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# Sorting a dictionary by value in Python

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schedule Jul 1, 2022
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To sort a dictionary by value in Python use the `sorted(~)` built-in function.

WARNING

This only works with Python 3.7+ as before this dictionaries did not preserve insertion order.

# Example

Consider the following dictionary:

``` d1 = {"C": 3, "B": 2, "A": 1} ```

## Ascending order

To sort the dictionary in ascending order of value:

``` dict(sorted(d1.items(), key=lambda item:item[1])) {'A': 1, 'B': 2, 'C': 3} ```

Let us explain the above by breaking it down into a few parts. The iterable we are passing to `sorted(~)` for sorting is `d1.items()`:

``` dict_items([('C', 3), ('B', 2), ('A', 1)]) ```

We sort the above iterable according to the lambda function provided as the `key`. Here the lambda function states that we want to sort on the second item `item[1]` within each tuple `item`. This means we sort the tuples according to their second items (`3`, `2`, and `1`) in ascending order.

Finally we pass the result to `dict()` constructor to create the dictionary that we return.

## Descending order

To sort the dictionary in descending order of value:

``` dict(sorted(d1.items(), reverse=True, key=lambda item:item[1])) {"C": 3, "B": 2, "A": 1} ```

By passing `reverse=True` we sort in descending order.

Published by Arthur Yanagisawa
Edited by 0 others
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