數(shù)據(jù)分析中經(jīng)常會(huì)使用到數(shù)據(jù)擬合，本文中將闡述如何實(shí)現(xiàn)一元以及多元的線性擬合以及多項(xiàng)式擬合，本文中只涉及實(shí)現(xiàn)方式，不涉及理論知識(shí)。
模型擬合中涉及的誤差評(píng)估方法如下所示：
import numpy as np
def stdError_func(y_test, y):
return np.sqrt(np.mean((y_test - y) ** 2))
def R2_1_func(y_test, y):
return 1 - ((y_test - y) ** 2).sum() / ((y.mean() - y) ** 2).sum()
def R2_2_func(y_test, y):
y_mean = np.array(y)
y_mean[:] = y.mean()
return 1 - stdError_func(y_test, y) / stdError_func(y_mean, y)
一元線性擬合
import pandas as pd
import numpy as np
from sklearn.preprocessing import PolynomialFeatures
from sklearn import linear_model
filename = "E:/data.csv"
df= pd.read_csv(filename)
x = np.array(df.iloc[:,0].values)
y = np.array(df.iloc[:,5].values)
cft = linear_model.LinearRegression()
cft.fit(x[:,np.newaxis], y) #模型將x變成二維的形式, 輸入的x的維度為[None, 1]
print("model coefficients", cft.coef_)
print("model intercept", cft.intercept_)
predict_y = cft.predict(x[:,np.newaxis])
strError = stdError_func(predict_y, y)
R2_1 = R2_1_func(predict_y, y)
R2_2 = R2_2_func(predict_y, y)
score = cft.score(x[:,np.newaxis], y) ##sklearn中自帶的模型評(píng)估，與R2_1邏輯相同
print(' strError={:.2f}, R2_1={:.2f}, R2_2={:.2f}, clf.score={:.2f}'.format(
strError,R2_1,R2_2,score))
結(jié)果輸出為：
model coefficients [-31.2375]
model intercept 7.415750000000001
strError=1.11, R2_1=0.28, R2_2=0.15, clf.score=0.28
模型擬合的表達(dá)式為：
y = 7.415750000000001 +(-31.2375) * x
從擬合的均方誤差和得分來(lái)看效果不佳
多元線性擬合
import pandas as pd
import numpy as np
from sklearn.preprocessing import PolynomialFeatures
from sklearn import linear_model
filename = "E:/data.csv"
df= pd.read_csv(filename)
x = np.array(df.iloc[:,0:4].values)
y = np.array(df.iloc[:,5].values)
cft = linear_model.LinearRegression()
print(x.shape)
cft.fit(x, y) #
print("model coefficients", cft.coef_)
print("model intercept", cft.intercept_)
predict_y = cft.predict(x)
strError = stdError_func(predict_y, y)
R2_1 = R2_1_func(predict_y, y)
R2_2 = R2_2_func(predict_y, y)
score = cft.score(x, y) ##sklearn中自帶的模型評(píng)估，與R2_1邏輯相同
print('strError={:.2f}, R2_1={:.2f}, R2_2={:.2f}, clf.score={:.2f}'.format(
strError,R2_1,R2_2,score))
結(jié)果輸出為：
model coefficients [-31.2375 17.74375 44.325 5.7375 ]
model intercept 0.5051249999999978
strError=0.58, R2_1=0.80, R2_2=0.56, clf.score=0.80
模型擬合的表達(dá)式為：
y = 0.5051249999999978 +(-31.2375) * x11 + 17.74375 *x2 + 44.325 * x3 + 5.7375 * x4
從擬合的均方誤差和得分來(lái)看在之前的基礎(chǔ)上有所提升
一元多項(xiàng)式擬合
以三次多項(xiàng)式為例
import pandas as pd
import numpy as np
from sklearn.preprocessing import PolynomialFeatures
from sklearn import linear_model
filename = "E:/data.csv"
df= pd.read_csv(filename)
x = np.array(df.iloc[:,0].values)
y = np.array(df.iloc[:,5].values)
poly_reg =PolynomialFeatures(degree=3) #三次多項(xiàng)式
X_ploy =poly_reg.fit_transform(x[:, np.newaxis])
print(X_ploy.shape)
lin_reg_2=linear_model.LinearRegression()
lin_reg_2.fit(X_ploy,y)
predict_y = lin_reg_2.predict(X_ploy)
strError = stdError_func(predict_y, y)
R2_1 = R2_1_func(predict_y, y)
R2_2 = R2_2_func(predict_y, y)
