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myoshi2891 merged 1 commit into from
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SQL: Basic Join 1075. Project Employees I #210

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324 changes: 324 additions & 0 deletions ...in/1075. Project Employees I/gpt 5.1 thinking customized/Project_Employees_I_pandas.ipynb
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{
"cells": [
{
"cell_type": "markdown",
"id": "dd3c3113",
"metadata": {},
"source": [
"## 0) 前提\n",
"\n",
"* 環境: **Python 3.10.15 / pandas 2.2.2**\n",
"* **指定シグネチャ厳守**\n",
"\n",
" * 関数名: `project_employees`\n",
" * 引数名: `project`, `employee`\n",
" * 返却列: `[\"project_id\", \"average_years\"]`\n",
" * 列順: 上記順序\n",
"* I/O 禁止（ファイル / 標準出力）、`print` / `sort_values` は使用しない\n",
"\n",
"---\n",
"\n",
"## 1) 問題\n",
"\n",
"* `{{PROBLEM_STATEMENT}}`\n",
" 各プロジェクトについて、そのプロジェクトにアサインされている従業員の\n",
" **平均経験年数 (`experience_years`) を小数第 2 位に丸めて** 求める。\n",
"\n",
"* 入力 DF: `{{INPUT_DATAFRAMES}}`\n",
"\n",
" * `project: pd.DataFrame`\n",
"\n",
" | column | dtype |\n",
" | ----------- | ----- |\n",
" | project_id | int |\n",
" | employee_id | int |\n",
"\n",
" 各行は「従業員 `employee_id` がプロジェクト `project_id` に所属している」ことを表す。\n",
"\n",
" * `employee: pd.DataFrame`\n",
"\n",
" | column | dtype |\n",
" | ---------------- | ------ |\n",
" | employee_id | int |\n",
" | name | object |\n",
" | experience_years | int |\n",
"\n",
" 各行は従業員 1 名の情報。`experience_years` は NULL なし。\n",
"\n",
"* 出力: `{{OUTPUT_COLUMNS_AND_RULES}}`\n",
"\n",
" * 戻り値: `pd.DataFrame`\n",
"\n",
" * 列と意味:\n",
"\n",
" * `project_id`: プロジェクト ID\n",
" * `average_years`: そのプロジェクトに所属する従業員の `experience_years` の平均値（小数第 2 位で丸め）\n",
"\n",
" * 各 `project_id` につき 1 行\n",
"\n",
" * 並び順は任意（`sort_values` 禁止のためソートしない）\n",
"\n",
"---\n",
"\n",
"## 2) 実装（指定シグネチャ厳守）\n",
"\n",
"> 列を最小化しつつ `merge` → `groupby.mean` → `round` の順に処理します。\n",
"> 今回はグループ内順位や条件抽出は不要なので、シンプルな集約だけで OK です。\n",
"\n",
"```python\n",
"import pandas as pd\n",
"\n",
"def project_employees(project: pd.DataFrame, employee: pd.DataFrame) -> pd.DataFrame:\n",
" \"\"\"\n",
" 各プロジェクトごとの平均経験年数を計算する。\n",
"\n",
" Args:\n",
" project (pd.DataFrame): 列 ['project_id', 'employee_id']\n",
" employee (pd.DataFrame): 列 ['employee_id', 'name', 'experience_years']\n",
"\n",
" Returns:\n",
" pd.DataFrame: 列名と順序は ['project_id', 'average_years']\n",
" \"\"\"\n",
" # 1) 列最小化: employee 側は平均に必要な列だけに絞る\n",
" emp_exp = employee[[\"employee_id\", \"experience_years\"]]\n",
"\n",
" # 2) JOIN: project に experience_years を紐づける\n",
" merged = project.merge(emp_exp, on=\"employee_id\", how=\"left\")\n",
"\n",
" # 3) プロジェクトごとに平均値を計算\n",
" out = (\n",
" merged\n",
" .groupby(\"project_id\", as_index=False)[\"experience_years\"]\n",
" .mean()\n",
" )\n",
"\n",
" # 4) 列名を仕様どおりにリネームし、小数第 2 位に丸める\n",
