problemeasyalgorithmsleetcode-2889leetcode 2889leetcode2889

Reshape Data- Pivot

EasyUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

DataFrame weather +-------------+--------+ | Column Name | Type | +-------------+--------+ | city | object | | month | object | | temperature | int | +-------------+--------+

Write a solution to pivot the data so that each row represents temperatures for a specific month, and each city is a separate column.

The result format is in the following example.

Examples

Example 1

Input:
+--------------+----------+-------------+
| city         | month    | temperature |
+--------------+----------+-------------+
| Jacksonville | January  | 13          |
| Jacksonville | February | 23          |
| Jacksonville | March    | 38          |
| Jacksonville | April    | 5           |
| Jacksonville | May      | 34          |
| ElPaso       | January  | 20          |
| ElPaso       | February | 6           |
| ElPaso       | March    | 26          |
| ElPaso       | April    | 2           |
| ElPaso       | May      | 43          |
+--------------+----------+-------------+
Output:
+----------+--------+--------------+
| month    | ElPaso | Jacksonville |
+----------+--------+--------------+
| April    | 2      | 5            |
| February | 6      | 23           |
| January  | 20     | 13           |
| March    | 26     | 38           |
| May      | 43     | 34           |
+----------+--------+--------------+
Explanation: The table is pivoted, each column represents a city, and each row represents a specific month.

## Solution

### Method 1 - Pandas Pivot

#### Intuition
We want to pivot the DataFrame so that each city becomes a column and each month is a row, with temperature as the value.

#### Approach
Use pandas `pivot` to reshape the DataFrame, then reset the index and sort by month for a clean output.

#### Code

{{< code_tabs >}}
##### Python (pandas)
```python
def pivot_weather(weather: pd.DataFrame) -> pd.DataFrame:
    result = weather.pivot(index='month', columns='city', values='temperature').reset_index()
    # Optional: sort by month for consistent output
    result = result.sort_values('month').reset_index(drop=True)
    return result

Complexity

  • ⏰ Time complexity: O(n) — where n is the number of rows in the DataFrame.
  • 🧺 Space complexity: O(n) — for the output DataFrame.

Comments