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Use sorted(d) to get a list of a dictionary’s keys in ascending order. To build a new dictionary in key order, use {key: d[key] for key in sorted(d)}. To build one ordered by values, sort the key-value pairs with dict(sorted(d.items(), key=lambda item: item[1])). Add reverse=True for descending order.
Sort a dictionary by key
Passing a dictionary to sorted() sorts its keys and returns a list, not a dictionary:
scores = {"Mina": 91, "Dev": 78, "Alex": 91}
keys = sorted(scores)
print(keys) # ['Alex', 'Dev', 'Mina']If you need a mapping whose entries are inserted in key order, create a new dictionary from that key list:
by_key = {key: scores[key] for key in sorted(scores)}
print(by_key) # {'Alex': 91, 'Dev': 78, 'Mina': 91}Python dictionaries preserve insertion order, so iterating over by_key follows the order in which the sorted keys were inserted. This guarantee applies from Python 3.7 onward; it preserves insertion order but does not sort a dictionary automatically. Python’s built-in types documentation describes dictionary ordering.
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Sort a dictionary by value
Use items() to get key-value pairs, and set the sort key to the value at index 1. Wrap the result in dict() to produce a new dictionary:
by_value = dict(sorted(scores.items(), key=lambda item: item[1]))
print(by_value) # {'Dev': 78, 'Mina': 91, 'Alex': 91}To sort from highest value to lowest, pass reverse=True:
by_value_desc = dict(
sorted(scores.items(), key=lambda item: item[1], reverse=True)
)sorted() returns a new list of sorted items; it does not rearrange the original dictionary in place. Building a new dictionary from those items gives you a mapping with the sorted insertion sequence.
Choose how ties are ordered
Python’s sort is stable: pairs with equal sort values keep their relative order from the input items. If you want an explicit secondary rule, include it in a tuple key. This example sorts by value, then by key:
by_value_then_key = dict(
sorted(scores.items(), key=lambda item: (item[1], item[0]))
)The secondary key works when the values and keys being compared have compatible ordering. For a concise value key, you can use operator.itemgetter(1) instead of a lambda:
from operator import itemgetter
by_value = dict(sorted(scores.items(), key=itemgetter(1)))The Python Sorting HOW TO documents stable sorting, tuple-based sort keys, and itemgetter().
Handle text and values that cannot be compared
- Case-insensitive text keys: use
key=str.casefold, for examplesorted(names, key=str.casefold). This normalizes case for comparison; it is not locale-aware alphabetical sorting. - Locale-aware text: use
locale.strxfrm()orlocale.strcoll()when the desired ordering depends on the active locale. - Mixed or special values: values of incompatible types may raise
TypeErrorwhen Python tries to compare them.Nonecannot be ordered against ordinary numbers, and NaN values do not participate in ordinary numeric ordering consistently. Define a deliberate key or handle these values separately. - String conversion: converting mixed values to strings can make them sortable, but the result is lexical rather than necessarily numeric; for example, text ordering may not match numeric ordering.
The sorting guide covers sort keys and special comparison cases.
Quick Recap
Quick reference
| Goal | Code | Result |
|---|---|---|
| Get keys in ascending order | sorted(d) |
A list of keys |
| Create a dictionary in ascending key order | {k: d[k] for k in sorted(d)} |
A new dictionary |
| Create a dictionary in ascending value order | dict(sorted(d.items(), key=lambda item: item[1])) |
A new dictionary |
| Sort values in descending order | dict(sorted(d.items(), key=lambda item: item[1], reverse=True)) |
A new dictionary, highest value first |
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