I have fitted a GLM and failing to plot the model using ggplot-CodePudding

I have made a GLM and am trying to plot the model using ggplot using the following code. I think i need to add the type argument so that my model doesn't just produce a horizontal flat line. However i am struggling to fit it using the type argument as i run into this error message;

 Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = 
 object$xlevels) : 
 object is not a matrix

Here is my dataset and code used to get this if anyone knows the fix for this

my data (first 10 rows);

 aids
   cases quarter  date
 1      2       1 83.00
 2      6       2 83.25
 3     10       3 83.50
 4      8       4 83.75
 5     12       1 84.00
 6      9       2 84.25
 7     28       3 84.50
 8     28       4 84.75
 9     36       1 85.00
 10    32       2 85.25

my code used to create the model and plot

 model3 = glm(cases ~ date,
          data = aids,
          family = poisson(link='log'))


 #plotting the model (“link”, “response”, “terms”)
 plot_predictions <- function(model, type = 'response') {

   #make predictions
    preds <- predict(model, df)

   #plot
   df %>% 
    ggplot(aes(date, cases))  
    geom_point()  
    geom_line(aes(date, preds), col = 'red')  
    ggtitle("Model 2 - Poisson GLM predicting cases")  
    theme(plot.title = element_text(hjust = 0.5, size = 12, face = 'bold'))
 }

 plot_predictions(model3, aids)

dput output

dput(head(aids, 10))
structure(list(cases = c(2, 6, 10, 8, 12, 9, 28, 28, 36, 32), 
quarter = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
2L), .Label = c("1", "2", "3", "4"), class = "factor"), date = 
c(83, 
83.25, 83.5, 83.75, 84, 84.25, 84.5, 84.75, 85, 85.25)), 
row.names = c(NA, 
10L), class = "data.frame")

CodePudding user response：

Here's one that puts the predictions on the right scale:

library(tidyverse)
aids <- structure(list(cases = c(2, 6, 10, 8, 12, 9, 28, 28, 36, 32), 
               quarter = structure(c(1L, 2L, 3L, 4L, 1L, 2L, 3L, 4L, 1L, 
                                     2L), .Label = c("1", "2", "3", "4"), class = "factor"), date = 
                 c(83, 
                   83.25, 83.5, 83.75, 84, 84.25, 84.5, 84.75, 85, 85.25)), 
          row.names = c(NA, 
                        10L), class = "data.frame")
model3 = glm(cases ~ date,
             data = aids,
             family = poisson(link='log'))


plot_predictions <- function(model, df, type = 'response') {
  require(tidyverse)
  #make predictions
  preds <- predict(model, df, type= type)
  
  #plot
  df %>% 
    ggplot(aes(date, cases))  
    geom_point()  
    geom_line(aes(date, preds), col = 'red')  
    ggtitle("Model 2 - Poisson GLM predicting cases")  
    theme(plot.title = element_text(hjust = 0.5, size = 12, face = 'bold'))
}

plot_predictions(model3, aids)

^{Created on 2022-05-25 by the}

You can even get the standard error to show by omitting se = FALSE:

ggplot(aids, aes(date, cases))  
    geom_segment(aes(xend = date, yend = 0), color = "deepskyblue4")  
    geom_point(size = 3)  
    geom_smooth(col = 'red3', fill = "red3", method = glm, alpha = 0.1,
                method.args = list(family = poisson(link = 'log')))  
    ggtitle("Model 2 - Poisson GLM predicting cases")  
    theme_minimal(base_size = 16)  
    theme(plot.title = element_text(hjust = 0.5, face = 'bold'))

CodePudding user response：

You can also do this using broom::augment to add the predictions of your model to the data and plot that:

model3 %>%
    broom::augment() %>%
    ggplot(aes(date, cases))  
    geom_point()  
    geom_line(aes(date, exp(.fitted)), col = 'red')  
    ggtitle("Model 2 - Poisson GLM predicting cases")  
    theme(plot.title = element_text(hjust = 0.5, size = 12, face = 'bold'))