多項式回歸模型（Office Prices）

題目：

分析：還是上次的房價**題目，指明要用多項式回歸擬合。在多元多項式擬合時候，目標函式表示如下

對其目標函式求偏導得到

很容易寫出**。

**：

#coding:utf-8

import math
class data:
def __init__(self):
self.x = 
self.y = 0.0
def makematrix(row, col, fill = 0.0):
mat = 
for i in range(row):
return mat
def wx(d, w, b):
res = 0.0
for k in range(len(d.x)):
for j in range(b + 1):
res += w[k][j] * math.pow(d.x[k], j)
return res
def gradient(d, w, f, b, alpha):
for k in range(f):
for j in range(b + 1):
t1, t2 = 0.0, 0.0
for i in range(len(d)):
t1 += (wx(d[i], w, b) - d[i].y) * math.pow(d[i].x[k], j)
w[k][j] -= alpha * t1
def getvalues(d, w, b):
res = 0.0
for i in range(len(d)):
tmp = wx(d[i], w, b)
res += 0.5 * (d[i].y - tmp) * (d[i].y - tmp)
return res
def iterator(d, w, f, b):
alpha = 0.003
delta = 0.5
oldval = getvalues(d, w, b)
gradient(d, w, f, b, alpha)
newval = getvalues(d, w, b)
while abs(oldval - newval) > delta:
oldval = newval
gradient(d, w, f, b, alpha)
newval = getvalues(d, w, b)
def main():
while true:
try:
f, n = map(int, raw_input().split())
d = 
b = 5
w = makematrix(f, b + 1)
for i in range(0, n):
t = data()
t.x = map(float, raw_input().split())
t.y = t.x.pop()
iterator(d, w, f, b)
n = int(raw_input())
for i in range(0, n):
t = data()
t.x = map(float, raw_input().split())
print '%.2f'% wx(t, w, b)
except eoferror:
break
if __name__ == '__main__':
main()

**：

#include #include #include #include #include #include #define vector vector

using namespace std;
struct data
;double wx(const data& d, const vector& w)
void gradient(const vector& d, vector&w, double alpha)
}double getvalues(const vector& d, vectorw)
return res;
}void iterator(const vector& d, vector&w)
}vectorgetfeatures(vectorx)
int main()
t.x = getfeatures(t.x);
features = t.x.size();
scanf("%lf", &_y);
t.y = _y;
d.push_back(t);
}		for(int i = 0; i < features; i++)
w.push_back(0);
iterator(d, w);
d.clear();
scanf("%d", &n);
for(int i = 0; i < n; i++)
t.x = getfeatures(t.x);
printf("%.2lf\n", wx(t, w));
}	}return 0;
}

題解：

sklearn官網：

sklearn源**：

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