python - 在绘制多元线性回归模型的成本与时期时获得空图
问题描述
我正在学习机器学习并尝试在汽车价格数据集上实施多元线性回归来预测未来汽车的价格。
这是我的代码
In [2]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
In [3]:
temp = pd.read_csv('car_price.csv')
In [4]:
temp.columns
Out[4]:
Index(['name', 'year', 'selling_price', 'km_driven', 'fuel', 'seller_type',
'transmission', 'owner', 'mileage', 'engine', 'max_power', 'torque',
'seats'],
dtype='object')
In [5]:
data = temp[['year', 'selling_price', 'km_driven', 'fuel', 'seller_type',
'transmission', 'owner', 'mileage', 'engine', 'max_power', 'torque',
'seats'
]]
In [6]:
data['Current_Year'] = 2020
In [7]:
data.head()
Out[7]:
year selling_price km_driven fuel seller_type transmission owner mileage engine max_power torque seats Current_Year
0 2014 450000 145500 Diesel Individual Manual First Owner 23.4 kmpl 1248 CC 74 bhp 190Nm@ 2000rpm 5.0 2020
1 2014 370000 120000 Diesel Individual Manual Second Owner 21.14 kmpl 1498 CC 103.52 bhp 250Nm@ 1500-2500rpm 5.0 2020
2 2006 158000 140000 Petrol Individual Manual Third Owner 17.7 kmpl 1497 CC 78 bhp 12.7@ 2,700(kgm@ rpm) 5.0 2020
3 2010 225000 127000 Diesel Individual Manual First Owner 23.0 kmpl 1396 CC 90 bhp 22.4 kgm at 1750-2750rpm 5.0 2020
4 2007 130000 120000 Petrol Individual Manual First Owner 16.1 kmpl 1298 CC 88.2 bhp 11.5@ 4,500(kgm@ rpm) 5.0 2020
In [8]:
data['# Years'] = data['Current_Year'] - data['year']
In [9]:
to_drop = ['Current_Year','year','torque','max_power','seller_type','owner']
data.drop(to_drop, inplace = True, axis = 1)
In [10]:
data.head()
Out[10]:
selling_price km_driven fuel transmission mileage engine seats # Years
0 450000 145500 Diesel Manual 23.4 kmpl 1248 CC 5.0 6
1 370000 120000 Diesel Manual 21.14 kmpl 1498 CC 5.0 6
2 158000 140000 Petrol Manual 17.7 kmpl 1497 CC 5.0 14
3 225000 127000 Diesel Manual 23.0 kmpl 1396 CC 5.0 10
4 130000 120000 Petrol Manual 16.1 kmpl 1298 CC 5.0 13
In [11]:
data['engine']= data['engine'].str.replace('[^\d.]', '',regex = True).astype(float)
In [12]:
data['mileage'] = data['mileage'].str.replace('[^\d.]', '',regex = True).astype(float)
In [13]:
data.head()
Out[13]:
selling_price km_driven fuel transmission mileage engine seats # Years
0 450000 145500 Diesel Manual 23.40 1248.0 5.0 6
1 370000 120000 Diesel Manual 21.14 1498.0 5.0 6
2 158000 140000 Petrol Manual 17.70 1497.0 5.0 14
3 225000 127000 Diesel Manual 23.00 1396.0 5.0 10
4 130000 120000 Petrol Manual 16.10 1298.0 5.0 13
In [14]:
data.replace(to_replace = ['Diesel','Petrol','LPG','CNG'],value=[1,2,3,4],inplace = True)
In [15]:
data.head()
Out[15]:
selling_price km_driven fuel transmission mileage engine seats # Years
0 450000 145500 1 Manual 23.40 1248.0 5.0 6
1 370000 120000 1 Manual 21.14 1498.0 5.0 6
2 158000 140000 2 Manual 17.70 1497.0 5.0 14
3 225000 127000 1 Manual 23.00 1396.0 5.0 10
