py-07-pandas-top-n

# Top-N rows per group (pandas)

Implement **`solution.py`** with:

```python
import pandas as pd

def top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:
    ...
```

Given a DataFrame `df`, return a new DataFrame containing, for each group defined
by `group_col`, the **top `n` rows ranked by `value_col` in descending order**.

Requirements:

- All original columns must be preserved (do not drop, rename, or reorder columns).
- Within each group, rows are ordered by `value_col` **descending**. Groups that
  have fewer than `n` rows contribute all of their rows.
- The result is ordered by group, and within each group by `value_col` descending.
  Group order follows the order in which each group first appears in `df`.
- Ties in `value_col` may be broken arbitrarily, but the number of rows returned
  per group must be exactly `min(n, group_size)`.
- The returned DataFrame must use a clean `RangeIndex` (`0..len-1`) — call
  `reset_index(drop=True)` on the result.
- Do not mutate the input `df`.

Use pandas (e.g. `sort_values` + `groupby(...).head(n)`).

Example:

```python
df = pd.DataFrame({
    "team": ["a", "a", "a", "b", "b"],
    "name": ["x", "y", "z", "p", "q"],
    "score": [10, 30, 20, 5, 15],
})
top_n_per_group(df, "team", "score", 2)
#   team name  score
# 0    a    y     30
# 1    a    z     20
# 2    b    q     15
# 3    b    p      5
```

import pandas as pd
from solution import top_n_per_group


def base_df():
    return pd.DataFrame({
        "team": ["a", "a", "a", "b", "b"],
        "name": ["x", "y", "z", "p", "q"],
        "score": [10, 30, 20, 5, 15],
    })


def test_basic_top2():
    df = base_df()
    out = top_n_per_group(df, "team", "score", 2)
    assert list(out.columns) == ["team", "name", "score"]
    assert len(out) == 4
    a = out[out["team"] == "a"]
    assert list(a["score"]) == [30, 20]
    assert list(a["name"]) == ["y", "z"]
    b = out[out["team"] == "b"]
    assert list(b["score"]) == [15, 5]


def test_index_is_rangeindex():
    df = base_df()
    out = top_n_per_group(df, "team", "score", 2)
    assert list(out.index) == [0, 1, 2, 3]


def test_group_smaller_than_n():
    df = base_df()
    out = top_n_per_group(df, "team", "score", 10)
    # all rows kept (3 in a, 2 in b)
    assert len(out) == 5
    assert (out["team"] == "a").sum() == 3
    assert (out["team"] == "b").sum() == 2


def test_n_equals_one():
    df = base_df()
    out = top_n_per_group(df, "team", "score", 1)
    assert len(out) == 2
    assert set(out["name"]) == {"y", "q"}


def test_does_not_mutate_input():
    df = base_df()
    snapshot = df.copy()
    _ = top_n_per_group(df, "team", "score", 2)
    pd.testing.assert_frame_equal(df, snapshot)


def test_group_order_follows_first_appearance():
    df = pd.DataFrame({
        "g": ["z", "a", "z", "a"],
        "v": [1, 2, 3, 4],
    })
    out = top_n_per_group(df, "g", "v", 1)
    # "z" appears first, so it should come first
    assert list(out["g"]) == ["z", "a"]
    assert list(out["v"]) == [3, 4]


def test_preserves_extra_columns():
    df = pd.DataFrame({
        "team": ["a", "a", "b"],
        "score": [1, 2, 3],
        "extra": ["p", "q", "r"],
    })
    out = top_n_per_group(df, "team", "score", 1)
    assert list(out.columns) == ["team", "score", "extra"]
    a = out[out["team"] == "a"].iloc[0]
    assert a["extra"] == "q"

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.

```python
import pandas as pd

def top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:
    # Sort by group column and then by value column in descending order
    sorted_df = df.sort_values([group_col, value_col], ascending=[True, False])
    
    # Get top n rows per group
    result = sorted_df.groupby(group_col).head(n)
    
    # Reset index to get clean RangeIndex
    return result.reset_index(drop=True)
```

