5 Best Ways to Return an Array with the Number of Nonoverlapping Occurrences of Substring in Python

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πŸ’‘ Problem Formulation: In Python, it’s common to need a count of how many times a nonoverlapping substring occurs within a string. For instance, given the input string “banana” and the substring “ana”, the desired output is an array [1] since “ana” occurs nonoverlapping once in “banana”.

Method 1: Using the count() method

This method involves the built-in str.count() function, which returns the number of nonoverlapping occurrences of the substring. To return an array, we simply encapsulate the result within a list. The count() method is case-sensitive and does not use regular expressions. It’s an efficient and straightforward approach to count the occurrences of a substring.

Here’s an example:

def count_substrings(s, sub):
    return [s.count(sub)]

# Example usage
print(count_substrings("banana", "ana"))

Output:

[1]

This code snippet defines a function count_substrings() that takes a string s and a substring sub as arguments. It counts the nonoverlapping occurrences of sub in s using the count() method and returns the result encapsulated in a list. In our example, “ana” occurs once in “banana”.

Method 2: Using re.findall() with a Regular Expression

The re.findall() method from Python’s regular expression module allows us to find all nonoverlapping occurrences of a pattern. By using a regex pattern that matches the substring, we can return the length of the resulting list as the count. This method is powerful when the substring has special pattern requirements.

Here’s an example:

import re

def count_substrings_regex(s, sub):
    pattern = re.escape(sub)  # Escape special regex characters in sub
    return [len(re.findall(pattern, s))]

# Example usage
print(count_substrings_regex("banana", "ana"))

Output:

[1]

Here, we use the re.findall() function with the escaped substring as our pattern to search through the base string. Using len() on the result gives us how many times the substring occurs nonoverlapping in the base string, which is then returned as a single-element list.

Method 3: Using re.finditer() and a Loop

Similar to findall(), the re.finditer() method returns an iterator yielding match objects over nonoverlapping occurrences. This method is more memory efficient for large strings, as it does not store all matches. The count is aggregated using a loop over the iterator.

Here’s an example:

import re

def count_substrings_iter(s, sub):
    pattern = re.escape(sub)
    return [sum(1 for _ in re.finditer(pattern, s))]

# Example usage
print(count_substrings_iter("banana", "ana"))

Output:

[1]

The function count_substrings_iter() uses re.finditer() to create an iterator for all matches of the substring. It then uses a generator expression to count the matches nonoverlapping and encapsulates the result inside a list. The substring “ana” is found once without overlapping in the string “banana”.

Method 4: Using a Loop to Manually Search

Without the help of the re module, we can manually iterate through the string and count the nonoverlapping occurrences of the substring. This method involves detailed handling of the indices and is useful when you wish to avoid regular expressions.

Here’s an example:

def count_substrings_manual(s, sub):
    count = 0
    i = 0
    while i <= len(s) - len(sub):
        if s[i:i+len(sub)] == sub:
            count += 1
            i += len(sub)  # jump past the substring
        else:
            i += 1
    return [count]

# Example usage
print(count_substrings_manual("banana", "ana"))

Output:

[1]

The count_substrings_manual() function searches through the base string using a while-loop. It compares slices of the string with the substring and skips the length of the substring if a match is found to avoid overlapping. The overall count of nonoverlapping occurrences is then returned as an array.

Bonus One-Liner Method 5: Using List Comprehension and str.count()

For a more Pythonic and concise approach, we can combine list comprehension with the str.count() method to achieve the same result in a single line of code, which is particularly useful for short and simple scripts.

Here’s an example:

count_substrings_oneliner = lambda s, sub: [s.count(sub)]

# Example usage
print(count_substrings_oneliner("banana", "ana"))

Output:

[1]

We’ve encapsulated the counting function within a lambda to create a one-liner that’s easy to read and write. The count_substrings_oneliner() lambda function takes the string and substring and directly returns the count inside a list. It’s compact, but its readability may be less clear for complex counting situations.

Summary/Discussion

  • Method 1: Using count(). Simple and straightforward. Limited to literal substring searches.
  • Method 2: Using re.findall(). Flexible for complex patterns. Slightly more overhead due to regex.
  • Method 3: Using re.finditer() and a Loop. Memory efficient for large datasets. Requires understanding of iterators.
  • Method 4: Using a Loop to Manually Search. Gives fine control over search process. Verbose and more prone to errors.
  • Bonus Method 5: One-Liner with str.count(). Elegant and Pythonic for simple cases. Not suitable for patterns or complex logic.