4 130000 120000 2 Manual 16.10 1298.0 5.0 13
In [16]:
data.replace(to_replace = ['Manual','Automatic'],value=[1.0,2.0],inplace = True)
In [17]:
data.head()
Out[17]:
selling_price km_driven fuel transmission mileage engine seats # Years
0 450000 145500 1 1.0 23.40 1248.0 5.0 6
1 370000 120000 1 1.0 21.14 1498.0 5.0 6
2 158000 140000 2 1.0 17.70 1497.0 5.0 14
3 225000 127000 1 1.0 23.00 1396.0 5.0 10
4 130000 120000 2 1.0 16.10 1298.0 5.0 13
In [18]:
data.head()
Out[18]:
selling_price km_driven fuel transmission mileage engine seats # Years
0 450000 145500 1 1.0 23.40 1248.0 5.0 6
1 370000 120000 1 1.0 21.14 1498.0 5.0 6
2 158000 140000 2 1.0 17.70 1497.0 5.0 14
3 225000 127000 1 1.0 23.00 1396.0 5.0 10
4 130000 120000 2 1.0 16.10 1298.0 5.0 13
In [ ]:
In [21]:
data = (data - data.mean())/data.std()
X = data.iloc[:,1:8]
ones = np.ones([X.shape[0],1])
X = np.concatenate((ones,X),axis=1)
y = data.iloc[:,0:1].values
theta = np.zeros([1,8])
print(X)
def computeCost(X,y,theta):
tobesummed = np.power(((X @ theta.T)-y),2)
return np.sum(tobesummed)/(2 * len(X))
def gradientDescent(X,y,theta,iters,alpha):
cost = np.zeros(iters)
for i in range(iters):
theta = theta - (alpha/len(X)) * np.sum(X * (X @ theta.T - y), axis=0)
cost[i] = computeCost(X, y, theta)
return theta,cost
alpha = 0.01
iters = 1000
g,cost = gradientDescent(X,y,theta,iters,alpha)
print(g)
finalCost = computeCost(X,y,g)
print(finalCost)
fig, ax = plt.subplots()
ax.plot(np.arange(iters), cost, 'r')
ax.set_xlabel('Iterations')
ax.set_ylabel('Cost')
ax.set_title('Error vs. Training Epoch')
[[ 1. 1.33828022 -0.86972865 ... -0.41797619 -0.43426926
-0.04846121]
[ 1. 0.88735626 -0.86972865 ... 0.07813794 -0.43426926
-0.04846121]
[ 1. 1.24102211 0.95315801 ... 0.07615349 -0.43426926
1.92965648]
...
[ 1. 0.88735626 -0.86972865 ... -0.41797619 -0.43426926
1.18786235]
[ 1. -0.79255652 -0.86972865 ... -0.12427662 -0.43426926
0.1988035 ]
[ 1. -0.79255652 -0.86972865 ... -0.12427662 -0.43426926
0.1988035 ]]
[[nan nan nan nan nan nan nan nan]]
nan
Out[21]:
Text(0.5, 1.0, 'Error vs. Training Epoch')
In [ ]:
当绘制成本与纪元时，我得到的是空图，而在打印成本值时，我得到的数据丢失了“nan”
我似乎无法理解我要去哪里错了。
解决方案
您的数据总共有 663 个空值，因此出现错误，
data.isnull().values.sum()
663
进行了平均插补并替换了所有 NaN。
data = data.fillna(data.mean())
data.isnull().values.sum()
0
然后执行剩下的代码，
alpha = 0.01
iters = 1000
g,cost = gradientDescent(X,y,theta,iters,alpha)
print(g)
[[ 3.37617073e-16 -1.03407822e-01 -7.06228959e-02 3.63774873e-01
2.51480688e-03 4.51868433e-01 -2.02133329e-01 -2.67955488e-01]]
finalCost = computeCost(X,y,g)
print(finalCost)
0.21914950571622366
阴谋：
fig, ax = plt.subplots()
ax.plot(np.arange(iters), cost, 'r')
ax.set_xlabel('Iterations')
ax.set_ylabel('Cost')
ax.set_title('Error vs. Training Epoch')
注意：：我认为你不应该像下面这样取整个数据的平均值，
data = (data - data.mean())/data.std()
目标变量selling_price应从缩放中排除。