Index by name
Input
items = [{'id':1,'name':'a'}, {'id':2,'name':'b'}, {'id':3,'name':'a'}]; key_name = 'name'
Output
{'a': [0, 2], 'b': [1]}
Value 'a' occurs at indices 0 and 2, 'b' at index 1.
Full lesson preview
Build an index mapping from a list
Create an index mapping values of a specified key to lists of positions they appear at in a list of dicts.
Python practice16 minList & Dictionary PatternsIntermediateLast updated March 20, 2026
Problem statement
Given a list of dictionaries items and a string key_name, implement build_index(items, key_name) that scans items and collects the zero-based indices of each dictionary that contains key_name. The function should skip dictionaries that do not contain key_name. The returned dictionary maps each distinct value (items[i][key_name]) to a list of indices (in increasing order) where that value appears. To keep the output deterministic, insert keys in the returned dict ordered by str(key).
Task
Write build_index(items, key_name) that returns a deterministic dictionary mapping each distinct value found at key_name to a list of indices where that value occurs in items.
Examples
Input format
Two parameters: items (list of dicts) and key_name (string).
Output format
A dict mapping each found value to a list of integer indices.
Constraints
- Skip entries that do not contain key_name. - Do not assume values are comparable; use str(value) for ordering keys in the returned dict while keeping the original values as keys. - The lists of indices must be in ascending order.
Samples
Sample 1
Input
[{'type':'x'}, {'other':2}, {'type':'y'}], 'type'
Output
{'x': [0], 'y': [2]}
Second dict lacks 'type' and is ignored.