python图正态分布

问题:python图正态分布

给定一个均值和方差,是否有一个简单的函数调用可以绘制正态分布?

Given a mean and a variance is there a simple function call which will plot a normal distribution?


回答 0

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()

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()


回答 1

我认为没有一个函数可以在一个调用中完成所有这些操作。但是,您可以在中找到高斯概率密度函数scipy.stats

因此,我想出的最简单方法是:

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:

I don’t think there is a function that does all that in a single call. However you can find the Gaussian probability density function in scipy.stats.

So the simplest way I could come up with is:

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()

Sources:


回答 2

使用seaborn代替我正在使用均值= 5 std = 3的1000值的seaborn的distplot

value = np.random.normal(loc=5,scale=3,size=1000)
sns.distplot(value)

您将获得正态分布曲线

Use seaborn instead i am using distplot of seaborn with mean=5 std=3 of 1000 values

value = np.random.normal(loc=5,scale=3,size=1000)
sns.distplot(value)

You will get a normal distribution curve


回答 3

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,所以我还是想更改一下:

x = np.linspace(-3 * sigma, 3 * sigma, 100)

对此:

x = np.linspace(-3 * sigma + mean, 3 * sigma + mean, 100)

Unutbu answer is correct. But because our mean can be more or less than zero I would still like to change this :

x = np.linspace(-3 * sigma, 3 * sigma, 100)

to this :

x = np.linspace(-3 * sigma + mean, 3 * sigma + mean, 100)

回答 4

如果您喜欢使用逐步方法,可以考虑以下解决方案

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()

If you prefer to use a step by step approach you could consider a solution like follows

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()

回答 5

我刚刚回到这个问题,我不得不安装scipy,因为MatplotlibDeprecationWarning: scipy.stats.norm.pdf在尝试上述示例时,matplotlib.mlab给了我错误消息。现在的示例是:

%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()

I have just come back to this and I had to install scipy as matplotlib.mlab gave me the error message MatplotlibDeprecationWarning: scipy.stats.norm.pdf when trying example above. So the sample is now:

%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()

回答 6

我相信设置高度很重要,因此创建了以下功能:

def my_gauss(x, sigma=1, h=1, mid=0):
    from math import exp, pow
    variance = pow(sdev, 2)
    return h * exp(-pow(x-mid, 2)/(2*variance))

其中sigma,标准偏差h为,高度mid为平均值。

这是使用不同高度和偏差的结果:

I believe that is important to set the height, so created this function:

def my_gauss(x, sigma=1, h=1, mid=0):
    from math import exp, pow
    variance = pow(sdev, 2)
    return h * exp(-pow(x-mid, 2)/(2*variance))

Where sigma is the standard deviation, h is the height and mid is the mean.

Here is the result using different heights and deviations:


回答 7

您可以轻松获得CDF。所以通过cdf pdf

    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()

you can get cdf easily. so pdf via 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()