score = lin_reg_2.score(X_ploy, y) ##sklearn中自帶的模型評(píng)估，與R2_1邏輯相同
print("model coefficients", lin_reg_2.coef_)
print("model intercept", lin_reg_2.intercept_)
print('degree={}: strError={:.2f}, R2_1={:.2f}, R2_2={:.2f}, clf.score={:.2f}'.format(
3, strError,R2_1,R2_2,score))
輸出結(jié)果
model coefficients [ 0. 990.64583333 -11906.25 44635.41666667]
model intercept -20.724999999999117
degree=3: strError=1.08, R2_1=0.32, R2_2=0.17, clf.score=0.32
對(duì)應(yīng)的函數(shù)表達(dá)式為 -20.724999999999117 + [0, 990.64583333, -11906.25, 44635.41666667] *[1， x, x^2, x.^3].T = -20.724999999999117 + 990.64583333 * x + ( -11906.25) * x^2 + 44635.41666667 * x^3
多元多項(xiàng)式擬合
import pandas as pd
import numpy as np
from sklearn.preprocessing import PolynomialFeatures
from sklearn import linear_model
filename = "E:/data.csv"
df= pd.read_csv(filename)
x = np.array(df.iloc[:,0:4].values)
y = np.array(df.iloc[:,5].values)
poly_reg =PolynomialFeatures(degree=2) #三次多項(xiàng)式
X_ploy =poly_reg.fit_transform(x)
lin_reg_2=linear_model.LinearRegression()
lin_reg_2.fit(X_ploy,y)
predict_y = lin_reg_2.predict(X_ploy)
strError = stdError_func(predict_y, y)
R2_1 = R2_1_func(predict_y, y)
R2_2 = R2_2_func(predict_y, y)
score = lin_reg_2.score(X_ploy, y) ##sklearn中自帶的模型評(píng)估，與R2_1邏輯相同
print("coefficients", lin_reg_2.coef_)
print("intercept", lin_reg_2.intercept_)
print('degree={}: strError={:.2f}, R2_1={:.2f}, R2_2={:.2f}, clf.score={:.2f}'.format(
3, strError,R2_1,R2_2,score))
函數(shù)輸出結(jié)果為：
coefficients [ 0. 332.28129937 -19.9240981 -9.10607925
-191.05593023 -287.93919929 -912.11402936 -1230.21922184
-207.90033986 99.03441748 190.26204994 433.25169929
273.13674555 257.66550523 344.92652936]
intercept 4.35175537840722
degree=3: strError=0.23, R2_1=0.97, R2_2=0.82, clf.score=0.97
代碼中輸入的自變量是一個(gè)包含四個(gè)變量的輸入， 對(duì)應(yīng)coefficients輸出的是長(zhǎng)度為15的向量， 其中對(duì)應(yīng)到的變量分別為 variable_X = [1, x1, x2, x3, x4, x1?x1, x1?x2, x1?x3, x1?x4, x2?x2, x2?x3, x2?x4, x3?x3, x3?x4, x4?x4]
對(duì)應(yīng)的方程式為： intercept+coefficients?variableX.T
intercept+coefficients?variableX?.T
代碼中涉及到的數(shù)據(jù)集如下：
a,b,c,d,e
0.06,0.2,0.02,0.1,0.340
0.1,0.28,0.02,0.14,0.370
0.12,0.32,0.02,0.16,0.377
0.08,0.24,0.02,0.12,0.383
0.08,0.32,0.04,0.1,0.383
0.12,0.28,0.03,0.1,0.393
0.1,0.24,0.05,0.1,0.385
0.06,0.32,0.05,0.14,0.362
0.12,0.2,0.05,0.12,0.320
0.06,0.28,0.04,0.12,0.393
0.08,0.28,0.05,0.16,0.402
0.08,0.2,0.03,0.14,0.349
0.1,0.2,0.04,0.16,0.335
0.1,0.32,0.03,0.12,0.387
0.12,0.24,0.04,0.14,0.390
0.06,0.24,0.03,0.16,0.315
指數(shù)函數(shù)和冪函數(shù)擬合參照網(wǎng)址：
https://blog.csdn.net/kl28978113/article/details/88818885
參考鏈接：
https://blog.csdn.
net/weixin_44794704/article/details/89246032
https://blog.csdn.net/bxg1065283526/article/details/80043049
https://www.cnblogs.com/Lin-Yi/p/8975638.html
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版權(quán)聲明：本文為CSDN博主「qq_33511693」的原創(chuàng)文章，遵循CC 4.0 BY-SA版權(quán)協(xié)議，轉(zhuǎn)載請(qǐng)附上原文出處鏈接及本聲明。
原文鏈接：https://blog.csdn.net/qq_33511693/java/article/details/105169807
