{
 "cells": [
  {
   "cell_type": "markdown",
   "id": "70f93a61",
   "metadata": {},
   "source": [
    "## Тема “Вычисления с помощью Numpy”"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "870fedf1",
   "metadata": {},
   "source": [
    "###### Задание 1\n",
    "Импортируйте библиотеку Numpy и дайте ей псевдоним np.\n",
    "Создайте массив Numpy под названием a размером 5x2, то есть состоящий из 5 строк и 2 столбцов. Первый столбец должен содержать числа 1, 2, 3, 3, 1, а второй - числа 6, 8, 11, 10, 7. Будем считать, что каждый столбец - это признак, а строка - наблюдение. Затем найдите среднее значение по каждому признаку, используя метод mean массива Numpy. Результат запишите в массив mean_a, в нем должно быть 2 элемента.\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 249,
   "id": "6b380c8a",
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 250,
   "id": "e456de05",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[ 1,  6],\n",
       "       [ 2,  8],\n",
       "       [ 3, 11],\n",
       "       [ 3, 10],\n",
       "       [ 1,  7]])"
      ]
     },
     "execution_count": 250,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a = np.array([[1, 2, 3, 3, 1],[ 6, 8, 11, 10, 7]]).T\n",
    "a"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 251,
   "id": "a03ab87d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([2. , 8.4])"
      ]
     },
     "execution_count": 251,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "mean_a = np.mean(a, axis=0)\n",
    "mean_a"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1055a5cc",
   "metadata": {},
   "source": [
    "###### Задание 2\n",
    "Вычислите массив a_centered, отняв от значений массива “а” средние значения соответствующих признаков, содержащиеся в массиве mean_a. Вычисление должно производиться в одно действие. Получившийся массив должен иметь размер 5x2"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 252,
   "id": "f8e2828c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([[-1. , -2.4],\n",
       "       [ 0. , -0.4],\n",
       "       [ 1. ,  2.6],\n",
       "       [ 1. ,  1.6],\n",
       "       [-1. , -1.4]])"
      ]
     },
     "execution_count": 252,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a_centered = np.vstack((a[:,0]-mean_a[0], a[:,1]-mean_a[1])).T\n",
    "a_centered"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 253,
   "id": "433c0d3c",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "(5, 2)"
      ]
     },
     "execution_count": 253,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a_centered.shape"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "1b41f215",
   "metadata": {},
   "source": [
    "###### Задание 3\n",
    "Найдите скалярное произведение столбцов массива a_centered. В результате должна получиться величина a_centered_sp. Затем поделите a_centered_sp на N-1, где N - число наблюдений."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 254,
   "id": "5b9124f0",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "8.0"
      ]
     },
     "execution_count": 254,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a_centered_sp = a_centered[:,0] @ a_centered[:,1]\n",
    "a_centered_sp"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 255,
   "id": "09840068",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.0"
      ]
     },
     "execution_count": 255,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "a_centered_sp / (a.shape[0] - 1)"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "eb344eb7",
   "metadata": {},
   "source": [
    "###### Задание 4**\n",
    "Число, которое мы получили в конце задания 3 является ковариацией двух признаков, содержащихся в массиве “а”. В задании 4 мы делили сумму произведений центрированных признаков на N-1, а не на N, поэтому полученная нами величина является несмещенной оценкой ковариации."
