我正在尝试编写一个使用梯度下降法的简单线性回归的基本代码。这里是我的代码。

linear = function(x,y,lr){theta0 = 0theta1 = 0m=length(x)hypo = theta0 +theta1*xcostt = cost(hypo , y)prev_cost = 1000000while (prev_cost > cost){prev_cost = costtheta0 = theta0 - (lr/m)*(hypo - y)theta1 = theta1 - (lr/m)*(hypo - y)*xhypo = theta0 + theta1*xNew_cost = cost(hypo , y)if(New_cost < cost){cost = New_cost}}theta = c(theta0 , theta1)return( theta )}  cost = function(hypo , y){interm = (hypo - y)^2interm1 = sum(interm)interm2 = interm1/(2 * m)return(interm2)  }

但是当我用数据测试它时，会生成一个警告消息。

There were 50 or more warnings (use warnings() to see the first 50)

然后停止运行。代码哪里出了问题？当我使用warnings()时，我得到以下信息：

Warning messages:1: In while (prev_cost > cost) { ... :the condition has length > 1 and only the first element will be used 

lr = 0.01，这是学习率。这是x和y数据的截图

回答：

对我来说这个运行得很好。它产生了一个长度是length(x)两倍的向量 – 这是你想要的结果吗？

linear = function(x,y,lr){    theta0 = 0    theta1 = 0    m=length(x)    hypo = theta0 +theta1*x    costt = cost(hypo , y, m)    prev_cost = 1000000    while (prev_cost > costt)    {        prev_cost = costt        theta0 = theta0 - (lr/m)*(hypo - y)        theta1 = theta1 - (lr/m)*(hypo - y)*x        hypo = theta0 + theta1*x        New_cost = cost(hypo , y, m)        if(New_cost < costt)        {            costt = New_cost        }    }    theta = c(theta0 , theta1)    return( theta )}  cost = function(hypo , y, m){    interm = (hypo - y)^2    interm1 = sum(interm)    interm2 = interm1/(2 * m)    return(interm2)  }x <- rnorm(80)y <- rnorm(80)lr <- 0.01linear(x, y, lr)