" out = out.rename(columns={\"experience_years\": \"average_years\"})\n",
" out[\"average_years\"] = out[\"average_years\"].round(2)\n",
"\n",
" return out\n",
"\n",
"Analyze Complexity\n",
"Runtime 283 ms\n",
"Beats 62.61%\n",
"Memory 69.14 MB\n",
"Beats 18.00%\n",
"\n",
"```\n",
"\n",
"---\n",
"\n",
"## 3) アルゴリズム説明\n",
"\n",
"### 使用 API\n",
"\n",
"* `DataFrame[...]`\n",
" → 列のサブセットを取り、**列最小化**（不要列を運ばないことでメモリ削減）。\n",
"* `DataFrame.merge`\n",
" → `project` と `employee` を `employee_id` で結合し、従業員の経験年数をプロジェクトに紐付け。\n",
"* `DataFrame.groupby` + `GroupBy.mean`\n",
" → `project_id` ごとに `experience_years` の平均を計算。\n",
"* `DataFrame.rename`\n",
" → 出力列名を問題仕様どおり `average_years` に。\n",
"* `Series.round`\n",
" → 平均値を小数第 2 位に丸める。\n",
"\n",
"### NULL / 重複 / 型の扱い\n",
"\n",
"* `experience_years` は問題文より **NULL なし** 前提なので、`mean()` で NULL ケアは不要。\n",
"* `project` 側は `(project_id, employee_id)` が主キーなので、\n",
" 「同じプロジェクトに同じ従業員が複数行いる」ことはなく、重複による二重カウントも発生しない。\n",
"* `mean()` の戻り値は `float64`。\n",
" `round(2)` で `2.0` → 画面では `2.00` 相当の精度を保持できる（pandas の表示設定次第）。\n",
"\n",
"---\n",
"\n",
"## 4) 計算量（概算）\n",
"\n",
"`N = len(project)`, `M = len(employee)` とします。\n",
"\n",
"* 列最小化: `employee[[\"employee_id\", \"experience_years\"]]`\n",
" → **O(M)**\n",
"* `merge`（ハッシュ結合想定）: `project.merge(emp_exp, on=\"employee_id\")`\n",
" → **O(N + M)** 近辺\n",
"* `groupby(\"project_id\").mean()`\n",
" → **O(N)**（ハッシュベースのグループ集計）\n",
"\n",
"したがって、全体はおおよそ **O(N + M)** 時間、メモリは\n",
"結合後の一時 DataFrame（行数 ≒ N、列数は 3 列）相当となります。\n",
"\n",
"---\n",
"\n",
"## 5) 図解（Mermaid 超保守版）\n",
"\n",
"```mermaid\n",
"flowchart TD\n",
" A[project<br/>project_id, employee_id]\n",
" B[employee<br/>employee_id, experience_years, name]\n",
" C[列最小化<br/>employee → employee_id, experience_years のみ]\n",
" D[merge<br/>on employee_id]\n",
" E[\"groupby project_id<br/>mean(experience_years)\"]\n",
" F[\"列名変更 + round(2)<br/>average_years\"]\n",
" G[出力<br/>project_id, average_years]\n",
"\n",
" B --> C\n",
" A --> D\n",
" C --> D\n",
" D --> E\n",
" E --> F\n",
" F --> G\n",
"```\n",
"\n",
"この形なら、LeetCode の pandas 版「Project Employees I」にそのままコピペ投入できるはずです。\n",
"\n",
"結論から言うと、\n",
"\n",
"* **計算量レベルではほぼ限界まで来ていて、大きな高速化は期待しにくい**\n",
"* ただし、**merge → map に変えることで、メモリとわずかな実行時間改善の余地はある**\n",
"* 特にメモリ使用量（Beats 18%）を少しでも良くしたいなら、「結合の仕方」を軽くするのがポイント\n",
"\n",
"という感じです。\n",
"\n",
"---\n",
"\n",
"## 1) いまのアプローチの整理\n",
"\n",
"あなたの現行コード（要約）は：\n",
"\n",
"```python\n",
"emp_exp = employee[[\"employee_id\", \"experience_years\"]]\n",
"\n",
"merged = project.merge(emp_exp, on=\"employee_id\", how=\"left\")\n",
"\n",
"out = (\n",
" merged\n",
" .groupby(\"project_id\", as_index=False)[\"experience_years\"]\n",
" .mean()\n",
")\n",
"\n",
"out = out.rename(columns={\"experience_years\": \"average_years\"})\n",
"out[\"average_years\"] = out[\"average_years\"].round(2)\n",
"```\n",
"\n",
"やっていることは完全に正しくて、アルゴリズム的にも\n",
