在 R 中绘制逻辑回归中的两条曲线

Question

我在 R (glm) 中运行逻辑回归。然后我设法绘制结果。我的代码如下：

 temperature.glm = glm(Response~Temperature, data=mydata,family=binomial)

 plot(mydata$Temperature,mydata$Response, ,xlab="Temperature",ylab="Probability of Response")
 curve(predict(temperature.glm,data.frame(Temperature=x),type="resp"),add=TRUE, col="red")
 points(mydata$Temperature,fitted(temperature.glm),pch=20)
 title(main="Response-Temperature with Fitted GLM Logistic Regression Line") 

我的问题是：

1. 如何在一张图中绘制两条逻辑回归曲线？
2. 我从其他统计软件中获得了这两个系数。如何创建随机数据，插入这两组 coef（Set 1 和 Set 2），然后生成两条逻辑回归曲线？

模型：

                   SET 1
 (Intercept)     -88.4505
 Temperature       2.9677

                  SET 2
 (Intercept)    -88.585533
 Temperature      2.972168

mydata 有 2 列和约 700 行。

Response Temperature 
1 29.33 
1 30.37 
1 29.52 
1 29.66 
1 29.57 
1 30.04 
1 30.58 
1 30.41 
1 29.61 
1 30.51 
1 30.91 
1 30.74 
1 29.91 
1 29.99 
1 29.99 
1 29.99 
1 29.99 
1 29.99 
1 29.99 
1 30.71 
0 29.56 
0 29.56 
0 29.56 
0 29.56 
0 29.56 
0 29.57 
0 29.51

Answer 1

1. 要绘制曲线，您只需定义响应和预测变量之间的关系，并指定您希望绘制曲线的预测变量值的范围。例如：

   dat <- structure(list(Response = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
     1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 
     0L, 0L), Temperature = c(29.33, 30.37, 29.52, 29.66, 29.57, 30.04, 
     30.58, 30.41, 29.61, 30.51, 30.91, 30.74, 29.91, 29.99, 29.99, 
     29.99, 29.99, 29.99, 29.99, 30.71, 29.56, 29.56, 29.56, 29.56, 
     29.56, 29.57, 29.51)), .Names = c("Response", "Temperature"), 
     class = "data.frame", row.names = c(NA, -27L))
   
   temperature.glm <- glm(Response ~ Temperature, data=dat, family=binomial)
   
   plot(dat$Temperature, dat$Response, xlab="Temperature", 
        ylab="Probability of Response")
   curve(predict(temperature.glm, data.frame(Temperature=x), type="resp"), 
         add=TRUE, col="red")
   # To add an additional curve, e.g. that which corresponds to 'Set 1':
   curve(plogis(-88.4505 + 2.9677*x), min(dat$Temperature), 
         max(dat$Temperature), add=TRUE, lwd=2, lty=3)
   legend('bottomright', c('temp.glm', 'Set 1'), lty=c(1, 3), 
          col=2:1, lwd=1:2, bty='n', cex=0.8)
   

   在上面的第二个curve 调用中，我们说逻辑函数定义了x 和y 之间的关系。 plogis(z) 的结果等同于评估1/(1+exp(-z)) 时获得的结果。 min(dat$Temperature) 和max(dat$Temperature) 参数定义了x 的范围，y 应该被评估。我们不需要告诉函数x 指的是温度；当我们指定应针对该范围的预测值评估响应时，这是隐含的。

2. 如您所见，curve 函数允许您绘制曲线，而无需模拟预测变量（例如温度）数据。如果您仍然需要这样做，例如要绘制符合特定模型的伯努利试验的一些模拟结果，您可以尝试以下操作：

   n <- 100 # size of random sample
   
   # generate random temperature data (n draws, uniform b/w 27 and 33)
   temp <- runif(n, 27, 33)
   
   # Define a function to perform a Bernoulli trial for each value of temp, 
   #   with probability of success for each trial determined by the logistic
   #   model with intercept = alpha and coef for temperature = beta.
   # The function also plots the outcomes of these Bernoulli trials against the 
   #   random temp data, and overlays the curve that corresponds to the model
   #   used to simulate the response data.
   sim.response <- function(alpha, beta) {
     y <- sapply(temp, function(x) rbinom(1, 1, plogis(alpha + beta*x)))  
     plot(y ~ temp, pch=20, xlab='Temperature', ylab='Response')
     curve(plogis(alpha + beta*x), min(temp), max(temp), add=TRUE, lwd=2)    
     return(y)
   }
   

   例子：

   # Simulate response data for your model 'Set 1'
   y <- sim.response(-88.4505, 2.9677)
   
   # Simulate response data for your model 'Set 2'
   y <- sim.response(-88.585533, 2.972168)
   
   # Simulate response data for your model temperature.glm
   # Here, coef(temperature.glm)[1] and coef(temperature.glm)[2] refer to
   #   the intercept and slope, respectively
   y <- sim.response(coef(temperature.glm)[1], coef(temperature.glm)[2])
   

   下图显示了上面第一个示例生成的图，即温度随机向量的每个值的单个伯努利试验的结果，以及描述模拟数据的模型的曲线。