[
  {
    "total": 7,
    "answer": "```python\nimport pandas as pd\n\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Sort by group column and then by value column in descending order\n    sorted_df = df.sort_values([group_col, value_col], ascending=[True, False])\n    \n    # Get top n rows per group\n    result = sorted_df.groupby(group_col).head(n)\n    \n    # Reset index to get clean RangeIndex\n    return result.reset_index(drop=True)\n```",
    "passed": 6,
    "test_error": ".....F.                                                                  [100%]\n=================================== FAILURES ===================================\n__________________ test_group_order_follows_first_appearance ___________________\n\n    def test_group_order_follows_first_appearance():\n        df = pd.DataFrame({\n            \"g\": [\"z\", \"a\", \"z\", \"a\"],\n            \"v\": [1, 2, 3, 4],\n        })\n        out = top_n_per_group(df, \"g\", \"v\", 1)\n        # \"z\" appears first, so it should come first\n>       assert list(out[\"g\"]) == [\"z\", \"a\"]\nE       AssertionError: assert ['a', 'z'] == ['z', 'a']\nE         \nE         At index 0 diff: 'a' != 'z'\nE         Use -v to get more diff\n\ntest_top_n.py:61: AssertionError\n=========================== short test summary info ============================\nFAILED test_top_n.py::test_group_order_follows_first_appearance - AssertionEr...\n1 failed, 6 passed in 0.20s"
  },
  {
    "total": 7,
    "answer": "The issue is that when we sort the DataFrame first by group column and then by value column, the group order gets disrupted. We need to preserve the original group appearance order while still sorting within groups by value.\n\nLet me fix this:\n\n```python\nimport pandas as pd\n\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Create a copy to avoid mutating the input\n    df_copy = df.copy()\n    \n    # Sort by value column in descending order\n    df_copy = df_copy.sort_values(value_col, ascending=False)\n    \n    # Get the first occurrence of each group to preserve original order\n    group_order = df_copy.drop_duplicates(group_col, keep='first')[group_col]\n    \n    # Group by the group column and get top n rows for each group\n    result = df_copy.groupby(group_col).head(n)\n    \n    # Reorder groups according to their first appearance in the original dataframe\n    result['group_order'] = pd.Categorical(result[group_col], categories=group_order, ordered=True)\n    result = result.sort_values(['group_order', value_col], ascending=[True, False])\n    \n    # Remove the helper column and reset index\n    result = result.drop('group_order', axis=1).reset_index(drop=True)\n    \n    return result\n```\n\nWait, let me think of a simpler approach. The key insight is that we need to:\n1. Sort by value descending within groups\n2. Preserve group order based on first appearance\n\n```python\nimport pandas as pd\n\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Create a copy to avoid mutating the input\n    df_copy = df.copy()\n    \n    # Sort by value column in descending order\n    df_copy = df_copy.sort_values(value_col, ascending=False)\n    \n    # Get group order based on first appearance\n    unique_groups = df_copy[group_col].drop_duplicates(keep='first')\n    \n    # Group and get top n rows for each group\n    result = df_copy.groupby(group_col).head(n)\n    \n    # Reorder groups to match their first appearance order\n    result['group_order'] = result[group_col].map({group: i for i, group in enumerate(unique_groups)})\n    result = result.sort_values(['group_order', value_col], ascending=[True, False])\n    \n    # Remove helper column and reset index\n    result = result.drop('group_order', axis=1).reset_index(drop=True)\n    \n    return result\n```",
    "passed": 6,
    "test_error": ".....F.                                                                  [100%]\n=================================== FAILURES ===================================\n__________________ test_group_order_follows_first_appearance ___________________\n\n    def test_group_order_follows_first_appearance():\n        df = pd.DataFrame({\n            \"g\": [\"z\", \"a\", \"z\", \"a\"],\n            \"v\": [1, 2, 3, 4],\n        })\n        out = top_n_per_group(df, \"g\", \"v\", 1)\n        # \"z\" appears first, so it should come first\n>       assert list(out[\"g\"]) == [\"z\", \"a\"]\nE       AssertionError: assert ['a', 'z'] == ['z', 'a']\nE         \nE         At index 0 diff: 'a' != 'z'\nE         Use -v to get more diff\n\ntest_top_n.py:61: AssertionError\n=========================== short test summary info ============================\nFAILED test_top_n.py::test_group_order_follows_first_appearance - AssertionEr...\n1 failed, 6 passed in 0.17s"
  },
  {
    "total": 7,
    "answer": "Looking at this more carefully, I think I'm overcomplicating it. Let me trace through what should happen:\n\n1. Original df has groups in order: \"z\", \"a\", \"z\", \"a\" \n2. When we sort by value descending: [4, 3, 2, 1] -> rows with values 4, 3, 2, 1\n3. The group order should be preserved based on first appearance in original data\n\nLet me approach this differently - I'll use the fact that pandas groupby preserves the order of groups as they appear:\n\n```python\nimport pandas as pd\n\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Sort by value column in descending order\n    sorted_df = df.sort_values(value_col, ascending=False)\n    \n    # Group by the group column and get top n rows for each group\n    # This preserves the group order based on first appearance\n    result = sorted_df.groupby(group_col).head(n)\n    \n    # Reset index to get clean RangeIndex\n    return result.reset_index(drop=True)\n```\n\nWait, that's still not right. The issue is that when we sort by value first, we're changing the order of rows within groups. Let me re-read the requirements.\n\nActually, I think the problem is simpler - let me just make sure I understand what should happen:\n\n1. First, we need to identify which groups exist and in what order they appear\n2. Then for each group, get top n rows sorted by value descending  \n3. Combine them in group order\n\n```python\nimport pandas as pd\n\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Get the unique groups in order of first appearance\n    unique_groups = df[group_col].drop_duplicates().tolist()\n    \n    # Create a list to store results for each group\n    result_dfs = []\n    \n    # Process each group in order of first appearance\n    for group in unique_groups:\n        # Filter rows for this group\n        group_df = df[df[group_col] == group]\n        \n        # Sort by value descending\n        group_df = group_df.sort_values(value_col, ascending=False)\n        \n        # Take top n rows (or all if less than n)\n        group_df = group_df.head(n)\n        \n        result_dfs.append(group_df)\n    \n    # Concatenate all results\n    if result_dfs:\n        result = pd.concat(result_dfs, ignore_index=True)\n    else:\n        result = df.iloc[0:0].copy()  # Empty DataFrame with same columns\n    \n    return result\n```",
    "passed": 7,
    "test_error": ""
  }
]