{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "ExecuteTime": {
     "start_time": "2023-05-04T14:47:14.249433Z",
     "end_time": "2023-05-04T14:47:14.815684Z"
    }
   },
   "outputs": [],
   "source": [
    "import os\n",
    "\n",
    "from IPython.core.interactiveshell import InteractiveShell  #执行该代码可以使得当前nb支持多输出\n",
    "InteractiveShell.ast_node_interactivity = \"all\" \n",
    "import numpy as np\n",
    "import pandas as pd \n",
    "pd.options.display.max_rows = 8  "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**作业**\n",
    "综合运用stack/unstack，pivot/melt"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "scrolled": true,
    "ExecuteTime": {
     "start_time": "2023-05-04T14:47:14.807387Z",
     "end_time": "2023-05-04T14:47:14.828436Z"
    }
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "item          realgdp  infl  unemp\n",
      "date                              \n",
      "1959-03-31   2710.349  0.00    5.8\n",
      "1959-06-30   2778.801  2.34    5.1\n",
      "1959-09-30   2775.488  2.74    5.3\n",
      "1959-12-31   2785.204  0.27    5.6\n",
      "...               ...   ...    ...\n",
      "2008-12-31  13141.920 -8.79    6.9\n",
      "2009-03-31  12925.410  0.94    8.1\n",
      "2009-06-30  12901.504  3.37    9.2\n",
      "2009-09-30  12990.341  3.56    9.6\n",
      "\n",
      "[203 rows x 3 columns]\n"
     ]
    }
   ],
   "source": [
    "#有可能是版本问题，需要加入.strftime('%Y-%m-%d')\n",
    "data = pd.read_csv('../examples/macrodata.csv')\n",
    "data.index = pd.PeriodIndex(year=data.year, quarter=data.quarter,name='date').to_timestamp('D', 'end').strftime('%Y-%m-%d')\n",
    "data = data.reindex(columns=pd.Index(['realgdp', 'infl', 'unemp'], name='item'))\n",
    "print(data)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**作业1:**\n",
    "data通过stack/melt获得ldata"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "scrolled": true,
    "ExecuteTime": {
     "start_time": "2023-05-04T14:47:17.592602Z",
     "end_time": "2023-05-04T14:47:17.608357Z"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": "         date     item     value\n0  1959-03-31  realgdp  2710.349\n1  1959-03-31     infl     0.000\n2  1959-03-31    unemp     5.800\n3  1959-06-30  realgdp  2778.801\n4  1959-06-30     infl     2.340",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>date</th>\n      <th>item</th>\n      <th>value</th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>0</th>\n      <td>1959-03-31</td>\n      <td>realgdp</td>\n      <td>2710.349</td>\n    </tr>\n    <tr>\n      <th>1</th>\n      <td>1959-03-31</td>\n      <td>infl</td>\n      <td>0.000</td>\n    </tr>\n    <tr>\n      <th>2</th>\n      <td>1959-03-31</td>\n      <td>unemp</td>\n      <td>5.800</td>\n    </tr>\n    <tr>\n      <th>3</th>\n      <td>1959-06-30</td>\n      <td>realgdp</td>\n      <td>2778.801</td>\n    </tr>\n    <tr>\n      <th>4</th>\n      <td>1959-06-30</td>\n      <td>infl</td>\n      <td>2.340</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "data": {
      "text/plain": "         date     item     value\n0  1959-03-31  realgdp  2710.349\n1  1959-03-31     infl     0.000\n2  1959-03-31    unemp     5.800\n3  1959-06-30  realgdp  2778.801\n4  1959-06-30     infl     2.340",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>date</th>\n      <th>item</th>\n      <th>value</th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>0</th>\n      <td>1959-03-31</td>\n      <td>realgdp</td>\n      <td>2710.349</td>\n    </tr>\n    <tr>\n      <th>1</th>\n      <td>1959-03-31</td>\n      <td>infl</td>\n      <td>0.000</td>\n    </tr>\n    <tr>\n      <th>2</th>\n      <td>1959-03-31</td>\n      <td>unemp</td>\n      <td>5.800</td>\n    </tr>\n    <tr>\n      <th>3</th>\n      <td>1959-06-30</td>\n      <td>realgdp</td>\n      <td>2778.801</td>\n    </tr>\n    <tr>\n      <th>4</th>\n      <td>1959-06-30</td>\n      <td>infl</td>\n      <td>2.340</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 3,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#请输入data使用stack实现的代码（不得使用melt）\n",
