Count users per language
Input
group_frequency([{'lang':'py','user':'a'},{'lang':'py','user':'b'},{'lang':'js','user':'a'}],'lang','user')
Output
{'py': {'a': 1, 'b': 1}, 'js': {'a': 1}}
Group by lang, then count user occurrences in each language.
Full lesson preview
Build a grouped frequency report with Counter and defaultdict
Group records by a key and produce frequency counts for items using Counter and defaultdict.
Python practice25 minModules & Standard LibraryAdvancedLast updated March 23, 2026
Problem statement
Write group_frequency(records, group_by, count_field=None) that aggregates counts from a list of record dicts.
- records: list of dict-like records.
- group_by (str): key name used to group records; if missing in a record, group under None.
- count_field (str|None): if None, count records per group and return a mapping group -> int. If provided, count occurrences of the value at count_field per group and return mapping group -> dict(value -> count).
Return regular Python dicts (not Counters) so the output prints as plain dicts. Use collections.Counter and collections.defaultdict internally to simplify aggregation.
Task
Use collections.Counter and defaultdict to aggregate counts by group, handling missing keys gracefully.
Examples
Input format
A list of dict records, group_by key, and optional count_field key.
Output format
If count_field is None: dict mapping group -> int. If count_field provided: dict mapping group -> dict mapping item -> count.
Constraints
- Use collections.Counter and collections.defaultdict.
- Missing group_by keys map to group None.
- Missing count_field keys within a group count as None.
- Return plain dicts (convert Counters to dicts) for predictable string representation.
Samples
Sample 1
Input
group_frequency([{'lang':'py'},{'lang':'py'},{'lang':'js'}],'lang')
Output
{'py': 2, 'js': 1}
Counts per group when count_field is omitted.