繪製R邏輯迴歸中的兩條曲線

Question

我正在R（glm）中運行邏輯迴歸。然後我設法繪製結果。我的代碼如下：

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") 

我的問題是：

1. 我怎麼能在一個圖形繪製2條Logistic迴歸曲線？
2. 我從其他統計軟件中得到了這兩個係數。我怎樣才能創建隨機數據，插入這兩套coef（Set 1和Set 2），然後產生兩個邏輯迴歸曲線？

該機型：

    SET 1 
(Intercept)  -88.4505 
Temperature  2.9677 

        SET 2 
(Intercept) -88.585533 
Temperature  2.972168 

mydata是在2列和〜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. 要繪製一條曲線，你只需要定義響應和預測之間的關係，並指定要爲其該曲線繪製的預測值的範圍。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curve電話，我們都在講邏輯函數定義x和y之間的關係。 plogis(z)的結果與評估1/(1+exp(-z))時的結果相同。參數min(dat$Temperature)和max(dat$Temperature)定義了x的範圍，其中y應該被評估。我們不需要告訴x指的是溫度;當我們指定應該爲預測值的範圍評估響應時，這是隱含的。

   

2. 正如可以看到的，curve功能可以繪製的曲線，而無需模擬預測器（例如，溫度）的數據。如果你仍然需要這樣做，例如繪製符合特定型號的伯努利試驗的一些模擬的結果，那麼可以嘗試以下方法：

   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) 
   } 
   

   實例：

   # 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]) 
   

   下圖顯示了由上面的第一實施例製造的積即對溫度的隨機向量的每個值進行單個伯努利試驗的結果，以及描述從中模擬數據的模型的曲線。