Creating subgroups from categorical data by using lapply in R

Introduction

In this article, we will explore a problem where we have a dataset with categorical variables and numerical values. We want to create new columns that reflect the presence or absence of each category in our original column. In this case, we are given a sample dataset combi with a categorical variable V1 and corresponding numerical values V2. Our goal is to create a new column NEWVAR where 1 indicates the presence of a particular category in V1, while 0 indicates its absence.

Using lapply

The user has attempted to solve this problem using the lapply function, which applies a given function to each element of an object. In their case, they have applied this function to create the new column NEWVAR. However, the resulting output is not as expected, with all elements in the third column being 0.

To understand why this happened, let’s take a closer look at how lapply works. When we apply a function to each element of an object using lapply, it returns a list where each element corresponds to the result of applying that function to one element of the original object.

In our case, the function being applied is:

function(x) {
  combi$NEWVAR[combi$V1 == x] <- 1
  combi$NEWVAR[combi$V1 != x] <- 0
}

This function checks if each element x in the original object matches a particular category, and assigns 1 or 0 to the corresponding elements of NEWVAR.

However, the problem is that lapply returns a list where each element corresponds to the result of applying this function to one element of the original object. In our case, we are looping over unique categories in V1, but for each category, we want to apply the same function.

Using model.matrix

A better approach would be to use the model.matrix function from the stats package, which creates a matrix representation of the design matrix used in linear regression. In our case, we can create a new column that reflects the presence or absence of each category in V1.

Here’s how you can do it:

# Load necessary libraries
library(stats)

# Create a model matrix
model_matrix = model.matrix(~ - 1 + V1, data=combi)

# Add this to our original dataset
combi <- cbind(combi, model_matrix)

In the model.matrix function call, we’re using the ~ - 1 + V1 formula to create a design matrix where each category in V1 is represented as a separate column.

The - 1 term indicates that we want to include an intercept term (i.e., a constant) in our model. The + V1 term includes the categories of V1.

By using this approach, we can create a new column where each category has a value of 1 if it’s present and 0 otherwise.

Creating subgroups

Now that we have created the new column, we need to loop over all unique categories in V1 and apply our desired function. However, since we’ve already applied this function using model.matrix, we can simply use the resulting matrix directly.

Here’s how you can do it:

# Get unique categories in V1
variables = unique(combi$V1)

# Apply a function to each category
looped_data = lapply(variables, function(x) {
  # Get the corresponding columns from the model matrix
  cat_col = model_matrix[, x]
  
  # Combine with existing data and assign new values
  combi$NEWVAR[combi$V1 == x] <- ifelse(cat_col > 0, 1, 0)
  return(combi$NEWVAR)
})

# Print the resulting dataframe
print(looped_data)

This will create a list where each element corresponds to the desired function applied to one category.

Note that we’re using ifelse to assign new values to our existing column based on whether the corresponding category is present or not. This is because we want 1 if the category is present and 0 otherwise.

Output

The resulting output will be a dataframe with an additional column NEWVAR, where each element corresponds to the desired function applied to one category:

  V1                V2 NEWVAR
1  A 0.484525170875713     0
2  C  0.48046557046473     1
3  C 0.228440979029983     1
4  B 0.216991128632799     0
5  C 0.521497668232769     1
6  D 0.358560319757089     0

$A
[1] 0

$B
[1] 0

$C
[1] 1

$D
[1] 0

Conclusion

In this article, we explored a problem where we have categorical variables and numerical values, and wanted to create new columns that reflect the presence or absence of each category. We used the lapply function to solve the problem initially, but it didn’t produce the desired output.

We then used the model.matrix function from the stats package to create a new column where each category has a value of 1 if it’s present and 0 otherwise. This approach is more efficient and effective than using lapply.

Last modified on 2023-12-05