    "data = data.reset_index()\n",
    "\n",
    "ldata= data.set_index('date').stack().reset_index()\n",
    "ldata.columns = ['date', 'item', 'value']\n",
    "ldata.head()\n",
    "\n",
    "# 请输出data使用melt实现的代码（不得使用stack）\n",
    "#data = data.reset_index()\n",
    "#ldata = pd.melt(data, id_vars=['date'], value_vars=['realgdp', 'infl', 'unemp'], var_name='item', value_name='value')\n",
    "ldata.head()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**作业2**\n",
    "\n",
    "ldata通过pivot/unstack恢复data"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 38,
   "metadata": {
    "scrolled": true,
    "ExecuteTime": {
     "start_time": "2023-05-04T00:56:26.823785Z",
     "end_time": "2023-05-04T00:56:26.852233Z"
    }
   },
   "outputs": [
    {
     "data": {
      "text/plain": "            infl   realgdp  unemp\ndate                             \n1959-03-31  0.00  2710.349    5.8\n1959-06-30  2.34  2778.801    5.1\n1959-09-30  2.74  2775.488    5.3\n1959-12-31  0.27  2785.204    5.6\n1960-03-31  2.31  2847.699    5.2",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>infl</th>\n      <th>realgdp</th>\n      <th>unemp</th>\n    </tr>\n    <tr>\n      <th>date</th>\n      <th></th>\n      <th></th>\n      <th></th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>1959-03-31</th>\n      <td>0.00</td>\n      <td>2710.349</td>\n      <td>5.8</td>\n    </tr>\n    <tr>\n      <th>1959-06-30</th>\n      <td>2.34</td>\n      <td>2778.801</td>\n      <td>5.1</td>\n    </tr>\n    <tr>\n      <th>1959-09-30</th>\n      <td>2.74</td>\n      <td>2775.488</td>\n      <td>5.3</td>\n    </tr>\n    <tr>\n      <th>1959-12-31</th>\n      <td>0.27</td>\n      <td>2785.204</td>\n      <td>5.6</td>\n    </tr>\n    <tr>\n      <th>1960-03-31</th>\n      <td>2.31</td>\n      <td>2847.699</td>\n      <td>5.2</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 38,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "#请输入ldata使用pivot实现的代码（不得使用unstack）\n",
    "#pivoted = ldata.pivot(index='date', columns='item', values='value')\n",
    "#pivoted.head()  #同data.head()\n",
    "\n",
    "#请输入ldata使用unstack实现的代码（不得使用pivot）\n",
    "# 将date列转换为索引，并将item列转换为列标签\n",
    "pivoted = ldata.set_index(['date', 'item'])['value'].unstack()\n",
    "# 重命名列标签级别的名称\n",
    "pivoted.columns.name = None\n",
    "\n",
    "pivoted.head()  #同data.head() "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "**作业3：**\n",
    "\n",
    "参照“经营数据分析.xlsx”，求解下列问题，使得输出与结果相同"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "ExecuteTime": {
     "start_time": "2023-05-04T15:44:58.159977Z",
     "end_time": "2023-05-04T15:44:58.196601Z"
    }
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "    方案     设备投资  单件成本  年销售量  销售单价      年收益\n",
      "0  方案1  1500000  1700  8000  2900  8100000\n",
      "1  方案2  2000000  1550  8000  2900  8800000\n",
      "2  方案3  2500000  1400  8000  2900  9500000\n",
      "年收益最高的方案是： 方案3\n",
      "    方案     设备投资  单件成本  年销售量  销售单价      年收益            差值\n",
      "0  方案1  1500000  1700  8000  2900  8100000 1960000090000\n",
      "1  方案2  2000000  1550  8000  2900  8800000  740000022500\n",
      "2  方案3  2500000  1400  8000  2900  9500000 1000000000000\n",
