python实现简单线性回归

Posted 2021-02-11 mambakb

之前推导了一元线性回归和多元线性回归，今天就用python来实现一下一元线性回归

先看下之前推导的结果

  ， 

第一种是用循环迭代的计算方法。这里的x，y是numpy中的array类型

def sum(x):
    sum1 = 0
    for i in x:
        sum1 += i
    return sum1 
def sub(x,y):
    ret = []
    for i in range(len(x)):
        ret.append(x[i] - y)
    return np.array(ret)
def mean(num):
    sum = 0
    for i in num:
        sum += i
    return sum / len(num)
def multiply(x,y):
    ret = []
    for i in range(len(x)):
        ret.append(x[i]*y[i])
    return np.array(ret)
def square(x):
    ret = []
    for i in range(len(x)):
        ret.append (x[i] * x[i])
    return np.array(ret)def linearRegression(x,y):
    length = len(x)
    t1 = time()
    x_mean1 = mean(x)
    a = sum(multiply(y,sub(x,x_mean1))) / (sum(square(x)) - sum(x) ** 2 / length)
    sum1 = 0
    for i in range(length):
        sum1 += (y[i] - a * x[i])
    b = sum1 / length

 

第二种人是用vectorization的方法

def linearRegression_(x,y):
    length = len(x)
    x_mean = x.mean()
    a = (y * (x - x_mean)).sum() / ((x ** 2).sum() - x.sum() ** 2 / length)
    b = (y - a * x).sum() / length

为了比较二者的性能，这里我们随机生成10000条数据，分别统计两种方法运行的时间

x=np.random.randint(0,100,10000)
y=np.random.randint(0,100,10000)

t1 = time() 
linearRegression(x,y)
t2 = time()
print(t2 - t1)

t1 = time()
linearRegression_(x,y)
t2 = time()

 

得到二者的结果

0.1349632740020752
0.0009996891021728516

 

上面的是循环计算所需的时间，下面的是vectorization所需得时间。很明显vectorization要优于循环计算