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:
    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])
    result = df_sorted.groupby(group_col, sort=False).head(n)
    result = result.reset_index(drop=True)
    return result
```

The task is to implement a function that returns the top N rows per group based on a value column in descending order.

Key requirements:
1. All original columns must be preserved
2. Within each group, rows ordered by `value_col` descending
3. Result ordered by group (first appearance order), then by `value_col` descending
4. Groups with fewer than n rows contribute all their rows
5. Clean RangeIndex (reset_index(drop=True))
6. Don't mutate input df

The example shows:
- Team "a" has scores [10, 30, 20], top 2 are 30 and 20 (y and z)
- Team "b" has scores [5, 15], top 2 are 15 and 5 (q and p)
- Result is ordered by team first appearance (a then b), then by score descending

Approach:
1. Sort the DataFrame by `group_col` and `value_col` (descending)
2. Use `groupby` with `head(n)` to get top n per group
3. Reset the index

Let me think about the order preservation:
- "Group order follows the order in which each group first appears in df"
- This means I need to preserve the first appearance order of groups

One approach:
```python
def top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:
    # Sort by group_col first (to preserve order), then by value_col descending
    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])
    
    # Group by and take top n
    result = df_sorted.groupby(group_col, sort=False).head(n)
    
    # Reset index
    result = result.reset_index(drop=True)
    
    return result
```

Wait, let me check the example more carefully:


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

After sorting by team (ascending) and score (descending):
- a: 30, 20, 10 (y, z, x)
- b: 15, 5 (q, p)

Taking top 2 per group:
- a: y (30), z (20)
- b: q (15), p (5)

This matches the expected output. The `sort=False` parameter in groupby preserves the order of groups as they appear in the sorted DataFrame.

I should verify that `sort_values` maintains the original group order when sorting by group_col ascending. Since I'm sorting by team first (ascending), groups appear in their first appearance order. Then within each group, I sort by score descending.

Let me trace through:
1. Original df: team order is a, a, a, b, b
2. After sort_values([group_col, value_col], ascending=[True, False]):
   - First by team ascending: all a's come before all b's
   - Then by score descending within each team
   - Result: a (30, 20, 10), b (15, 5)
3. groupby with sort=False preserves this order
4. head(2) takes top 2 from each group
5. reset_index(drop=True) gives clean RangeIndex

This should work correctly.

```python
import pandas as pd

def top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:
    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])
    result = df_sorted.groupby(group_col, sort=False).head(n)
    result = result.reset_index(drop=True)
    return result
```

I need to verify the sort behavior more carefully. When sorting by multiple columns with ascending=[True, False], the first column sorts ascending and the second sorts descending within each group of the first column.

For the example:
- Original: a(10), a(30), a(20), b(5), b(15)
- After sort by team ascending, score descending: a(30), a(20), a(10), b(15), b(5)
- groupby with sort=False preserves this order
- head(2) gives a(30), a(20), b(15), b(5)
- reset_index(drop=True) gives 0, 1, 2, 3

This matches the expected output.

I should also consider edge cases like when n is larger than group size (head handles this), when n is 0 (returns empty), and when there are ties (handled arbitrarily as specified).

One concern: does sort_values preserve the original order for equal values? The documentation says it's stable, so rows with equal values maintain their relative order. But since ties can be broken arbitrarily, this shouldn't matter for correctness.

Let me finalize the solution:

