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denis-on 0fc98a6122 dg_les_2
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{
"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": [
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"</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": {
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"</style>\n",
"<table border=\"1\" class=\"dataframe\">\n",
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" <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": {
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"<table border=\"1\" class=\"dataframe\">\n",
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" <tr style=\"text-align: right;\">\n",
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" <th>min_price</th>\n",
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" <th>mean_price</th>\n",
" </tr>\n",
" <tr>\n",
" <th>author_name</th>\n",
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" <th></th>\n",
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" </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": {
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" <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",
" vertical-align: middle;\n",
" }\n",
"\n",
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" vertical-align: top;\n",
" }\n",
"\n",
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"</style>\n",
"<table border=\"1\" class=\"dataframe\">\n",
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" <th></th>\n",
" <th></th>\n",
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" </tr>\n",
" <tr>\n",
" <th>author_name</th>\n",
" <th>cover</th>\n",
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" <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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"<table border=\"1\" class=\"dataframe\">\n",
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" <th></th>\n",
" <th></th>\n",
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" </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",
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"</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",
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" .dataframe tbody tr th:only-of-type {\n",
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" </tr>\n",
" <tr>\n",
" <th>author_name</th>\n",
" <th>cover</th>\n",
" <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()"
]
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3 (ipykernel)",
"language": "python",
"name": "python3"
},
"language_info": {
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"version": 3
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"pygments_lexer": "ipython3",
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"nbformat": 4,
"nbformat_minor": 5
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