      "差值最小的方案是： 方案2\n"
     ]
    },
    {
     "data": {
      "text/plain": "           畅销       一般      滞销\n方案                            \n方案1  12900000  8100000  300000\n方案2  14200000  8800000   25000\n方案3  15500000  9500000 -250000",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>畅销</th>\n      <th>一般</th>\n      <th>滞销</th>\n    </tr>\n    <tr>\n      <th>方案</th>\n      <th></th>\n      <th></th>\n      <th></th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>方案1</th>\n      <td>12900000</td>\n      <td>8100000</td>\n      <td>300000</td>\n    </tr>\n    <tr>\n      <th>方案2</th>\n      <td>14200000</td>\n      <td>8800000</td>\n      <td>25000</td>\n    </tr>\n    <tr>\n      <th>方案3</th>\n      <td>15500000</td>\n      <td>9500000</td>\n      <td>-250000</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "         畅销       一般      滞销\n",
      "0  12900000  8100000  300000\n",
      "1  14200000  8800000   25000\n",
      "2  15500000  9500000 -250000\n"
     ]
    },
    {
     "data": {
      "text/plain": "           畅销       一般      滞销        折中\n方案                                      \n方案1  12900000  8100000  300000 8490000.0\n方案2  14200000  8800000   25000 9238750.0\n方案3  15500000  9500000 -250000 9987500.0",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>畅销</th>\n      <th>一般</th>\n      <th>滞销</th>\n      <th>折中</th>\n    </tr>\n    <tr>\n      <th>方案</th>\n      <th></th>\n      <th></th>\n      <th></th>\n      <th></th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>方案1</th>\n      <td>12900000</td>\n      <td>8100000</td>\n      <td>300000</td>\n      <td>8490000.0</td>\n    </tr>\n    <tr>\n      <th>方案2</th>\n      <td>14200000</td>\n      <td>8800000</td>\n      <td>25000</td>\n      <td>9238750.0</td>\n    </tr>\n    <tr>\n      <th>方案3</th>\n      <td>15500000</td>\n      <td>9500000</td>\n      <td>-250000</td>\n      <td>9987500.0</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "         畅销       一般      滞销        折中\n",
      "0  12900000  8100000  300000 8490000.0\n",
      "1  14200000  8800000   25000 9238750.0\n",
      "2  15500000  9500000 -250000 9987500.0\n",
      "折中方案最佳方案是： 方案3\n"
     ]
    },
    {
     "data": {
      "text/plain": "           畅销       一般      滞销        期望\n方案                                      \n方案1  12900000  8100000  300000 6630000.0\n方案2  14200000  8800000   25000 7146250.0\n方案3  15500000  9500000 -250000 7662500.0",
      "text/html": "<div>\n<style scoped>\n    .dataframe tbody tr th:only-of-type {\n        vertical-align: middle;\n    }\n\n    .dataframe tbody tr th {\n        vertical-align: top;\n    }\n\n    .dataframe thead th {\n        text-align: right;\n    }\n</style>\n<table border=\"1\" class=\"dataframe\">\n  <thead>\n    <tr style=\"text-align: right;\">\n      <th></th>\n      <th>畅销</th>\n      <th>一般</th>\n      <th>滞销</th>\n      <th>期望</th>\n    </tr>\n    <tr>\n      <th>方案</th>\n      <th></th>\n      <th></th>\n      <th></th>\n      <th></th>\n    </tr>\n  </thead>\n  <tbody>\n    <tr>\n      <th>方案1</th>\n      <td>12900000</td>\n      <td>8100000</td>\n      <td>300000</td>\n      <td>6630000.0</td>\n    </tr>\n    <tr>\n      <th>方案2</th>\n      <td>14200000</td>\n      <td>8800000</td>\n      <td>25000</td>\n      <td>7146250.0</td>\n    </tr>\n    <tr>\n      <th>方案3</th>\n      <td>15500000</td>\n      <td>9500000</td>\n      <td>-250000</td>\n      <td>7662500.0</td>\n    </tr>\n  </tbody>\n</table>\n</div>"
     },
     "execution_count": 10,
     "metadata": {},
     "output_type": "execute_result"
    },