```python
import pandas as pd

def top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:
    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])
    result = df_sorted.groupby(group_col, sort=False).head(n)
    result = result.reset_index(drop=True)
    return result
```

This should work correctly for all the requirements.

[
  {
    "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    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    result = result.reset_index(drop=True)\n    return result\n```",
    "passed": 6,
    "reasoning": "The task is to implement a function that returns the top N rows per group based on a value column in descending order.\n\nKey requirements:\n1. All original columns must be preserved\n2. Within each group, rows ordered by `value_col` descending\n3. Result ordered by group (first appearance order), then by `value_col` descending\n4. Groups with fewer than n rows contribute all their rows\n5. Clean RangeIndex (reset_index(drop=True))\n6. Don't mutate input df\n\nThe example shows:\n- Team \"a\" has scores [10, 30, 20], top 2 are 30 and 20 (y and z)\n- Team \"b\" has scores [5, 15], top 2 are 15 and 5 (q and p)\n- Result is ordered by team first appearance (a then b), then by score descending\n\nApproach:\n1. Sort the DataFrame by `group_col` and `value_col` (descending)\n2. Use `groupby` with `head(n)` to get top n per group\n3. Reset the index\n\nLet me think about the order preservation:\n- \"Group order follows the order in which each group first appears in df\"\n- This means I need to preserve the first appearance order of groups\n\nOne approach:\n```python\ndef top_n_per_group(df: pd.DataFrame, group_col: str, value_col: str, n: int) -> pd.DataFrame:\n    # Sort by group_col first (to preserve order), then by value_col descending\n    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])\n    \n    # Group by and take top n\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    \n    # Reset index\n    result = result.reset_index(drop=True)\n    \n    return result\n```\n\nWait, let me check the example more carefully:\n\n\n```python\ndf = pd.DataFrame({\n    \"team\": [\"a\", \"a\", \"a\", \"b\", \"b\"],\n    \"name\": [\"x\", \"y\", \"z\", \"p\", \"q\"],\n    \"score\": [10, 30, 20, 5, 15],\n})\n```\n\nAfter sorting by team (ascending) and score (descending):\n- a: 30, 20, 10 (y, z, x)\n- b: 15, 5 (q, p)\n\nTaking top 2 per group:\n- a: y (30), z (20)\n- b: q (15), p (5)\n\nThis matches the expected output. The `sort=False` parameter in groupby preserves the order of groups as they appear in the sorted DataFrame.\n\nI should verify that `sort_values` maintains the original group order when sorting by group_col ascending. Since I'm sorting by team first (ascending), groups appear in their first appearance order. Then within each group, I sort by score descending.\n\nLet me trace through:\n1. Original df: team order is a, a, a, b, b\n2. After sort_values([group_col, value_col], ascending=[True, False]):\n   - First by team ascending: all a's come before all b's\n   - Then by score descending within each team\n   - Result: a (30, 20, 10), b (15, 5)\n3. groupby with sort=False preserves this order\n4. head(2) takes top 2 from each group\n5. reset_index(drop=True) gives clean RangeIndex\n\nThis should work correctly.\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    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    result = result.reset_index(drop=True)\n    return result\n```\n\nI need to verify the sort behavior more carefully. When sorting by multiple columns with ascending=[True, False], the first column sorts ascending and the second sorts descending within each group of the first column.\n\nFor the example:\n- Original: a(10), a(30), a(20), b(5), b(15)\n- After sort by team ascending, score descending: a(30), a(20), a(10), b(15), b(5)\n- groupby with sort=False preserves this order\n- head(2) gives a(30), a(20), b(15), b(5)\n- reset_index(drop=True) gives 0, 1, 2, 3\n\nThis matches the expected output.\n\nI should also consider edge cases like when n is larger than group size (head handles this), when n is 0 (returns empty), and when there are ties (handled arbitrarily as specified).\n\nOne concern: does sort_values preserve the original order for equal values? The documentation says it's stable, so rows with equal values maintain their relative order. But since ties can be broken arbitrarily, this shouldn't matter for correctness.\n\nLet me finalize the solution:\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    df_sorted = df.sort_values([group_col, value_col], ascending=[True, False])\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    result = result.reset_index(drop=True)\n    return result\n```\n\nThis should work correctly for all the requirements.\n",
    "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 0.25s"
  },
  {
    "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 value_col descending within each group, preserving group order\n    df_sorted = df.sort_values(value_col, ascending=False)\n    # Group by group_col without sorting groups, then take top n per group\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    # Reset index to get clean RangeIndex\n    result = result.reset_index(drop=True)\n    return result\n```",
    "passed": 6,
    "reasoning": "The issue is that when I use `sort_values([group_col, value_col], ascending=[True, False])`, the group column is sorted alphabetically, which puts \"a\" before \"z\". But the requirement is that groups should appear in the order they first appear in the original DataFrame.\n\nThe problem is that `sort_values` on the group column sorts it alphabetically, not by first appearance order.\n\nTo preserve the order of first appearance, I need to:\n1. First sort by group_col to get groups together\n2. Then sort by value_col descending within each group\n3. But I need to preserve the original order of groups\n\nActually, looking at the test case:\n```python\ndf = pd.DataFrame({\n    \"g\": [\"z\", \"a\", \"z\", \"a\"],\n    \"v\": [1, 2, 3, 4],\n})\n```\n\nThe first appearance order is: \"z\" (index 0), then \"a\" (index 1).\n\nSo the output should have \"z\" group first, then \"a\" group.\n\nThe issue is that `sort_values([group_col, value_col], ascending=[True, False])` sorts the group column alphabetically, which puts \"a\" before \"z\".\n\nI need to preserve the order of first appearance. One way to do this is to use `groupby(...).head(n)` without sorting the group column, but then I need to make sure the groups are in the right order.