第十四周python作业：pandas seaborn练习

Part 1：

For each of the four datasets…

Compute the mean and variance of both x and y 

Compute the correlation coefficient between x and y 

Computethe linear regression line: y=β0+β1x+ϵy=β0+β1x+ϵ (hint: use statsmodels andlook at the Statsmodels notebook)

Part 2：

Using Seaborn, visualize all four datasets.

hint: use sns.FacetGrid combined with plt.scatter 

    Part1 代码：

import random  
import numpy as np  
import scipy as sp  
import pandas as pd  
import matplotlib.pyplot as plt  
import seaborn as sns  

import statsmodels.api as sm  
import statsmodels.formula.api as smf 

sns.set_context("talk")  
anscombe = pd.read_csv('anscombe.csv') 

mx = anscombe.groupby('dataset').mean().x  
my = anscombe.groupby('dataset').mean().y  
vx = anscombe.groupby('dataset').var().x  
vy = anscombe.groupby('dataset').var().y  

print('x mean : \n', mx, '\n')  
print('y mean : \n', my, '\n')  
print('x var : \n', vx, '\n')  
print('y var : \n', vy, '\n')  

cor = anscombe.groupby('dataset').corr()  
print('correlation : \n', cor, '\n')  

for a in [anscombe[anscombe.dataset == i] for i in ['I', 'II', 'III', 'IV']]:  
    s_x = sm.add_constant(np.array(a.x))  
    s_y = np.array(a.y)  
    beta_pair = sm.OLS(s_y, s_x).fit()  
    print('β1, β0 = ', beta_pair.params) 

运行结果：

x mean : 
 dataset
I      9.0
II     9.0
III    9.0
IV     9.0
Name: x, dtype: float64 

y mean : 
 dataset
I      7.500909
II     7.500909
III    7.500000
IV     7.500909
Name: y, dtype: float64 

x var : 
 dataset
I      11.0
II     11.0
III    11.0
IV     11.0
Name: x, dtype: float64 

y var : 
 dataset
I      4.127269
II     4.127629
III    4.122620
IV     4.123249
Name: y, dtype: float64 

correlation : 
                   x         y
dataset                      
I       x  1.000000  0.816421
        y  0.816421  1.000000
II      x  1.000000  0.816237
        y  0.816237  1.000000
III     x  1.000000  0.816287
        y  0.816287  1.000000
IV      x  1.000000  0.816521
        y  0.816521  1.000000 

β1, β0 =  [3.00009091 0.50009091]
β1, β0 =  [3.00090909 0.5       ]
β1, β0 =  [3.00245455 0.49972727]
β1, β0 =  [3.00172727 0.49990909]

Part2 代码：

import random  
import numpy as np  
import scipy as sp  
import pandas as pd  
import matplotlib.pyplot as plt  
import seaborn as sns  

import statsmodels.api as sm  
import statsmodels.formula.api as smf 

sns.set_context("talk")  
anscombe = pd.read_csv('anscombe.csv') 

mx = anscombe.groupby('dataset').mean().x  
my = anscombe.groupby('dataset').mean().y  
vx = anscombe.groupby('dataset').var().x  
vy = anscombe.groupby('dataset').var().y  

print('x mean : \n', mx, '\n')  
print('y mean : \n', my, '\n')  
print('x var : \n', vx, '\n')  
print('y var : \n', vy, '\n')  

cor = anscombe.groupby('dataset').corr()  
print('correlation : \n', cor, '\n')  

for a in [anscombe[anscombe.dataset == i] for i in ['I', 'II', 'III', 'IV']]:  
    s_x = sm.add_constant(np.array(a.x))  
    s_y = np.array(a.y)  
    beta_pair = sm.OLS(s_y, s_x).fit()  
    print('β1, β0 = ', beta_pair.params) 
temp = sns.FacetGrid(data=anscombe, col='dataset', col_wrap=4)  
temp.map(plt.scatter, 'x', 'y')  
plt.show()  

运行结果图：