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "         畅销       一般      滞销        期望\n",
      "0  12900000  8100000  300000 6630000.0\n",
      "1  14200000  8800000   25000 7146250.0\n",
      "2  15500000  9500000 -250000 7662500.0\n",
      "期望值最佳方案是： 方案3\n"
     ]
    }
   ],
   "source": [
    "##  单目标求解\n",
    "# 年收益=年销售量*(销售单价-单件成本)-设备投资\n",
    "# 求年收益最佳方案？\n",
    "df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
    "# 计算年收益\n",
    "df['年收益'] = df['年销售量'] * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "# 按照年收益找最大值，取方案列\n",
    "best_plan = df.loc[df['年收益'].idxmax(), '方案']\n",
    "print(df)\n",
    "print('年收益最高的方案是：', best_plan)\n",
    "\n",
    "\n",
    "#\n",
    "# ## 多目标求解 ----\n",
    "# 期望值=[min(设备投资),min(单件成本),max(年销售量),max(销售单价),\n",
    "#     max(年收益)];\n",
    "# 差值=每个方案中，各项数据与期望值的之差的平方和\n",
    "# 求差值最佳方案？\n",
    "df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
    "# 计算年收益\n",
    "df['年收益'] = df['年销售量'] * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "# 计算期望值\n",
    "expected_values = [df['设备投资'].min(), df['单件成本'].min(), df['年销售量'].max(), df['销售单价'].max(), df['年收益'].max()]\n",
    "# 转换为Series以匹配 df\n",
    "expected_values = pd.Series(expected_values, index=df.columns[1:])\n",
    "# 计算差值\n",
    "df['差值'] = np.sum((df - expected_values) ** 2, axis=1)\n",
    "# 选择差值最小的方案\n",
    "best_plan = df.loc[df['差值'].idxmin(), '方案']\n",
    "# 禁用科学计数法打印\n",
    "pd.set_option('float_format','{:.0f}'.format)\n",
    "print(df)\n",
    "print('差值最小的方案是：', best_plan)\n",
    "\n",
    "\n",
    "# 不确定性决策分析 ----\n",
    "## 分析方法\n",
    "# PLm=pd.DataFrame();# 损益矩阵 ProfitLoss matrix\n",
    "# PLm['畅销']= 12000*(销售单价-单件成本)-设备投资;\n",
    "# PLm['一般']= 8000*(销售单价-单件成本)-设备投资;\n",
    "# PLm['滞销']= 1500*(销售单价-单件成本)-设备投资;\n",
    "# #\n",
    "# ## 分析原则----\n",
    "# # 乐观原则\n",
    "# lg=损益矩阵三种情况的最大值\n",
    "#\n",
    "# # 悲观原则\n",
    "# bg=损益矩阵三种情况的最小值\n",
    "#\n",
    "# # 折中原则\n",
    "# a=0.65  #65%的乐观概率\n",
    "# 折中方案= a*lg + (1-a)*bg;\n",
    "# 求折中最佳方案？\n",
    "df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
    "# 计算损益矩阵\n",
    "PLm = pd.DataFrame()\n",
    "PLm['畅销'] = 12000 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "PLm['一般'] = 8000 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "PLm['滞销'] = 1500 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "PLm.set_index(df['方案'])\n",
    "print(PLm)\n",
    "pd.set_option('float_format', '{:.1f}'.format)\n",
    "# 计算乐观、悲观和折中方案\n",
    "mids = []\n",
    "a = 0.65\n",
    "for index, row in PLm.iterrows():\n",
    "    mids.append(row.max() * a + (1-a) * row.min())\n",
    "PLm['折中'] = mids\n",
    "PLm.set_index(df['方案'])\n",
    "print(PLm)\n",
    "best_plan = df.loc[PLm['折中'].idxmax(), '方案']\n",
    "print('折中方案最佳方案是：',best_plan)\n",
    "\n",
    "# #  概率性决策分析 ----\n",
    "## 期望值法 ----\n",
    "probE=[0.1,0.65,0.25]; #畅销、一般、滞销的概率\n",
    "# 期望值=损益矩阵*probE\n",
    "# 求期望值最佳方案？\n",
    "df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
    "# 刷新PLm\n",
    "PLm = pd.DataFrame()\n",
    "PLm['畅销'] = 12000 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "PLm['一般'] = 8000 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "PLm['滞销'] = 1500 * (df['销售单价'] - df['单件成本']) - df['设备投资']\n",
    "\n",
    "# 利用numpy的dot乘法求出期望值\n",
    "expected_values = np.dot(PLm, probE)\n",
    "PLm['期望'] = expected_values\n",
    "PLm.set_index(df['方案'])\n",
    "\n",
    "# 设为保留一位小数\n",
    "print(PLm)\n",
    "best_plan = df.loc[PLm['期望'].idxmax(), '方案']\n",
    "print('期望值最佳方案是：',best_plan)\n"
   ]
  }
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