\n\nActually, `groupby(...).head(n)` preserves the order of groups as they appear in the original DataFrame. But I need to sort by value_col descending within each group.\n\nLet me think about this differently:\n1. I need to sort by value_col descending within each group\n2. But I need to preserve the order of groups as they first appear\n\nOne approach:\n1. Use `groupby` with `sort=False` to preserve group order\n2. Sort by value_col descending within each group using `sort_values` on the grouped data\n\n\n3. Take the top n rows per group\n\nI realize `groupby(...).head(n)` already sorts by value_col descending by default. The key is to ensure the group order matches the first appearance order. I can use `groupby(..., sort=False)` to maintain the original group sequence, then apply `head(n)` to get the top rows.\n\nThe challenge is sorting by value_col while keeping the group order intact. I'll need to carefully manipulate the DataFrame to achieve this without disrupting the group sequence.\n\nMy current approach of sorting by group and value might be causing the group order to change. I need a method that preserves the original group appearance while still sorting values correctly.\n\nThe key is using `groupby(..., sort=False)` to maintain the original group order, then applying `head(n)` to select the top rows within each group. This should keep the group sequence intact while selecting the highest value rows.\n\nI'll test this approach to see if it resolves the group ordering issue while still selecting the top n rows per group.\n",
    "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": "```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_col descending to get top rows per group\n    df_sorted = df.sort_values(value_col, ascending=False)\n    # Group by group_col without sorting groups, then take top n per group\n    # This preserves the order groups first appear in the original df\n    result = df_sorted.groupby(group_col, sort=False).head(n)\n    # Reset index to get clean RangeIndex\n    result = result.reset_index(drop=True)\n    return result\n```",
    "passed": 6,
    "reasoning": "The issue is that when we sort by `value_col` descending first, we lose the original order of groups. The test expects groups to appear in the order they first appear in the original DataFrame.\n\nIn the test case:\n```python\ndf = pd.DataFrame({\n    \"g\": [\"z\", \"a\", \"z\", \"a\"],\n    \"v\": [1, 2, 3, 4],\n})\n```\n\nGroup \"z\" appears first (at index 0), then group \"a\" appears (at index 1). So the output should have \"z\" first, then \"a\".\n\nThe problem with my current solution is that `sort_values(value_col, ascending=False)` doesn't preserve the original group order.\n\nI need to:\n1. First identify the order of groups as they first appear in the DataFrame\n2. Then sort by value_col descending within each group\n3. Make sure the final result maintains the group order\n\nOne approach is to use `groupby().head()` which should preserve group order when `sort=False`, but the issue is that we need to sort by value_col first.\n\nActually, looking at the pandas documentation, `groupby(...).head(n)` should preserve the group order when `sort=False`. The issue might be that we're sorting the entire DataFrame before grouping, which changes the order.\n\nLet me think about this differently. I should:\n1. Sort by group_col first to maintain group order, then by value_col descending\n2. Then groupby with sort=False and head(n)\n\nWait, but if I sort by group_col first, that will put all groups in alphabetical order, not in order of first appearance.\n\nThe correct approach is:\n1. Use `groupby().head(n)` which internally handles the grouping\n2. But we need to ensure the groups are in the order they first appear\n\nActually, looking at pandas documentation more carefully, when you use `groupby(..., sort=False)`, the groups should maintain their original order. But the issue is that we're sorting the entire DataFrame before grouping.\n\n\nI see the core challenge is maintaining the original group order while selecting top rows. The solution requires a careful approach to preserve the initial group sequence. I'll need to track the first appearance of each group and ensure the final output respects this order.\n\nThe key is to use a method that doesn't disrupt the original group sequence when selecting top rows. This means avoiding a full DataFrame sort that might reorder groups unexpectedly.\n\nBy sorting the value column first and then grouping, I can ensure the top rows are selected while maintaining the original group order. The `sort=False` parameter in `groupby` will help preserve the initial group sequence.\n\nThe critical insight is that `sort_values` on the value column doesn't inherently change group order, but the subsequent `groupby` operation needs careful handling to maintain the original group sequence.\n\nI'll need to verify how pandas handles group ordering when using `head()` with `sort=False`, as this might be the key to solving the group order preservation challenge.\n\nThe test case reveals a critical nuance: when groups have different values, the selection method must respect the original group appearance order, not just the value ranking.\n\nMy current approach of sorting by value first and then grouping seems to be disrupting the expected group sequence. I need to find a way to select top rows per group while maintaining the original group order.\n\nThe key insight is that `groupby().head()` should preserve group order when `sort=False`, but the initial sorting might be interfering with this behavior. I'll need to carefully examine how pandas handles group ordering during the selection process.\n\nThe test case suggests that group order matters significantly, and my current implementation isn't respecting the original group sequence. I'll need to modify my approach to ensure the group order is preserved when selecting top rows.\n\nThe key insight is that `sort_values` on the value column might be disrupting the original group order. I should explore alternative methods that maintain the initial group sequence while still selecting the top rows per group.\n\nMy current implementation needs refinement to handle group ordering more precisely, potentially by using a different sorting or grouping strategy that respects the original DataFrame's structure.\n",
    "test_error": ""
  }
]