Exercises/PT-BR/Prova_A2_2017.ipynb

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   "source": [
    "A2 de Linguagens de Programação – 30/11/2017\n",
    "--\n",
    "Professor: Renato Rocha Souza  \n",
    "--"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Nome do Aluno(a):"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Entre com o seu nome completo: Aluno\n",
      "Olá, Aluno!\n",
      "Bom trabalho!\n"
     ]
    }
   ],
   "source": [
    "meu_nome = input('Entre com o seu nome completo: ')\n",
    "print('Olá, {}!'.format(meu_nome))\n",
    "print('Bom trabalho!')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "import numpy as np\n",
    "import pandas as pd\n",
    "import matplotlib.pyplot as plt\n",
    "import requests\n",
    "import random\n",
    "import re\n",
    "import string\n",
    "import time\n",
    "import pymysql\n",
    "import sqlite3\n",
    "import getpass\n",
    "import pandas.io.sql as psql\n",
    "import networkx as nx\n",
    "from IPython.display import clear_output\n",
    "%matplotlib inline"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão 1 (1,5 pontos)  \n",
    "\n",
    "\n",
    "Utilizando Python, crie um banco de dados no sqlite que guarde a seguinte estrutura, demonstrada no dataframe do Pandas:\n",
    "Use os tipos de dados mais adequados.  \n",
    "\n",
    "Veja:  \n",
    "https://sqlite.org/datatype3.html    \n",
    "https://sqlite.org/autoinc.html  \n",
    "https://www.tutorialspoint.com/sqlite/sqlite_update_query.htm  "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "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>Gender</th>\n",
       "      <th>Age</th>\n",
       "      <th>Salary before Stern</th>\n",
       "      <th>Expected salary</th>\n",
       "      <th>Candidate</th>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>id</th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "      <th></th>\n",
       "    </tr>\n",
       "  </thead>\n",
       "  <tbody>\n",
       "    <tr>\n",
       "      <th>0</th>\n",
       "      <td>Male</td>\n",
       "      <td>27</td>\n",
       "      <td>60000</td>\n",
       "      <td>110000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>1</th>\n",
       "      <td>Male</td>\n",
       "      <td>30</td>\n",
       "      <td>125000</td>\n",
       "      <td>125000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>2</th>\n",
       "      <td>Male</td>\n",
       "      <td>27</td>\n",
       "      <td>50000</td>\n",
       "      <td>120000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>3</th>\n",
       "      <td>Male</td>\n",
       "      <td>26</td>\n",
       "      <td>56000</td>\n",
       "      <td>100000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>4</th>\n",
       "      <td>Male</td>\n",
       "      <td>27</td>\n",
       "      <td>82000</td>\n",
       "      <td>100000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>5</th>\n",
       "      <td>Male</td>\n",
       "      <td>32</td>\n",
       "      <td>150000</td>\n",
       "      <td>150000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>6</th>\n",
       "      <td>Female</td>\n",
       "      <td>24</td>\n",
       "      <td>55000</td>\n",
       "      <td>100000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>7</th>\n",
       "      <td>Male</td>\n",
       "      <td>30</td>\n",
       "      <td>110000</td>\n",
       "      <td>150000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>8</th>\n",
       "      <td>Male</td>\n",
       "      <td>26</td>\n",
       "      <td>48000</td>\n",
       "      <td>80000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "    <tr>\n",
       "      <th>9</th>\n",
       "      <td>Male</td>\n",
       "      <td>26</td>\n",
       "      <td>120000</td>\n",
       "      <td>80000</td>\n",
       "      <td>Bush</td>\n",
       "    </tr>\n",
       "  </tbody>\n",
       "</table>\n",
       "</div>"
      ],
      "text/plain": [
       "    Gender  Age  Salary before Stern  Expected salary Candidate\n",
