Question

我正在尝试找到适合我的数据

的最佳分布曲线

y-axis = [0, 0, 0, 0, 0.24, 0.53, 0.49, 0.64, 0.54, 0.78, 0.59, 0.44, 
          0.34, 0.88, 0.2, 0.49, 0.39, 0.39, 0.29, 0.2, 0.05, 0.05, 
          0.25, 0.05, 0.1, 0.15, 0.1, 0.1, 0.1, 0, 0, 0, 0, 0]

y轴是在x轴时间仓中发生的事件的概率：

x-axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 
          12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 
          22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 
          32.0, 33.0, 34.0]

我在python跟随Fitting empirical distribution to theoretical ones with Scipy (Python)?

提供的示例中执行此操作

具体来说，我正在尝试重新创建名为“方差误差和（SSE）的分布拟合”的部分，其中您运行不同的分布以找到适合数据的分布。

如何修改该示例以使其在我的数据输入上工作？回答

根据Bill的回复更新版本，但现在尝试根据数据绘制拟合曲线并查看其中的内容：

%matplotlib inline
import matplotlib.pyplot as plt
import scipy
import scipy.stats
import numpy as np
from scipy.stats import gamma, lognorm, loglaplace
from scipy.optimize import curve_fit

x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0]
y_axis = [0, 0, 0, 0, 0.24, 0.53, 0.49, 0.64, 0.54, 0.78, 0.59, 0.44, 0.34, 0.88, 0.2, 0.49, 0.39, 0.39, 0.29, 0.2, 0.05, 0.05, 0.25, 0.05, 0.1, 0.15, 0.1, 0.1, 0.1, 0, 0, 0, 0, 0]

matplotlib.rcParams['figure.figsize'] = (16.0, 12.0)
matplotlib.style.use('ggplot')

def f(x, a, loc, scale):
    return gamma.pdf(x, a, loc, scale)

result, pcov = curve_fit(f, x_axis, y_axis)

# get curve shape, location, scale
shape = result[:-2]
loc = result[-2]
scale = result[-1]

# construct the curve
x = np.linspace(0, 36, 100)
y = f(x, *result)

plt.bar(x_axis, y_axis, width, alpha=0.75)
plt.plot(x, y, c='g')

Answer 1

您的情况与您引用的问题中处理的情况不同。你有数据点的纵坐标和横坐标，而不是通常的i.i.d.样品。我建议您使用scipy curve_fit。这是一个样本。

x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 32.0, 33.0, 34.0]
y_axis = [0, 0, 0, 0, 0.24, 0.53, 0.49, 0.64, 0.54, 0.78, 0.59, 0.44, 0.34, 0.88, 0.2, 0.49, 0.39, 0.39, 0.29, 0.2, 0.05, 0.05, 0.25, 0.05, 0.1, 0.15, 0.1, 0.1, 0.1, 0, 0, 0, 0, 0]

## y_axis values must be normalised
sum_ys = sum(y_axis)
y_axis = [_/sum_ys for _ in y_axis]
print (sum(y_axis))

from scipy.stats import gamma, norm
from scipy.optimize import curve_fit

def gamma_f(x, a, loc, scale):
    return gamma.pdf(x, a, loc, scale)

def norm_f(x, loc, scale):
    return norm.pdf(x, loc, scale)

fitting = norm_f

result = curve_fit(fitting, x_axis, y_axis)
print (result)

import matplotlib.pyplot as plt

plt.plot(x_axis, y_axis, 'ro')
plt.plot(x_axis, [fitting(_, *result[0]) for _ in x_axis], 'b-')
plt.axis([0,35,0,.5])
plt.show()

此版本显示了如何绘制一个图表，以便正常拟合数据。（伽马不合适。）法线只需要两个参数。一般而言，您只需要输出结果的第一部分，参数估计，形状，位置和比例。

(array([  2.3352639 ,  -3.08105104,  10.15024823]), array([[   5954.86532869,  -27818.92220973,  -19675.22421994],
       [ -27818.92220973,  133161.76500251,   90741.43608615],
       [ -19675.22421994,   90741.43608615,   66054.79087992]]))

请注意，伽玛分布的pdf也可以在scipy中使用，我认为，您可以使用其他代码来节省编码工作。

我从第一个代码中省略的最重要的事情是需要对y值进行归一化，即使它们总和为1，因为它们应该近似直方图高度。

Answer 2

我使用OpenTURNS平台尝试了您的示例这是我得到的。

在导入openturns和openturs.viewer.View进行绘图之后，我从与您相同的数据开始

    import openturns as ot
    from openturns.viewer import View

    x_axis = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0, 9.0, 10.0, 11.0, 
          12.0, 13.0, 14.0, 15.0, 16.0, 17.0, 18.0, 19.0, 20.0, 21.0, 
          22.0, 23.0, 24.0, 25.0, 26.0, 27.0, 28.0, 29.0, 30.0, 31.0, 
          32.0, 33.0, 34.0]

    y_axis = [0, 0, 0, 0, 0.24, 0.53, 0.49, 0.64, 0.54, 0.78, 0.59, 0.44, 
          0.34, 0.88, 0.2, 0.49, 0.39, 0.39, 0.29, 0.2, 0.05, 0.05, 
          0.25, 0.05, 0.1, 0.15, 0.1, 0.1, 0.1, 0, 0, 0, 0, 0]

第一步：我们可以定义相应的分布

    distribution = ot.UserDefined(ot.Sample([[s] for s in x_axis]), y_axis)
    graph = distribution.drawPDF()
    graph.setColors(["black"])
    graph.setLegends(["your input"])

在此阶段，如果您View(graph)，您将获得：

第二步：我们可以从获得的分布中得出样本

    sample = distribution.getSample(10000)

该样本将用于拟合任何种类的分布。我尝试使用WeibullMin和Gamma分布

    # WeibullMin Factory
    distribution2 = ot.WeibullMinFactory().build(sample)
    print(distribution2)
    graph2 = distribution2.drawPDF() ; graph2.setLegends(["Best WeibullMin"])
    >>> WeibullMin(beta = 8.83969, alpha = 1.48142, gamma = 4.76832)

    # Gamma Factory
    distribution3 = ot.GammaFactory().build(sample)
    print(distribution3)
    >>> Gamma(k = 2.08142, lambda = 0.25157, gamma = 4.9995)
    graph3 = distribution3.drawPDF() ; graph3.setLegends(["Best Gamma"]) ; 
    graph3.setColors(["blue"])

    # plotting all the results
    graph.add(graph2) ; graph.add(graph3)
    View(graph)

Answer 3

我认为这是计算平方误差总和的最好和最简单的方法：

#编写函数

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<块引用>

#now调用函数并获取结果

def SSE(y_true, y_pred):

     sse= np.sum((y_true-y_pred)**2)

     print(sse)

具有平方误差和（SSE）的Python分布拟合

3 个答案: