Easy

Have a go at it. The editor and the docs panel are open, and your code is saved as you type. Running it needs a free account — you’ll come back to exactly what you wrote.

groupby and Aggregate

~12 mincode completion

groupby is the single most valuable thing pandas does. It splits the rows into groups, applies a calculation to each group, and puts the answers back together.

df.groupby("city")["price"].mean()

Read it in three parts:

groupby("city") split the rows by the value in the city column

["price"] from each group, take the price column

.mean() and reduce it to one number

The result is a Series indexed by the group key. .reset_index() turns it back into a normal DataFrame with the key as a column, which is usually what you want next.

Other aggregations work the same way: .sum(), .count(), .max(), .std(), and .agg(["mean", "count"]) for several at once.

A detail worth knowing early: rows where the grouping key is NaN are dropped silently by default. If your group totals do not add up to the whole, that is usually why.

Your task:

Write mean_by_city(df) that returns the mean price for each city, as a dictionary mapping city name to mean price.

Use groupby, then convert with .to_dict().

Example Tests

Each city gets the mean of its own rows

Input: {"df":{"city":["a","a","b"],"price":[10,20,7]}}

Expected: {"a":15,"b":7}

A single city gives a single entry

Input: {"df":{"city":["x","x"],"price":[4,6]}}

Expected: {"x":5}

Every city appears exactly once in the result

Input: {"df":{"city":["a","b","c","a"],"price":[1,2,3,5]}}

Expected: {"a":3,"b":2,"c":3}

Python

import pandas as pd

def mean_by_city(df):
    """
    Average price per city.

    Args:
        df: a DataFrame with "city" and "price" columns

    Returns:
        A dict mapping city -> mean price.
    """
    # YOUR CODE HERE
    pass

def _run(df):
    import pandas as pd
    return mean_by_city(pd.DataFrame(df))

Run your code to see results

⌘↵ runs against the visible tests

Loading docs…