   ]
  },
  {
   "cell_type": "markdown",
   "id": "7c02908a",
   "metadata": {},
   "source": [
    "В этом задании проверьте получившееся число, вычислив ковариацию еще одним способом - с помощью функции np.cov. В качестве аргумента m функция np.cov должна принимать транспонированный массив “a”. В получившейся ковариационной матрице (массив Numpy размером 2x2) искомое значение ковариации будет равно элементу в строке с индексом 0 и столбце с индексом "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 256,
   "id": "d478c468",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/plain": [
       "2.0"
      ]
     },
     "execution_count": 256,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "np.cov(a.T)[0][1]"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "b0b213c8",
   "metadata": {},
   "source": [
    "## Тема “Работа с данными в Pandas”"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "101bae0d",
   "metadata": {},
   "source": [
    "###### Задание 1\n",
    "Импортируйте библиотеку Pandas и дайте ей псевдоним pd. Создайте датафрейм authors со столбцами author_id и author_name, в которых соответственно содержатся данные: [1, 2, 3] и ['Тургенев', 'Чехов', 'Островский'].\n",
    "Затем создайте датафрейм book cо столбцами author_id, book_title и price, в которых соответственно содержатся данные:  \n",
    "[1, 1, 1, 2, 2, 3, 3],\n",
    "['Отцы и дети', 'Рудин', 'Дворянское гнездо', 'Толстый и тонкий', 'Дама с собачкой', 'Гроза', 'Таланты и поклонники'],\n",
    "[450, 300, 350, 500, 450, 370, 290]."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 257,
   "id": "ebe7d756",
   "metadata": {},
   "outputs": [],
   "source": [
    "import pandas as pd"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 258,
   "id": "36c112a2",
   "metadata": {},
   "outputs": [],
   "source": [
    "authors = pd.DataFrame({\n",
    "    \"author_id\": [1, 2, 3],\n",
    "    \"author_name\": ['Тургенев', 'Чехов', 'Островский']\n",
    "})\n",
    "book = pd.DataFrame({\n",
    "    \"author_id\": [1, 1, 1, 2, 2, 3, 3],\n",
    "    \"book_title\": ['Отцы и дети', 'Рудин', 'Дворянское гнездо', 'Толстый и тонкий', 'Дама с собачкой', 'Гроза', 'Таланты и поклонники'],\n",
    "    \"price\": [450, 300, 350, 500, 450, 370, 290]\n",
    "})"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "23d8fcee",
   "metadata": {},
   "source": [
    "###### Задание 2\n",
    "Получите датафрейм authors_price, соединив датафреймы authors и books по полю author_id."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 259,
   "id": "1bf02209",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "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>author_id</th>\n",
       "      <th>author_name</th>\n",
       "      <th>book_title</th>\n",
       "      <th>price</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Отцы и дети</td>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Рудин</td>\n",
       "      <td>300</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Дворянское гнездо</td>\n",
       "      <td>350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Толстый и тонкий</td>\n",
       "      <td>500</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Дама с собачкой</td>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>5</th>\n",
       "      <td>3</td>\n",
       "      <td>Островский</td>\n",
       "      <td>Гроза</td>\n",
       "      <td>370</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>6</th>\n",
       "      <td>3</td>\n",
       "      <td>Островский</td>\n",
       "      <td>Таланты и поклонники</td>\n",
       "      <td>290</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "   author_id author_name            book_title  price\n",
       "0          1    Тургенев           Отцы и дети    450\n",
       "1          1    Тургенев                 Рудин    300\n",
       "2          1    Тургенев     Дворянское гнездо    350\n",
       "3          2       Чехов      Толстый и тонкий    500\n",
       "4          2       Чехов       Дама с собачкой    450\n",
       "5          3  Островский                 Гроза    370\n",
       "6          3  Островский  Таланты и поклонники    290"
      ]
     },
     "execution_count": 259,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "authors_price = pd.merge(authors, book, on='author_id', how='inner')\n",
    "authors_price"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "c2eaa126",
   "metadata": {},
   "source": [
    "###### Задание 3\n",
    "Создайте датафрейм top5, в котором содержатся строки из authors_price с пятью самыми дорогими книгами."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 260,
   "id": "52d7df3e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "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>author_id</th>\n",
       "      <th>author_name</th>\n",
       "      <th>book_title</th>\n",
       "      <th>price</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Дворянское гнездо</td>\n",
       "      <td>350</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>5</th>\n",
       "      <td>3</td>\n",
       "      <td>Островский</td>\n",
       "      <td>Гроза</td>\n",
       "      <td>370</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Отцы и дети</td>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Дама с собачкой</td>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Толстый и тонкий</td>\n",
