How to Fix in R: Contrasts can be applied only to factors with 2 or more levels.

In this article, we will discuss how we can fix “contrasts can be applied only to factors with 2 or more levels” error in the R programming language.

Contrasts can be applied only to factors with 2 or more levels:

It is a common error produced by the R compiler. The complete form of this error is given below:

Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : 

  contrasts can be applied only to factors with 2 or more levels

Such an error is produced by the R compiler when we try to fit a regression model with the help of the predictor variable that is either a character or factor and contains only one unique value.

When this error might occur:

R

# Create a data frame
dataframe <- data.frame(parameter1=c(8, 1, 13, 24, 9),
                 parameter2=as.factor(6),
                 parameter3=c(17, 9, 18, 13, 12),
                 parameter4=c(12, 21, 32, 4, 19))

# Print the dataframe
dataframe

Output:

As you can see in the above code, parameter2 is a factor has contains only a unique value. If we try to fit a multiple regression model in which parameter2 serves the purpose of a predictor variable then the compiler produces the following error:

Example:

Here, the  R compiler produces this kind of error because parameter2 contains only one unique value in it that is equal to 6. Since no variation exists at all in this predictor variable hence R compiler couldn’t fit a regression model effectively.

R

# Create a data frame
dataframe <- data.frame(parameter1=c(8, 1, 13, 24, 9),
                 parameter2=as.factor(6),
                 parameter3=c(17, 9, 18, 13, 12),
                 parameter4=c(12, 21, 32, 4, 19))


# Try to fit a multiple regression model
model <- lm(parameter4 ~ parameter1 + parameter2 +
            parameter3, data=dataframe)

Output:

How to fix this error:

Note that we can use the sapply() and lapply() functions together to count the number of unique values present in each of the predictor variables.

Example:

Here, Using lapply() function we can even print the values present in the individual predictor variables.

R

# Create a data frame
dataframe <- data.frame(parameter1=c(8, 1, 13, 24, 9),
                 parameter2=as.factor(6),
                 parameter3=c(17, 9, 18, 13, 12),
                 parameter4=c(12, 21, 32, 4, 19))

# Find the unique values for each variable
sapply(lapply(dataframe, unique), length)

Output:

Output

Example:

Now from the below code, we can see that parameter2 contains only one unique value. Hence, we can fix this error by simply removing parameter2 from the regression model.

R

# Create a data frame
dataframe <- data.frame(parameter1=c(8, 1, 13, 24, 9),
                 parameter2=as.factor(6),
                 parameter3=c(17, 9, 18, 13, 12),
                 parameter4=c(12, 21, 32, 4, 19))

# Find the unique values for each variable
lapply(dataframe[c('parameter1', 'parameter2',
                   'parameter3', 'parameter4')],
       unique)

Output:

Example:

Hence, by removing parameter2, the program compiled successfully without any error.

R

# Create a data frame
dataframe <- data.frame(parameter1=c(8, 1, 13, 24, 9),
                 parameter2=as.factor(6),
                 parameter3=c(17, 9, 18, 13, 12),
                 parameter4=c(12, 21, 32, 4, 19))

# Fit regression model using all the predictor variables
# except parameter2
model <- lm(parameter4 ~ parameter1 + parameter3, data=dataframe)

# Display model summary
summary(model)

Output: