给定均值和方差,是否有一个简单的函数调用可以绘制正态分布?
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10 回答
273
import matplotlib.pyplot as plt
import numpy as np
import scipy.stats as stats
import math
mu = 0
variance = 1
sigma = math.sqrt(variance)
x = np.linspace(mu - 3*sigma, mu + 3*sigma, 100)
plt.plot(x, stats.norm.pdf(x, mu, sigma))
plt.show()
于 2012-04-13T09:25:09.900 回答
58
我认为没有一个函数可以在一次调用中完成所有这些。但是,您可以在 中找到高斯概率密度函数scipy.stats
。
所以我能想到的最简单的方法是:
import numpy as np
import matplotlib.pyplot as plt
from scipy.stats import norm
# Plot between -10 and 10 with .001 steps.
x_axis = np.arange(-10, 10, 0.001)
# Mean = 0, SD = 2.
plt.plot(x_axis, norm.pdf(x_axis,0,2))
plt.show()
资料来源:
于 2012-04-13T09:29:36.937 回答
19
使用 seaborn 代替我使用 seaborn 的 distplot,mean=5 std=3 of 1000 个值
value = np.random.normal(loc=5,scale=3,size=1000)
sns.distplot(value)
你会得到一个正态分布曲线
于 2018-08-09T03:43:51.147 回答
13
如果您更喜欢使用逐步的方法,您可以考虑如下解决方案
import numpy as np
import matplotlib.pyplot as plt
mean = 0; std = 1; variance = np.square(std)
x = np.arange(-5,5,.01)
f = np.exp(-np.square(x-mean)/2*variance)/(np.sqrt(2*np.pi*variance))
plt.plot(x,f)
plt.ylabel('gaussian distribution')
plt.show()
于 2017-12-11T23:18:46.590 回答
11
Unutbu 的答案是正确的。但是因为我们的平均值可以大于或小于零,所以我仍然想改变它:
x = np.linspace(-3 * sigma, 3 * sigma, 100)
对此:
x = np.linspace(-3 * sigma + mean, 3 * sigma + mean, 100)
于 2015-02-07T10:21:31.920 回答
1
我刚刚回到这个问题,我不得不安装 scipy,因为 matplotlib.mlabMatplotlibDeprecationWarning: scipy.stats.norm.pdf
在尝试上面的示例时给了我错误消息。所以现在的样本是:
%matplotlib inline
import math
import matplotlib.pyplot as plt
import numpy as np
import scipy.stats
mu = 0
variance = 1
sigma = math.sqrt(variance)
x = np.linspace(mu - 3*sigma, mu + 3*sigma, 100)
plt.plot(x, scipy.stats.norm.pdf(x, mu, sigma))
plt.show()
于 2018-07-05T09:34:34.287 回答
0
您可以轻松获得 cdf。所以pdf通过cdf
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate
import scipy.stats
def setGridLine(ax):
#http://jonathansoma.com/lede/data-studio/matplotlib/adding-grid-lines-to-a-matplotlib-chart/
ax.set_axisbelow(True)
ax.minorticks_on()
ax.grid(which='major', linestyle='-', linewidth=0.5, color='grey')
ax.grid(which='minor', linestyle=':', linewidth=0.5, color='#a6a6a6')
ax.tick_params(which='both', # Options for both major and minor ticks
top=False, # turn off top ticks
left=False, # turn off left ticks
right=False, # turn off right ticks
bottom=False) # turn off bottom ticks
data1 = np.random.normal(0,1,1000000)
x=np.sort(data1)
y=np.arange(x.shape[0])/(x.shape[0]+1)
f2 = scipy.interpolate.interp1d(x, y,kind='linear')
x2 = np.linspace(x[0],x[-1],1001)
y2 = f2(x2)
y2b = np.diff(y2)/np.diff(x2)
x2b=(x2[1:]+x2[:-1])/2.
f3 = scipy.interpolate.interp1d(x, y,kind='cubic')
x3 = np.linspace(x[0],x[-1],1001)
y3 = f3(x3)
y3b = np.diff(y3)/np.diff(x3)
x3b=(x3[1:]+x3[:-1])/2.
bins=np.arange(-4,4,0.1)
bins_centers=0.5*(bins[1:]+bins[:-1])
cdf = scipy.stats.norm.cdf(bins_centers)
pdf = scipy.stats.norm.pdf(bins_centers)
plt.rcParams["font.size"] = 18
fig, ax = plt.subplots(3,1,figsize=(10,16))
ax[0].set_title("cdf")
ax[0].plot(x,y,label="data")
ax[0].plot(x2,y2,label="linear")
ax[0].plot(x3,y3,label="cubic")
ax[0].plot(bins_centers,cdf,label="ans")
ax[1].set_title("pdf:linear")
ax[1].plot(x2b,y2b,label="linear")
ax[1].plot(bins_centers,pdf,label="ans")
ax[2].set_title("pdf:cubic")
ax[2].plot(x3b,y3b,label="cubic")
ax[2].plot(bins_centers,pdf,label="ans")
for idx in range(3):
ax[idx].legend()
setGridLine(ax[idx])
plt.show()
plt.clf()
plt.close()
于 2018-04-22T04:36:59.787 回答
0
import math
import matplotlib.pyplot as plt
import numpy
import pandas as pd
def normal_pdf(x, mu=0, sigma=1):
sqrt_two_pi = math.sqrt(math.pi * 2)
return math.exp(-(x - mu) ** 2 / 2 / sigma ** 2) / (sqrt_two_pi * sigma)
df = pd.DataFrame({'x1': numpy.arange(-10, 10, 0.1), 'y1': map(normal_pdf, numpy.arange(-10, 10, 0.1))})
plt.plot('x1', 'y1', data=df, marker='o', markerfacecolor='blue', markersize=5, color='skyblue', linewidth=1)
plt.show()
于 2020-10-02T10:48:55.380 回答
0
对我来说,如果您尝试绘制特定的 pdf,这非常有效
theta1 = {
"a": 0.5,
"cov" : 1,
"mean" : 0
}
x = np.linspace(start = 0, stop = 1000, num = 1000)
pdf = stats.norm.pdf(x, theta1['mean'], theta1['cov']) + theta2['a']
sns.lineplot(x,pdf)
于 2022-03-02T09:11:22.130 回答