Intro

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Sorting and Taking the Top Rows

~9 mincode completion

df.sort_values("score")                      # ascending
df.sort_values("score", ascending=False)     # descending
df.sort_values(["city", "score"])            # by city, then score within city
df.head(3)                                   # the first 3 rows

sort_values returns a new frame. It does not sort in place unless you pass inplace=True, and you generally should not: reassigning is clearer and avoids a whole category of "why did my other variable change" confusion.

nlargest does sort-then-head in one step and is faster on a big frame, because it does not have to sort everything:

df.nlargest(3, "score")

One thing to know about the index: after sorting, the row labels come along for the ride, so your top row might be labelled 47. That is correct behaviour, not a bug. .reset_index(drop=True) renumbers if you need tidy labels.

Your task:

Write top_scores(df, n) that returns the n rows with the highest score, highest first.

Example Tests

The two highest scores come back, highest first

Input: {"n":2,"df":{"name":["a","b","c"],"score":[70,95,82]}}

Expected: [95,82]

Asking for one row gives just the single best

Input: {"n":1,"df":{"name":["a","b"],"score":[10,20]}}

Expected: [20]

Asking for more rows than exist returns everything available

Input: {"n":10,"df":{"name":["a"],"score":[5]}}

Expected: [5]

Python

import pandas as pd

def top_scores(df, n):
    """
    The n highest-scoring rows.

    Args:
        df: a DataFrame with a "score" column
        n: how many rows to return

    Returns:
        A DataFrame of n rows, ordered from highest score down.
    """
    # YOUR CODE HERE
    pass

def _scores(df, n):
    import pandas as pd
    return [float(v) for v in top_scores(pd.DataFrame(df), n)["score"]]

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