answer:Here's the Python function to count the occurrences of each unique word across a collection of texts:```pythonimport stringdef count_word_occurrences(texts): word_counts = {} for text in texts: # Remove punctuation from the text text_no_punct = ''.join(char for char in text if char not in string.punctuation) # Split the text into words words = text_no_punct.split() # Count the occurrences of each word for word in words: word_counts[word] = word_counts.get(word, 0) + 1 return word_counts```To explain the function:1. We import the `string` module to access the `string.punctuation` constant, which contains a string of all punctuation characters.2. The `count_word_occurrences` function takes a list of strings `texts` as input.3. We initialize an empty dictionary `word_counts` to store the word counts.4. We iterate over each text document in `texts`.5. For each text, we remove punctuation characters using a list comprehension and the `string.punctuation` constant.6. We split the text (without punctuation) into words using the `split()` method.7. For each word in the text, we update the `word_counts` dictionary by incrementing the count for that word. If the word is not in the dictionary yet, we initialize its count to 1 using the `get` method with a default value of 0.8. After iterating over all texts, we return the `word_counts` dictionary containing the word counts.The function handles any number of documents and any length of text. It also considers words to be case-sensitive, as per the requirement.

answer:def is_valid_citation(citation): import re pattern = r"^(d+)s+([A-Z.]+)s+(d+)" match = re.match(pattern, citation) if match: volume, reporter, page = match.groups() return True else: return False

answer:def analyze_text(text, keywords): # Convert text to lowercase and remove punctuation text = ''.join(c for c in text.lower() if c.isalnum() or c == ' ') # Split text into words words = text.split() # Initialize a dictionary to store word counts word_counts = {keyword: 0 for keyword in keywords} # Count occurrences of keywords for word in words: if word in word_counts: word_counts[word] += 1 return word_counts

answer:Here's the Python function to solve the given problem:```pythondef filter_long_episodes(episode_lengths): # Calculate the average episode length average_length = sum(episode_lengths) / len(episode_lengths) # Filter out episodes longer than the average long_episodes = [length for length in episode_lengths if length > average_length] return long_episodes```This function takes a list of integers `episode_lengths` as input, where each integer represents the length of an episode in minutes. It performs the following steps:1. Calculate the average episode length by summing up all the lengths and dividing by the total number of episodes.2. Use a list comprehension to filter out the episodes that are longer than the average length.3. Return the list of long episodes.To use this function, you can call it with a list of episode lengths, like this:```pythonepisode_lengths = [40, 55, 60, 35, 50]long_episodes = filter_long_episodes(episode_lengths)print(long_episodes) # Output: [55, 60, 50]```The output will be a list containing only the episodes that are longer than the average length.

answer:def count_keyword_frequency(titles, keyword): count = 0 for title in titles: count += title.lower().split().count(keyword.lower()) return count

answer:def calculate_average_gas_price(transactions): total_gas_price = 0 num_transactions = len(transactions) for transaction in transactions: total_gas_price += transaction["gasPrice"] if num_transactions > 0: average_gas_price = total_gas_price / num_transactions return float(average_gas_price) else: return 0.0