       "      <td>500</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "   author_id author_name         book_title  price\n",
       "2          1    Тургенев  Дворянское гнездо    350\n",
       "5          3  Островский              Гроза    370\n",
       "0          1    Тургенев        Отцы и дети    450\n",
       "4          2       Чехов    Дама с собачкой    450\n",
       "3          2       Чехов   Толстый и тонкий    500"
      ]
     },
     "execution_count": 260,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "top5  = (authors_price.sort_values('price')).tail(5)\n",
    "top5"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "d3beda82",
   "metadata": {},
   "source": [
    "###### Задание 4\n",
    "Создайте датафрейм authors_stat на основе информации из authors_price. В датафрейме authors_stat должны быть четыре столбца:\n",
    "author_name, min_price, max_price и mean_price,\n",
    "в которых должны содержаться соответственно имя автора, минимальная, максимальная и средняя цена на книги этого автора"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 261,
   "id": "ec5b3bc8",
   "metadata": {},
   "outputs": [],
   "source": [
    "authors_stat = authors_price.groupby(\"author_name\")\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 262,
   "id": "e3688ea4",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "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>min_price</th>\n",
       "      <th>max_price</th>\n",
       "      <th>mean_price</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>author_name</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Островский</th>\n",
       "      <td>370</td>\n",
       "      <td>290</td>\n",
       "      <td>330.000000</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Тургенев</th>\n",
       "      <td>450</td>\n",
       "      <td>300</td>\n",
       "      <td>366.666667</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Чехов</th>\n",
       "      <td>500</td>\n",
       "      <td>450</td>\n",
       "      <td>475.000000</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "             min_price  max_price  mean_price\n",
       "author_name                                  \n",
       "Островский         370        290  330.000000\n",
       "Тургенев           450        300  366.666667\n",
       "Чехов              500        450  475.000000"
      ]
     },
     "execution_count": 262,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "authors_stat_price = pd.concat([authors_stat.agg({\"price\": \"max\" }), authors_stat.agg({\"price\": \"min\" }), authors_stat.agg({\"price\": \"mean\" })], axis=1, ignore_index=True);\n",
    "authors_stat_price.columns= [\"min_price\",\"max_price\",\"mean_price\"]\n",
    "authors_stat_price"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "f9b0606e",
   "metadata": {},
   "source": [
    "###### Задание 5**\n",
    "Создайте новый столбец в датафрейме authors_price под названием cover, в нем будут располагаться данные о том, какая обложка у данной книги - твердая или мягкая. В этот столбец поместите данные из следующего списка:\n",
    "['твердая', 'мягкая', 'мягкая', 'твердая', 'твердая', 'мягкая', 'мягкая'].\n",
    "Просмотрите документацию по функции pd.pivot_table с помощью вопросительного знака.\n",
    "\n",
    "Для каждого автора посчитайте суммарную стоимость книг в твердой и мягкой обложке. Используйте для этого функцию pd.pivot_table. При этом столбцы должны называться \"твердая\" и \"мягкая\", а индексами должны быть фамилии авторов. Пропущенные значения стоимостей заполните нулями, при необходимости загрузите библиотеку Numpy.\n",
    "Назовите полученный датасет book_info и сохраните его в формат pickle под названием \"book_info.pkl\". Затем загрузите из этого файла датафрейм и назовите его book_info2. Удостоверьтесь, что датафреймы book_info и book_info2 идентичны."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 263,
   "id": "dfb4e399",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
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       "    }\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>author_id</th>\n",
       "      <th>author_name</th>\n",
       "      <th>book_title</th>\n",
       "      <th>price</th>\n",
       "      <th>cover</th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Отцы и дети</td>\n",
       "      <td>450</td>\n",
       "      <td>твердая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Рудин</td>\n",
       "      <td>300</td>\n",
       "      <td>мягкая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>1</td>\n",
       "      <td>Тургенев</td>\n",
       "      <td>Дворянское гнездо</td>\n",
       "      <td>350</td>\n",
       "      <td>мягкая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Толстый и тонкий</td>\n",
       "      <td>500</td>\n",
       "      <td>твердая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>2</td>\n",
       "      <td>Чехов</td>\n",
       "      <td>Дама с собачкой</td>\n",
       "      <td>450</td>\n",
       "      <td>твердая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>5</th>\n",
       "      <td>3</td>\n",
       "      <td>Островский</td>\n",
       "      <td>Гроза</td>\n",
       "      <td>370</td>\n",
       "      <td>мягкая</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>6</th>\n",
       "      <td>3</td>\n",
       "      <td>Островский</td>\n",