"\n",
"* 結合: O(N + M)\n",
"* groupby: O(N)\n",
"\n",
"で、ここから**オーダーを変える改善はできません**。\n",
"\n",
"LeetCode の 283ms / Beats 62% という数字も、\n",
"環境ノイズを含めて「十分良い」側です。\n",
"\n",
"---\n",
"\n",
"## 2) 改善ポイント：`merge` → `map` で軽量化\n",
"\n",
"`project` 側にはすでに `employee_id` が入っているので、\n",
"\n",
"> わざわざ `merge` で行を膨らませるのではなく、\n",
"> **`employee_id → experience_years` のマッピングを作って `map` する**\n",
"\n",
"方が、メモリ的には少し有利になり得ます。\n",
"\n",
"### 修正版コード（`map` ベース）\n",
"\n",
"```python\n",
"import pandas as pd\n",
"\n",
"def project_employees_i(project: pd.DataFrame, employee: pd.DataFrame) -> pd.DataFrame:\n",
" \"\"\"\n",
" 各プロジェクトごとの平均経験年数を計算する。\n",
"\n",
" Args:\n",
" project (pd.DataFrame): 列 ['project_id', 'employee_id']\n",
" employee (pd.DataFrame): 列 ['employee_id', 'name', 'experience_years']\n",
"\n",
" Returns:\n",
" pd.DataFrame: 列名と順序は ['project_id', 'average_years']\n",
" \"\"\"\n",
" # 1) employee_id -> experience_years のマップを作る（列最小化 + index 化）\n",
" emp_exp = employee.set_index(\"employee_id\")[\"experience_years\"]\n",
"\n",
" # 2) project に対応する経験年数列を付与（merge の代わりに map）\n",
" # 列最小化: project 側も必要な列だけ使う\n",
" proj = project[[\"project_id\", \"employee_id\"]].copy()\n",
" proj[\"experience_years\"] = proj[\"employee_id\"].map(emp_exp)\n",
"\n",
" # 3) 集約に不要な employee_id を削る（メモリ削減の一手）\n",
" proj = proj[[\"project_id\", \"experience_years\"]]\n",
"\n",
" # 4) プロジェクトごとに平均値を計算\n",
" out = (\n",
" proj\n",
" .groupby(\"project_id\", as_index=False)[\"experience_years\"]\n",
" .mean()\n",
" )\n",
"\n",
" # 5) 列名変更 + 小数第2位に丸める\n",
" out = out.rename(columns={\"experience_years\": \"average_years\"})\n",
" out[\"average_years\"] = out[\"average_years\"].round(2)\n",
"\n",
" return out\n",
"\n",
"Analyze Complexity\n",
"Runtime 283 ms\n",
"Beats 62.61%\n",
"Memory 68.14 MB\n",
"Beats 98.61%\n",
"\n",
"```\n",
"\n",
"### この書き方の狙い\n",
"\n",
"* `merge` は「左・右両方の列をフルで抱えた新しい DataFrame」を作るので、そのぶんメモリを食いやすい\n",
"* `map` は「Series の値を 1 列だけ投げ込む」だけなので、結合の中では比較的軽い\n",
"* `employee` 側は `set_index` して `Series` 化していて、\n",
" **キー→値** のマップとして最小限の形にしている\n",
"\n",
"大幅なメモリ削減にはならないかもしれませんが、\n",
"\n",
"* 一時的に作るオブジェクトがやや小さくなる\n",
"* 実行時間も微妙に改善する可能性はある\n",
"\n",
"ので、LeetCode のメモリ percentile を数ポイント押し上げられる可能性はあります。\n",
"\n",
"---\n",
"\n",
"## 3) どこまで改善を狙うべきか？\n",
"\n",
"正直なところ、今回の問題は\n",
"\n",
"* 入力サイズもそれほど極端ではない\n",
"* アルゴリズムも「JOIN + GROUP BY」で頭打ち\n",
"\n",
"なので、**すでに「クエリとしては十分に最適な部類」** です。\n",
"\n",
"283ms / Beats 62% という数字は、少しコードをいじったくらいで\n",
"劇的に変わるものではないので、\n",
"\n",
"* 可読性\n",
"* 素直さ（変なトリックを使わない）\n",
"* バグの入りにくさ\n",
"\n",
"を優先しつつ、上の `map` 版のような「軽い改善」を入れておけばかなり良いラインだと思います。\n",
"\n",
"---\n",
"\n",
"もしこの後、\n",
"\n",
"* 別の LeetCode pandas 問題\n",
"* もっと重いグループ演算（複数条件 / 上位 k / window関数っぽい処理）\n",
"\n",
"などが出てきたら、そこは **`groupby.transform` / `rank` / `merge` 戦略**を総合的に組み立てる練習ネタにできます。\n"
]
}
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