Error: 'Data' Must Be a Data Frame, or Other Object Coercible by 'Fortify()', Not an S3 Object with Class Date

Error: `data` must be a data frame, or other object coercible by `fortify()`, not an S3 object with class Date

Try this:
Workflow is as follows:

 library(tidyverse)
    library(lubridate)
    #`df` is `CPOD_Time` saved as `df<-as.data.frame(CPOD_Time)`
    df<-as.data.frame(CPOD_Time)

Then the plot:

df %>% 
      rename(Time=time,Number=number,Date=date) %>% 
      mutate(Date=str_replace_all(Date,"\\D","-"),Date=as.character(Date),
             Date=dmy(Date)) %>% 
    ggplot(aes(x=Date, y=Number, fill=Time)) + 
      geom_bar(stat="identity") +theme_bw() + 
geom_text(aes(label=Number), vjust=1.6,size=3.5) 

`data` must be a data frame, or other object coercible by `fortify()`, not an S3 object with class ranger

As per the ggplot2 documentation, you need to provide a data.frame() or object that can be converted (coerced) to a data.frame(). In this case, if you want to reproduce the plot above in ggplot2, you will need to manually set up the data frame yourself.

Below is an example of how you could set up the data to display the plot in ggplot2.

Data Frame

First we create a data.frame() with the variables that we want to plot. The easiest way to do this is to just group them all in as separate columns. Note that I have used the as.numeric() function to first coerce the predicted values to a vector, because they were previously a data.table row, and if you don't convert them they are maintained as rows.

ggplot_data <- data.frame(unique.death.times = r_fit$unique.death.times,
                      pred1 = as.numeric(pred[1,]),
                      pred2 = as.numeric(pred[2,]),
                      pred3 = as.numeric(pred[3,]))
head(ggplot_data)
## unique.death.times     pred1     pred2     pred3
## 1                  5 0.9986676 1.0000000 0.9973369
## 2                 11 0.9984678 1.0000000 0.9824642
## 3                 12 0.9984678 0.9998182 0.9764154
## 4                 13 0.9984678 0.9998182 0.9627118
## 5                 15 0.9731656 0.9959416 0.9527424
## 6                 26 0.9731656 0.9959416 0.9093876

Pivot the data

This format is still not ideal, because in order to plot the data and colour by the correct column (variable), we need to 'pivot' the data. We need to load the tidyr package for this.

library(tidyr)
ggplot_data <- ggplot_data %>% 
  pivot_longer(cols = !unique.death.times, 
  names_to = "category", values_to = "predicted.value")

Plotting

Now the data is in a form that makes it really easy to plot in ggplot2.

plot <- ggplot(ggplot_data, aes(x = unique.death.times, y = predicted.value, colour = category)) +
      geom_line()
plot

If you really want to match the look of the base plot, you can add theme_classic():

plot + theme_classic()

Additional notes

Note that this doesn't include 95% confidence intervals, so they would have to be calculated separately. Be aware though, that a 95% confidence interval is not just 95% of the y value at a given x value. There are calculations that will give you the correct values of the confidence interval, including functions built into R.

For a quick view of a trend line with prediction intervals, you can use the geom_smooth() function in ggplot2, but in this case it adds a loess curve by default, and the intervals provided by that function.

plot + theme_classic() + geom_smooth()

Error: `data` must be a data frame, or other object coercible by `fortify()`,...Did you accidentally pass `aes()` to the `data` argument?

We need to specify the column name as unquoted

library(dplyr)
library(ggplot2)
Social_Split %>%
   rownames_to_column('rn') %>%
   ggplot(aes(x = rn, y = V1)) + 
      geom_col()

-output

---

Or use barplot from base R

barplot(t(Social_Split))

data

Social_Split <- structure(list(V1 = c(220L, 213L, 73L, 3L, 44L, 116L, 10L, 104L, 
88L, 129L)), class = "data.frame", row.names = c("Facebook", 
"Instagram", "Linkedin", "None", "Quora", "Reddit", "Signal", 
"Snapchat", "TikTok", "Twitter"))

ERROR:`data` must be a data frame, or other object coercible by `fortify()`, not an S3 object with class reactiveExpr/reactive

d1 is not a dataframe but an expression that needs to be evaluated (reactive does not return a dataframe). You need to active the reactive element before using it in ggplot2

server <- function(session, input, output) {

    d1 <- reactive({
        data %>% group_by(`Detected State`) %>% 
                 count(`Detected District`) %>%
                 filter(`Detected State` == input$state)
        })

    output$stateplot <- renderPlot({
        ggplot(d1(), aes(x = `Detected District`, y = n)) +
             geom_bar(stat = "identity")
    })

}

ggplot2 & spline() in R - Error in : `data` must be a data frame, or other object coercible by `fortify()`

Possible way to do that:

ggplot(data, aes(x = x, y = y)) + 
  geom_point() + 
  geom_line(data = data.frame(spline(x, y))) #+
  #ggthemes::theme_base()

The problem is: spline returns list, you just should have convert it into data.frame and that's it.

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