python画经验分布函数,python统计绘

参考文献：
一篇绘制累积经验分布函数图像的博客

如何得到样本数据的经验分布函数？ from statsmodels.distributions.empirical_distribution import ECDFecdf = ECDF([3,3,1,4]) # 返回了一个分布函数，我是说数学书上的函数type(ecdf)Out[19]: statsmodels.distributions.empirical_distribution.ECDFecdf(3) # 往这个分布函数中输入自变量，会得到其分布函数值Out[20]: 0.75ecdf(1)Out[21]: 0.25help(ECDF) # 看看介绍，我也没太懂。。。Help on class ECDF in module statsmodels.distributions.empirical_distribution:class ECDF(StepFunction) | ECDF(x, side='right') | | Return the Empirical CDF of an array as a step function. | | Parameters | ---------- | x : array_like | Observations | side : {'left', 'right'}, optional | Default is 'right'. Defines the shape of the intervals constituting the | steps. 'right' correspond to [a, b) intervals and 'left' to (a, b]. | | Returns | ------- | Empirical CDF as a step function. | | Examples | -------- | >>> import numpy as np | >>> from statsmodels.distributions.empirical_distribution import ECDF | >>> | >>> ecdf = ECDF([3, 3, 1, 4]) | >>> | >>> ecdf([3, 55, 0.5, 1.5]) | array([ 0.75, 1. , 0. , 0.25]) | | Method resolution order: | ECDF | StepFunction | builtins.object | | Methods defined here: | | __init__(self, x, side='right') | Initialize self. See help(type(self)) for accurate signature. | | ---------------------------------------------------------------------- | Methods inherited from StepFunction: | | __call__(self, time) | Call self as a function. | | ---------------------------------------------------------------------- | Data descriptors inherited from StepFunction: | | __dict__ | dictionary for instance variables (if defined) | | __weakref__ | list of weak references to the object (if defined) 绘制样本的经验分布函数 方法一：

博客原文

# 这也是借鉴别人博客的代码，我再补充点自己的看法import numpy as npimport pandas as pdimport matplotlib.pyplot as pltdatas = np.array([64.3, 65.0, 65.0, 67.2, 67.3, 67.3, 67.3, 67.3, 68.0, 68.0, 68.8, 68.8, 68.8, 69.7, 69.7, 69.7, 70.3,70.4, 70.4, 70.4, 70.4, 70.4,70.4, 70.4, 71.2, 71.2, 71.2, 71.2, 72.0, 72.0, 72.0, 72.0, 72.0, 72.0, 72.0, 72.7, 72.7, 72.7, 72.7, 72.7, 72.7, 72.7, 73.5, 73.5, 73.5, 73.5, 73.5, 73.5, 73.5, 73.5, 73.5,73.5, 73.5, 74.3, 74.3, 74.3, 74.3, 74.3, 74.3, 74.3, 74.3, 74.7, 75.0, 75.0, 75.0, 75.0, 75.0, 75.0, 75.0, 75.4, 75.6, 75.8, 75.8, 75.8, 75.8, 75.8, 76.5, 76.5, 76.5, 76.5, 76.5, 76.5, 76.5, 77.2, 77.2,77.6, 78.0, 78.8, 78.8, 78.8, 79.5, 79.5, 79.5, 80.3, 80.5, 80.5, 81.2, 81.6, 81.6, 84.3]) #数据特征计算s = np.std(datas, ddof=1)#样本标准差xbar = np.mean(datas)#样本均值#数据可视化 画数据经验分布曲线图nt, bins, patches = plt.hist(datas, bins=10, histtype='step', cumulative=True, density=True, color='darkcyan')#datas是数据,bins是分组数plt.title('bins = 10')plt.savefig('经验函数分布图1.jpg', dpi=200)plt.show()#数据可视化 画数据经验分布曲线图nt, bins, patches = plt.hist(datas, bins=15, histtype='step', cumulative=True, density=True, color='darkcyan')#datas是数据,bins是分组数plt.title('bins = 15')#正态分布函数曲线拟合y = (1 / (np.sqrt(2 * np.pi) * s)) * np.exp(-0.5 * ((bins - xbar) ** 2 / s ** 2))y = y.cumsum()y = y / y[-1]plt.plot(bins, y, 'tomato', linewidth = 1.5, label = 'Theoretical')plt.savefig('经验函数分布图2.jpg', dpi=200)plt.show()# 这种方法有点奇怪，指定bins参数之后，就不能说绘制出来的经验分布函数是原样本的经验分布函数了# 自己试试看，图太多了不方便一一往上贴，看看返回数组plt.hist([3,3,1,4],histtype='step',cumulative=True,density=True)Out[25]: (array([0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 0.75, 0.75, 0.75, 1. ]), array([1. , 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7, 4. ]), [<matplotlib.patches.Polygon at 0x14fb356be20>])plt.hist([3,3,1,4],histtype='step',cumulative=True) # 没有density参数表示计数，而非计算频率Out[26]: (array([1., 1., 1., 1., 1., 1., 3., 3., 3., 4.]), array([1. , 1.3, 1.6, 1.9, 2.2, 2.5, 2.8, 3.1, 3.4, 3.7, 4. ]), [<matplotlib.patches.Polygon at 0x14fb4c47d90>]) plt.hist([3,3,1,4],histtype='step',cumulative=True,bins=4,density=True)Out[28]: (array([0.25, 0.25, 0.75, 1. ]), array([1. , 1.75, 2.5 , 3.25, 4. ]), [<matplotlib.patches.Polygon at 0x14fb6e5bb20>])plt.hist([3,3,1,4],histtype='step',cumulative=True,bins=3,density=True)Out[29]: (array([0.25, 0.25, 1. ]), array([1., 2., 3., 4.]), [<matplotlib.patches.Polygon at 0x14fb4c31850>])plt.hist([3,3,1,4],histtype='step',cumulative=True,bins=5,density=True)Out[30]: (array([0.25, 0.25, 0.25, 0.75, 1. ]), array([1. , 1.6, 2.2, 2.8, 3.4, 4. ]), [<matplotlib.patches.Polygon at 0x14fb4c37820>]) # 也就是说，bins参数是平均分原数据的份数。由原数据的最大值和最小值以及bins参数共同决定各个子区间的范围。 方法二： from statsmodels.distributions.empirical_distribution import ECDFx = [3,3,1,4]ecdf = ECDF([3,3,1,4])type(ecdf)Out[37]: statsmodels.distributions.empirical_distribution.ECDFy = ecdf(x) # 计算分布函数值x.sort()xOut[43]: [1, 3, 3, 4]y.sort()# 画阶梯函数之前一定要记得排序，不然就是乱七八糟的回字形plt.step(x,y)Out[45]: [<matplotlib.lines.Line2D at 0x14fc03b6850>]

这就是我想要的了