Understanding Factor Variables in R: Resolving the Error with Median Calculation

Understanding the Problem and Solution

The problem presented involves creating a prediction dataframe for a model that has two factor variables (VegeType) and one continuous variable (DistAgriLand). The goal is to plot model predictions for the first factor, Month. However, an error occurs when trying to create the prediction dataframe with VegeType as a factor.

Error Explanation

The error occurs because R’s factor function in R can only be used to create a factor with levels that already exist in the data. When creating the prediction dataframe, the code tries to add level 6 to the factor, but this level does not exist in the original data.

Solution Approach

To resolve the issue, we need to ensure that when creating VegeType, it includes all the necessary levels from the start. This can be achieved by using the levels argument when creating the factor.

Example Code and Explanation

# Load required libraries
library(dplyr)

# Create a sample dataframe with factor variables (VegeType) and continuous variable (DistAgriLand)
a = structure(list(Month = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 
11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 
12L, 12L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L, 10L, 10L, 12L, 12L, 
3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 
6L, 6L, 10L, 10L, 3L, 6L, 6L, 10L, 10L, 3L, 4L, 6L, 6L, 8L, 8L, 
10L, 10L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L), .Label = c("1", 
"2", "3", "4", "5", "7", "8", "9", "10", "11", "12"), class = "factor"), 
    VegeType = structure(c(6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
    6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
    6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
    6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
    6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 
    6L), .Label = c("1", "2", "3", "4", "5", "7", "8", "9", "10", 
    "11", "12"), class = "factor"), DistAgriLand = c(580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37), row.names = c(NA, 
100L), class = "data.frame")

Alternative Approach

The alternative approach is to create the factor with level 6 directly.

# Load required libraries
library(dplyr)

# Create a sample dataframe with factor variables (VegeType) and continuous variable (DistAgriLand)
a = structure(list(Month = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 
11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 
12L, 12L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L, 10L, 10L, 12L, 12L, 
3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 
6L, 6L, 10L, 10L, 3L, 6L, 6L, 10L, 10L, 3L, 4L, 6L, 6L, 8L, 8L, 
10L, 10L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L), .Label = c("1", 
"2", "3", "4", "5", "7", "8", "9", "10", "11", "12"), class = "factor"), 
    VegeType = levels(a$VegeType), DistAgriLand = c(580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37), row.names = c(NA, 
100L), class = "data.frame")

Final Approach

The final approach is to create the prediction dataframe using expand.grid and then setting the names of the columns.

# Load required libraries
library(dplyr)

# Create a sample dataframe with factor variables (VegeType) and continuous variable (DistAgriLand)
a = structure(list(Month = structure(c(9L, 9L, 9L, 9L, 9L, 9L, 9L, 
9L, 9L, 9L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 10L, 
10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 
11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 
12L, 12L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L, 10L, 10L, 12L, 12L, 
3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 3L, 3L, 6L, 6L, 10L, 10L, 3L, 
6L, 6L, 10L, 10L, 3L, 6L, 6L, 10L, 10L, 3L, 4L, 6L, 6L, 8L, 8L, 
10L, 10L, 12L, 12L, 3L, 4L, 6L, 6L, 8L, 8L), .Label = c("1", 
"2", "3", "4", "5", "7", "8", "9", "10", "11", "12"), class = "factor"), 
    VegeType = levels(a$VegeType), DistAgriLand = c(580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 580.5, 
    580.5, 580.5, 580.5, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 594.37, 
    594.37, 594.37, 594.37, 594.37, 594.37), row.names = c(NA, 
100L), class = "data.frame")

# Create the prediction dataframe
Preds.Month = setNames(
  cbind(
    expand.grid(1:12, 1:12),
    median(a$DistAgriLand,na.rm=T)
  ), c("Month", "VegeType", "DistAgriLand")
)

This final approach ensures that the prediction dataframe includes all necessary levels for both Month and VegeType.

Last modified on 2023-12-05