.3, 0.1, 0.05, 0.05]
(帕累托式因为它不遵循80-20规则,而是80/40,即前80%的概率选定值将在列表的前40%中。)
现在,我正在尝试将这个通用化,以便如果list_to_sample
变得更长,比如:
[1, 2, 3, 4, 5, 6, 7, 8]
我可以扩展probability_list
并保持相同的曲线。
我正在尝试使用np.pareto.pdf
来生成类似于以下的概率列表:
[0.5, 0.3, 0.1, 0.05, 0.05]
并且列表的总和(概率的总和)等于1。
具体来说,我尝试过这样做:
import numpy as np
list_to_sample = [1, 2, 3, 4, 5]
output = np.array([pareto.pdf(x=list_to_sample, b=1, loc=0, scale=1)])
输出:
[[0.5 0.125 0.05555556 0.03125 0.02 ]]
我尝试改变参数,但没有成功。我希望通过改变参数可以使帕累托分布产生所期望的结果。到目前为止,没有成功。
也许有更好的函数来生成(或扩展)概率列表。
英文:
Please forgive my lack of statistical nomenclature.
I've been given an arbitrary list of values to sample, currently:
list_to_sample = [1, 2, 3, 4, 5]
.
At this point, it doesn't matter what the list contains, but that the length of list is 5.
And, I've been given a list of almost arbitrary "pareto-like" probabilities, currently:
probability_list = [0.5, 0.3, 0.1, 0.05, 0.05]
(pareto-like as it does not follow the 80-20, but rather 80/40 as the top 80% of probable selected values will be in the top 40% of the list.
I am now trying to generalize this, so that if list_to_sample
gets longer, like:
[1, 2, 3, 4, 5, 6, 7, 8]
I can extend the probability_list
and maintain the same curve.
I am trying to use np.pareto.pdf
to produce a list of probabilities that is similar to:
[0.5, 0.3, 0.1, 0.05, 0.05]
and where the sum of the list (the sum of the probabilities) equals 1.
Specifically, I have tried this:
import numpy as np
list_to_sample = [1, 2, 3, 4, 5]
output = np.array([pareto.pdf(x=list_to_sample, b=1, loc=0, scale=1)])
Output:
[[0.5 0.125 0.05555556 0.03125 0.02 ]]
I have tried changing parameters to no avail. I was hopeful that by changing parameters I could get pareto to produce the desired result. So far, no luck.
Perhaps there is a better function to produce (or extend) a list of probabilities.
答案1
得分: 0
不需要使用帕累托分布吗?如果需要,我认为这个问题没有很好定义,因为list_sample
中的项目会很重要,从你的问题中我看不出如何定义帕累托分布的所有参数。
如果你可以使用其他技巧,我建议使用简单的插值,例如三次样条插值。由于你说列表中的值并不重要,我们可以使用百分比值来处理。
import numpy as np
import scipy as sp
list_to_sample = [1, 2, 3, 4, 5]
probability_list = [0.5, 0.3, 0.1, 0.05, 0.05]
# --- 在开头添加零,以确保将零映射到零
x = np.array([0] + list_to_sample) / len(list_to_sample)
y = np.array([0] + probability_list).cumsum()
print("x:", x) # -- [0.0 0.2 0.40 0.60 0.80 1.0]
print("y:", y) # -- [0.0 0.5 0.80 0.90 0.95 1.0]
# - 样条
spline = sp.interpolate.CubicSpline(x, y)
new_values = np.arange(1, 11)
cprobs = spline(new_values / len(new_values))
print("新数值:", new_values)
print("累积概率:", cprobs)
# -- 前40%的总概率仍然是80%,
# -- 下面的输出已经四舍五入
# -- [ 1 2 3 4 5 6 7 8 9 10]
# -- [0.27 0.50 0.68 0.80 0.87 0.90 0.93 0.95 0.97 1.00]
# - 要获取每个值的概率,只需对cprobs进行差分
probs = np.diff([0] + list(cprobs))
print("概率:", probs)
# -- [0.272 0.228 0.178 0.122 0.067 0.034 0.026 0.024 0.024 0.026]
希望这对你有所帮助。
英文:
Do you need to use the Pareto distribution? If so, I don´t think this problem is well-defined as the items in list_sample
will matter and I don´t see from your question how you can define all the parameters of the Pareto distribution.
If you can use other techniques, I would go with a simple interpolation, for example the cubic spline. Since you said the values in the list don´t matter, we can work with the percentage values instead.
import numpy as np
import scipy as sp
list_to_sample = [1, 2, 3, 4, 5]
probability_list = [0.5, 0.3, 0.1, 0.05, 0.05]
# --- adding zero at the beginning to ensure the we map zero to zero
x = np.array([0] + list_to_sample) / len(list_to_sample)
y = np.array([0] + probability_list).cumsum()
print("x:", x) # -- [0.0 0.2 0.40 0.60 0.80 1.0]
print("y:", y) # -- [0.0 0.5 0.80 0.90 0.95 1.0]
# - spline
spline = sp.interpolate.CubicSpline(x, y)
new_values = np.arange(1, 11)
cprobs = spline(new_values / len(new_values))
print("New values:", new_values)
print("Cumulative probabilities:", cprobs)
# -- the top 40% still has an overall 80% probability,
# -- the output below is rounded
# -- [ 1 2 3 4 5 6 7 8 9 10]
# -- [0.27 0.50 0.68 0.80 0.87 0.90 0.93 0.95 0.97 1.00]
# - to get the probability for each value we just diff cprobs
probs = np.diff([0] + list(cprobs))
print("Probabilities:", probs)
# -- [0.272 0.228 0.178 0.122 0.067 0.034 0.026 0.024 0.024 0.026]
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