395 lines
19 KiB
Plaintext
395 lines
19 KiB
Plaintext
{
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 1,
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"metadata": {
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"ExecuteTime": {
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"start_time": "2023-05-04T14:47:14.249433Z",
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"end_time": "2023-05-04T14:47:14.815684Z"
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}
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},
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"outputs": [],
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"source": [
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"import os\n",
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"\n",
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"from IPython.core.interactiveshell import InteractiveShell #执行该代码可以使得当前nb支持多输出\n",
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"InteractiveShell.ast_node_interactivity = \"all\" \n",
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"import numpy as np\n",
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"import pandas as pd \n",
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"pd.options.display.max_rows = 8 "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**作业**\n",
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"综合运用stack/unstack，pivot/melt"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"metadata": {
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"scrolled": true,
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"ExecuteTime": {
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"start_time": "2023-05-04T14:47:14.807387Z",
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"end_time": "2023-05-04T14:47:14.828436Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"item realgdp infl unemp\n",
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"date \n",
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"1959-03-31 2710.349 0.00 5.8\n",
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"1959-06-30 2778.801 2.34 5.1\n",
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"1959-09-30 2775.488 2.74 5.3\n",
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"1959-12-31 2785.204 0.27 5.6\n",
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"... ... ... ...\n",
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"2008-12-31 13141.920 -8.79 6.9\n",
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"2009-03-31 12925.410 0.94 8.1\n",
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"2009-06-30 12901.504 3.37 9.2\n",
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"2009-09-30 12990.341 3.56 9.6\n",
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"\n",
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"[203 rows x 3 columns]\n"
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]
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}
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],
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"source": [
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"#有可能是版本问题，需要加入.strftime('%Y-%m-%d')\n",
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"data = pd.read_csv('../examples/macrodata.csv')\n",
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"data.index = pd.PeriodIndex(year=data.year, quarter=data.quarter,name='date').to_timestamp('D', 'end').strftime('%Y-%m-%d')\n",
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"data = data.reindex(columns=pd.Index(['realgdp', 'infl', 'unemp'], name='item'))\n",
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"print(data)\n"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**作业1:**\n",
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"data通过stack/melt获得ldata"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {
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"scrolled": true,
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"ExecuteTime": {
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"start_time": "2023-05-04T14:47:17.592602Z",
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"end_time": "2023-05-04T14:47:17.608357Z"
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}
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},
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"outputs": [
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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},
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"#请输入data使用stack实现的代码（不得使用melt）\n",
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"data = data.reset_index()\n",
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"\n",
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"ldata= data.set_index('date').stack().reset_index()\n",
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"ldata.columns = ['date', 'item', 'value']\n",
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"ldata.head()\n",
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"\n",
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"# 请输出data使用melt实现的代码（不得使用stack）\n",
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"#data = data.reset_index()\n",
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"#ldata = pd.melt(data, id_vars=['date'], value_vars=['realgdp', 'infl', 'unemp'], var_name='item', value_name='value')\n",
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"ldata.head()"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**作业2**\n",
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"\n",
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"ldata通过pivot/unstack恢复data"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 38,
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"metadata": {
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"scrolled": true,
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"ExecuteTime": {
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"start_time": "2023-05-04T00:56:26.823785Z",
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"end_time": "2023-05-04T00:56:26.852233Z"
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}
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},
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"outputs": [
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 38,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"#请输入ldata使用pivot实现的代码（不得使用unstack）\n",
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"#pivoted = ldata.pivot(index='date', columns='item', values='value')\n",
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"#pivoted.head() #同data.head()\n",
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"\n",
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"#请输入ldata使用unstack实现的代码（不得使用pivot）\n",
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"# 将date列转换为索引，并将item列转换为列标签\n",
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"pivoted = ldata.set_index(['date', 'item'])['value'].unstack()\n",
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"# 重命名列标签级别的名称\n",
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"pivoted.columns.name = None\n",
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"\n",
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"pivoted.head() #同data.head() "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"**作业3：**\n",
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"\n",
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"参照“经营数据分析.xlsx”，求解下列问题，使得输出与结果相同"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"metadata": {
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"ExecuteTime": {
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||
"start_time": "2023-05-04T15:44:58.159977Z",
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"end_time": "2023-05-04T15:44:58.196601Z"
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}
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},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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" 方案 设备投资 单件成本 年销售量 销售单价 年收益\n",
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"0 方案1 1500000 1700 8000 2900 8100000\n",
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"1 方案2 2000000 1550 8000 2900 8800000\n",
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"2 方案3 2500000 1400 8000 2900 9500000\n",
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"年收益最高的方案是： 方案3\n",
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" 方案 设备投资 单件成本 年销售量 销售单价 年收益 差值\n",
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"0 方案1 1500000 1700 8000 2900 8100000 1960000090000\n",
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"1 方案2 2000000 1550 8000 2900 8800000 740000022500\n",
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"2 方案3 2500000 1400 8000 2900 9500000 1000000000000\n",
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"差值最小的方案是： 方案2\n"
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]
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},
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 10,
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"metadata": {},
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"output_type": "execute_result"
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},
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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" 畅销 一般 滞销\n",
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"0 12900000 8100000 300000\n",
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"1 14200000 8800000 25000\n",
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"2 15500000 9500000 -250000\n"
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]
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},
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 10,
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"metadata": {},
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"output_type": "execute_result"
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},
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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" 畅销 一般 滞销 折中\n",
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"0 12900000 8100000 300000 8490000.0\n",
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"1 14200000 8800000 25000 9238750.0\n",
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"2 15500000 9500000 -250000 9987500.0\n",
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"折中方案最佳方案是： 方案3\n"
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]
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},
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{
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"data": {
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"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",
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"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>"
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},
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"execution_count": 10,
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"metadata": {},
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||
"output_type": "execute_result"
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},
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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" 畅销 一般 滞销 期望\n",
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"0 12900000 8100000 300000 6630000.0\n",
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"1 14200000 8800000 25000 7146250.0\n",
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"2 15500000 9500000 -250000 7662500.0\n",
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"期望值最佳方案是： 方案3\n"
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]
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}
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],
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"source": [
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"## 单目标求解\n",
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"# 年收益=年销售量*(销售单价-单件成本)-设备投资\n",
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"# 求年收益最佳方案？\n",
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"df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
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"# 计算年收益\n",
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"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",
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"# 按照年收益找最大值，取方案列\n",
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"best_plan = df.loc[df['年收益'].idxmax(), '方案']\n",
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"print(df)\n",
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"print('年收益最高的方案是：', best_plan)\n",
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"\n",
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"\n",
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"#\n",
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"# ## 多目标求解 ----\n",
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"# 期望值=[min(设备投资),min(单件成本),max(年销售量),max(销售单价),\n",
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"# max(年收益)];\n",
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"# 差值=每个方案中，各项数据与期望值的之差的平方和\n",
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"# 求差值最佳方案？\n",
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"df = pd.read_excel(\"../examples/经营数据分析.xlsx\")\n",
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"# 计算年收益\n",
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"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",
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"# 计算期望值\n",
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"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",
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"# 转换为Series以匹配 df\n",
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"expected_values = pd.Series(expected_values, index=df.columns[1:])\n",
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"# 计算差值\n",
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"df['差值'] = np.sum((df - expected_values) ** 2, axis=1)\n",
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"# 选择差值最小的方案\n",
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"best_plan = df.loc[df['差值'].idxmin(), '方案']\n",
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||
"# 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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