Table of Contents
Splitting Text into a Word List Without Spaces
Problem
Algorithm
Frequency-Based Algorithm
Code Implementation
Results
Home Backend Development Python Tutorial How can we efficiently split a text string of concatenated words without spaces into individual words?

How can we efficiently split a text string of concatenated words without spaces into individual words?

Nov 04, 2024 am 10:48 AM

How can we efficiently split a text string of concatenated words without spaces into individual words?

Splitting Text into a Word List Without Spaces

Problem

Given a text string consisting of concatenated words without spaces:

Input: "tableapplechairtablecupboard..."
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How can we efficiently split this text into a list of individual words?

Output: ["table", "apple", "chair", "table", ["cupboard", ["cup", "board"]], ...]
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Algorithm

A simple approach is to iteratively find the longest possible word within the text. However, this can lead to suboptimal results.

Frequency-Based Algorithm

Instead, we can exploit the relative frequency of words in the language to improve accuracy:

  1. Model the Word Distribution: Assume words are independently distributed and follow Zipf's law, where word probability is inversely proportional to its rank.
  2. Define Word Cost: The cost of a word is defined as the logarithm of the inverse of its likelihood.
  3. Dynamic Programming Approach:

    • Initialize a cost array where the first element is 0.
    • For each character in the text, find the word that minimizes the total cost for characters up to that point.
    • Backtrack from the end to reconstruct the minimum-cost word sequence.

Code Implementation

<code class="python">from math import log

wordcost = {}  # Dictionary of word costs using Zipf's law

maxword = max(len(word) for word in wordcost)

def infer_spaces(s):
    cost = [0]
    for i in range(1, len(s) + 1):
        candidates = enumerate(reversed(cost[max(0, i - maxword):i]))
        c, k = min((wordcost.get(s[i - k - 1:i], 9e999) + c, k + 1) for k, c in candidates)
        cost.append(c)

    out = []
    i = len(s)
    while i > 0:
        c, k = best_match(i)
        assert c == cost[i]
        out.append(s[i - k:i])
        i -= k

    return " ".join(reversed(out))</code>
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Results

This algorithm is able to accurately segment text into a list of words, even in the absence of spaces.

Example:

Input: "tableapplechairtablecupboard..."
Output: ["table", "apple", "chair", "table", ["cupboard", ["cup", "board"]], ...]
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Optimizations:

  • Suffix Tree: By building a suffix tree from the word list, the candidate search can be accelerated.
  • Text Block Splitting: For large text inputs, the text can be split into blocks to minimize memory usage while maintaining accuracy.

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