       "      <td>Таланты и поклонники</td>\n",
       "      <td>290</td>\n",
       "      <td>мягкая</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "   author_id author_name            book_title  price    cover\n",
       "0          1    Тургенев           Отцы и дети    450  твердая\n",
       "1          1    Тургенев                 Рудин    300   мягкая\n",
       "2          1    Тургенев     Дворянское гнездо    350   мягкая\n",
       "3          2       Чехов      Толстый и тонкий    500  твердая\n",
       "4          2       Чехов       Дама с собачкой    450  твердая\n",
       "5          3  Островский                 Гроза    370   мягкая\n",
       "6          3  Островский  Таланты и поклонники    290   мягкая"
      ]
     },
     "execution_count": 263,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "authors_price[\"cover\"] = ['твердая', 'мягкая', 'мягкая', 'твердая', 'твердая', 'мягкая', 'мягкая']\n",
    "authors_price"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 264,
   "id": "f631cfe7",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
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       "\n",
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       "\n",
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       "        text-align: right;\n",
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       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th>price</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>author_name</th>\n",
       "      <th>cover</th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Островский</th>\n",
       "      <th>мягкая</th>\n",
       "      <td>660</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th rowspan=\"2\" valign=\"top\">Тургенев</th>\n",
       "      <th>мягкая</th>\n",
       "      <td>650</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>твердая</th>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Чехов</th>\n",
       "      <th>твердая</th>\n",
       "      <td>950</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "                     price\n",
       "author_name cover         \n",
       "Островский  мягкая     660\n",
       "Тургенев    мягкая     650\n",
       "            твердая    450\n",
       "Чехов       твердая    950"
      ]
     },
     "execution_count": 264,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "book_info = pd.pivot_table(authors_price, index = [\"author_name\",\"cover\"],  values=[\"price\"], aggfunc=np.sum, fill_value=0)\n",
    "book_info"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 271,
   "id": "d265050d",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
       "    .dataframe tbody tr th:only-of-type {\n",
       "        vertical-align: middle;\n",
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       "\n",
       "    .dataframe tbody tr th {\n",
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       "\n",
       "    .dataframe thead th {\n",
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       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th>price</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>author_name</th>\n",
       "      <th>cover</th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>Островский</th>\n",
       "      <th>мягкая</th>\n",
       "      <td>660</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th rowspan=\"2\" valign=\"top\">Тургенев</th>\n",
       "      <th>мягкая</th>\n",
       "      <td>650</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>твердая</th>\n",
       "      <td>450</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>Чехов</th>\n",
       "      <th>твердая</th>\n",
       "      <td>950</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "                     price\n",
       "author_name cover         \n",
       "Островский  мягкая     660\n",
       "Тургенев    мягкая     650\n",
       "            твердая    450\n",
       "Чехов       твердая    950"
      ]
     },
     "execution_count": 271,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "book_info.to_pickle(\"book_info.pkl\")\n",
    "book_info2 = pd.read_pickle(\"book_info.pkl\")\n",
    "book_info2"
   ]
  },
  {
   "cell_type": "markdown",
   "id": "3228ed7f",
   "metadata": {},
   "source": [
    "Найдем различие"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 272,
   "id": "f2e5762e",
   "metadata": {},
   "outputs": [
    {
     "data": {
      "text/html": [
       "<div>\n",
       "<style scoped>\n",
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       "\n",
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       "        text-align: right;\n",
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       "</style>\n",
       "<table border=\"1\" class=\"dataframe\">\n",
       "  <thead>\n",
       "    <tr style=\"text-align: right;\">\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th>price</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>author_name</th>\n",
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       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "Empty DataFrame\n",
       "Columns: [price]\n",
       "Index: []"
      ]
     },
     "execution_count": 272,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "book_info2[ ~book_info2.isin(book_info)].dropna()"
   ]
  }
 ],
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