       "id                                                             \n",
       "0     Male   27                60000           110000      Bush\n",
       "1     Male   30               125000           125000      Bush\n",
       "2     Male   27                50000           120000      Bush\n",
       "3     Male   26                56000           100000      Bush\n",
       "4     Male   27                82000           100000      Bush\n",
       "5     Male   32               150000           150000      Bush\n",
       "6   Female   24                55000           100000      Bush\n",
       "7     Male   30               110000           150000      Bush\n",
       "8     Male   26                48000            80000      Bush\n",
       "9     Male   26               120000            80000      Bush"
      ]
     },
     "execution_count": 4,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "df_tf = pd.read_excel('../../Dados/votesurvey.xls')\n",
    "df_tf.index.name = 'id'\n",
    "df_tf.head(10)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "conn = sqlite3.connect('votesurvey.sqlite')\n",
    "cur = conn.cursor()\n",
    "\n",
    "# seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão 2 (1.5 ponto)  \n",
    "\n",
    "Conecte ao banco SQLite \"iris_database.sqlite\"e responda, demonstrando os comandos SQL apropriados:  \n",
    "\n",
    "a) Há quantas \"Species\" diferentes na tabela \"iris\"?  \n",
    "b) Quantos registros totais há na tabela \"iris\"?  \n",
    "\n",
    "Considerando a superfície das pétalas e sépalas como retangular, responda:  \n",
    "\n",
    "c) Qual espécie (\"Species\") contém, em média, a maior área de pétalas?  \n",
    "d) Qual espécie (\"Species\") contém, em média, a maior área total (pétalas e sépalas)?  \n",
    "\n",
    "(Por curiosidade, as espécies de Iris contém 3 pétalas e 3 sépalas)  "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "conn = sqlite3.connect('../../Dados/iris_database.sqlite')\n",
    "cur = conn.cursor()\n",
    "\n",
    "# seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão 3 (3.0 pontos)\n",
    "\n",
    "Conecte ao banco Python/Dados/chinook.sqlite e responda:\n",
    "\n",
    "a) Utilizando um comando SQL e as tabelas \"invoices\", \"invoice_items\", \"customers\" e \"tracks\", mostre quais as faixas compradas por clientes que têm o primeiro nome começando com a letra \"M\".  \n",
    "b) Quais clientes contêm a letra \"o\" tanto no nome como no sobrenome?  \n",
    "c) Crie um histograma de faixas (\"tracks\") mais compradas, usando o matplotlib.  \n",
    "d) Qual o cliente (\"customers\") que teve gasto médio em compras (\"invoices\") o valor mais próximo do valor médio do total de compras de todos os clientes?  \n",
    "e) Escreva um script em Python + SQL para construir um dicionário tendo os nomes dos artistas (\"artists\") como chaves e uma lista dos álbuns (\"albuns\") respectivos como valores."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "conn = sqlite3.connect('../../Dados/chinook.sqlite')\n",
    "cur = conn.cursor()\n",
    "\n",
    "# seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão 4 (1.5 pontos) \n",
    "\n",
    "Ajude as funções abaixo (escritas em R) a se tornarem mais eficientes, convertendo-as para Python. Faça o script o mais próximo possível do original, sem procurar por programas semelhantes na web!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "cpf_dig_controle <- function(y){\n",
    "    v1 <- 0L\n",
    "    v2 <- 0L\n",
    "    z <- as.integer(rev(strsplit(as.character(y), \"\")[[1]]))\n",
    "    for(i in seq_along(z)){\n",
    "        v1 <- v1 + z[i]*(9 - (i %% 10))\n",
    "        v2 <- v2 + z[i]*(9 - ((i + 1) %% 10))\n",
    "    }\n",
    "    v1 <- (v1 %% 11) %% 10\n",
    "    v2 <- v2 + v1*9\n",
    "    v2 <- (v2 %% 11) %% 10\n",
    "    c(v1, v2)\n",
    "}\n",
    "\n",
    "cpf_gerar <- function(x){\n",
    "    g <- function(y, v){\n",
    "        z <- strsplit(as.character(y), \"\")[[1]]\n",
    "        z <- as.integer(c(z, v))\n",
    "        res <- sprintf(fmt = \"%d%d%d.%d%d%d.%d%d%d-%02d\",\n",
