Reference library

Dictionaries & sets

Key–value maps, uniqueness, counting, grouping, and fast lookups.

4 matches

Filter Dictionary Keys by Prefix in Python

Use a dict comprehension to build a new dictionary containing only keys that start with a given prefix.

dict-comprehension filtering dictionaries

def filter_dict_keys(data, prefix="temp_"):
    """
    Filter a dictionary by keeping only keys that start with a given prefix.
    Uses a dict comprehension to build a new dictionary.
    """
    if not isinstance(data, dict):
        raise ValueError("data must be a dictionary")
    return {key: value for key, valu…

13 0 Open

Dictionaries & sets easy

How to Filter a Dictionary by Predicate on Values in Python

This code defines a reusable function that builds a new dictionary containing only the items whose values satisfy a given predicate function.

dictionary filtering lambda

Python

def filter_dict_by_predicate(d, predicate):
    """Return a new dict with only items whose value passes the predicate."""
    return {k: v for k, v in d.items() if predicate(v)}


if __name__ == "__main__":
    scores = {"Alice": 85, "Bob": 42, "Charlie": 91, "Diana": 60}
    # Keep only values greater than or equal t…

14 0 Open

Dictionaries & sets easy

How to Filter a List of Dictionaries by Category in Python

Filter a list of dictionaries to include only records whose category is in an allowed set.

dictionary set filter

Python

def filter_data(records, categories):
    """Return only records whose category is in the allowed set."""
    allowed = set(categories)
    filtered = []
    for record in records:
        if record["category"] in allowed:
            filtered.append(record)
    return filtered


if __name__ == "__main__":
    data = …

12 0 Open

Dictionaries & sets easy

How to Remove Banned Words from a Set in Python

Filter a vocabulary set by removing banned words using the .difference() method.

sets set difference filtering

Python

vocabulary = {"apple", "banana", "cherry", "date", "elderberry"}
banned_words = {"banana", "date", "fig"}

# Remove banned words using set difference
allowed_words = vocabulary.difference(banned_words)

print("Original vocabulary:", sorted(vocabulary))
print("Banned words:", sorted(banned_words))
print("Allowed words …

16 0 Open

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