Given a mean and a variance is there a simple function call which will plot a normal distribution?
Answers:
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Method 1
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()

Method 2
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:
- http://www.johndcook.com/distributions_scipy.html
- http://docs.scipy.org/doc/scipy/reference/stats.html
- http://telliott99.blogspot.com/2010/02/plotting-normal-distribution-with.html
Method 3
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
Method 4
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()
Method 5
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)
Method 6
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(sigma, 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:
Method 7
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()
Method 8
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()
Method 9
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()
Method 10
For me, this worked pretty well if you are trying to plot a particular pdf
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)
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0