    "                             z[1], z[2], z[3], z[4], z[5],\n",
    "                             z[6], z[7], z[8], z[9], z[10])\n",
    "        res\n",
    "    }\n",
    "    v <- lapply(x, cpf_dig_controle)\n",
    "    result <- sapply(seq_along(x), function(i) g(x[i], v[[i]]))\n",
    "    result\n",
    "}\n",
    "\n",
    "cpf_validar <- function(x){\n",
    "    f <- function(y){\n",
    "        z <- unlist(strsplit(y, \"\\\\.\"))\n",
    "        v <- substr(z[3], 3, 6)\n",
    "        v <- as.integer(unlist(strsplit(v, \"-\")))\n",
    "        z[3] <- substr(z[3], 1, 2)\n",
    "        z <- as.integer(paste(z, collapse = \"\"))\n",
    "        list(x = z, v = v)\n",
    "    }\n",
    "    result <- lapply(x, f)\n",
    "    valido <- lapply(result, function(r) unlist(lapply(r$x, cpf_dig_controle)))\n",
    "    valido <- sapply(seq_along(valido), function(i)\n",
    "                    result[[i]]$v[1] == valido[[i]][1] & result[[i]]$v[2] == valido[[i]][2]\n",
    "                )\n",
    "    valido\n",
    "}\n",
    "\n",
    "cpf <- cpf_gerar(c(12345678, 87654321))\n",
    "cpf\n",
    "#[1] \"123.456.786-02\" \"876.543.210-07\"\n",
    "\n",
    "cpf_validar(cpf)\n",
    "#[1] TRUE TRUE"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "#seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão 5 (1.5 pontos) \n",
    "\n",
    "Por vezes precisamos passar mensagens secretas aos nossos colegas de classe. Converta o código abaixo para Python:    "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "#include <stdio.h>\n",
    "#include <stdlib.h>\n",
    "#include <string.h>\n",
    " \n",
    "#define caesar(x) rot(13, x)\n",
    "#define decaesar(x) rot(13, x)\n",
    "#define decrypt_rot(x, y) rot((26-x), y)\n",
    " \n",
    "void rot(int c, char *str)\n",
    "{\n",
    "    int l = strlen(str);\n",
    "    const char *alpha[2] = { \"abcdefghijklmnopqrstuvwxyz\", \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"};\n",
    " \n",
    "    int i;\n",
    "    for (i = 0; i < l; i++)\n",
    "    {\n",
    "        if (!isalpha(str[i]))\n",
    "            continue;\n",
    " \n",
    "        str[i] = alpha[isupper(str[i])][((int)(tolower(str[i])-'a')+c)%26];\n",
    "    }\n",
    "}\n",
    " \n",
    " \n",
    "int main()\n",
    "{\n",
    "    char str[] = \"This is a top secret text message!\";\n",
    " \n",
    "    printf(\"Original: %s\\n\", str);\n",
    "    caesar(str);\n",
    "    printf(\"Encrypted: %s\\n\", str);\n",
    "    decaesar(str);\n",
    "    printf(\"Decrypted: %s\\n\", str);\n",
    " \n",
    "    return 0;\n",
    "}"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [],
   "source": [
    "#seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Questão *\"bônus\"* (0,5 pontos)  \n",
    "\n",
    "Converta o código abaixo para Python:  "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "(defun divisible-by (number list)\n",
    "  \"return true if divisible by an element in the list\"\n",
    "  (loop for i in list\n",
    "       thereis (= (mod number i) 0)))\n",
    "\n",
    "(defun prime-list-generator (current max-value prime-list)\n",
    "  \"business end of p10\"\n",
    "  (if (= current max-value)\n",
    "      prime-list\n",
    "      (if (divisible-by current prime-list)\n",
    "      (prime-list-generator (+ current 1) max-value prime-list)\n",
    "      (prime-list-generator (+ current 1) max-value (nconc prime-list (list current))))))\n",
    "\n",
    "(defun p10 (n)\n",
    "  \"Find the sum of all the primes below n\"\n",
    "  (time (reduce #'+ (prime-list-generator 2 n '()))))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {},
   "outputs": [],
   "source": [
    "#seu código aqui"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Auto-avaliação (1 ponto)  \n",
    "Escreva sua auto-avaliação (mínimo de 3 linhas) e atribua a esta um valor entre 0 e 1, levando em consideração o aproveitamento na disciplina.  "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Edite esta célula"
   ]
  }
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