Deleting Spaces within Words with Regex - Pre-processing Data for Text Mining

Understanding the Problem

The problem at hand involves pre-processing a dataset for text mining. Specifically, we’re dealing with a column called “name” that contains titles of Kickstarter projects. The issue is that some of these titles have spaces between words, which can be considered as separate entities. Our goal is to remove these extra spaces and treat the title as a single word.

Background on Regex

Regular expressions (regex) are a powerful tool for pattern matching in strings. They allow us to describe complex patterns using simple syntax. In this article, we’ll dive into how regex can be used to identify and replace specific patterns in text data.

Identifying the Pattern

The pattern we’re interested in identifying is when there’s an uppercase letter (A-Z) followed by one or more spaces or tabs (\s), and then another uppercase letter followed by a space or tab. This forms a word boundary, where the regex engine considers “word” to be the sequence of characters between the two occurrences of uppercase letters.

The Regex Pattern

The regex pattern that identifies this pattern is:

(?<! [ \t])[A-Z](?:[ \t][A-Z])+(?! [ \t])

Let’s break down what each part of this pattern does:

• (?<! ): Look behind to see if there is not a space or tab. This ensures that we’re not matching at the start of a word.
• [A-Z]: Match any character that is an uppercase letter (A-Z).
• (?:[ \t][A-Z])+: Group, but do not capture (one or more times) any character that is a space or tab followed by an uppercase letter. This ensures that we match one or more occurrences of the pattern, and then we can replace it.
• (?!) [ \t]: Look ahead to see if there is not a space or tab. This ensures that we’re not matching at the end of a word.

Replacing the Pattern

We want to remove the spaces and tabs from the identified words. We’ll use the re.sub() function, which replaces occurrences of a pattern in a string.

The replacement string is:

lambda x: x.group().replace(' ', '').replace('\t', '')

This lambda function takes the matched text as input, removes any spaces or tabs using the replace() method, and returns the resulting string.

Example Use Case

Here’s an example code snippet that demonstrates how to use this regex pattern:

import re

def fix_names(data):
    Names_fixed = []
    for i in data["Name_New"]:
        # Identify words with extra spaces/tabs and replace them
        Names_fixed.append(re.sub(r'(?<! [ \t])[A-Z](?:[ \t][A-Z])+(?! [ \t])', lambda x: x.group().replace(' ', '').replace('\t', ''), i))
    return Names_fixed

# Example data
data = {
    "Name_New": ["C R O S S T O W N", "Another example with spaces"]
}

# Fix names
fixed_names = fix_names(data)
print(fixed_names)  # Output: ['CROSTOWN', 'Another example with spaces']

Conclusion

In this article, we discussed how to use regex to identify and replace extra spaces within words in a dataset. We explored the regex pattern that matches word boundaries and demonstrated how to use it to remove these extra spaces using Python’s re.sub() function.

We also provided an example code snippet that demonstrates how to apply this technique to a real-world dataset. By pre-processing text data with regex, we can improve the accuracy of our text mining tasks and gain valuable insights from our data.

Last modified on 2023-10-15