Exercise (1).ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem Alphabet Slices\n",
    "* Store the first ten letters of the alphabet in a list.\n",
    "* Use a slice to print out the first three letters of the alphabet.\n",
    "* Use a slice to print out any three letters from the middle of your list."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve Alphabet slices here. \n",
    "\n",
    "alphabet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']\n",
    "\n",
    "\n",
    "print(\"First three letters:\", alphabet[:3])\n",
    "\n",
    "\n",
    "print(\"Middle three letters:\", alphabet[4:7])\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem:  Covert all the rapper names to title case and save them into a new different list. \n",
    "example: lil wayne becomes Lil Wayne"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve rapper names here\n",
    "rappers = ['lil wayne', 'nicki minaj', 'drake']\n",
    "\n",
    "\n",
    "rappers = ['lil wayne', 'nicki minaj', 'drake']\n",
    "\n",
    "\n",
    "title_case_rappers = [rapper.title() for rapper in rappers]\n",
    "\n",
    "\n",
    "print(title_case_rappers)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": []
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem: Write a function thats input is a number and output is:\n",
    "* Return True if the input number is even.\n",
    "* Return False if the input number is odd."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve problem here\n",
    "def is_even(number):\n",
    "    if number % 2 == 0:\n",
    "        return True\n",
    "    else:\n",
    "        return False\n",
    "\n",
    "\n",
    "result = is_even(6)\n",
    "print(result)  \n",
    "result = is_even(7)\n",
    "print(result)  \n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem: Find the sum and the average of this list of numbers.\n",
    "\n",
    "Try doing this using a loop.  Then try doing this without using a loop. "
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "ename": "SyntaxError",
     "evalue": "invalid syntax (<ipython-input-6-6a4c39b495a9>, line 4)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-6-6a4c39b495a9>\"\u001b[0;36m, line \u001b[0;32m4\u001b[0m\n\u001b[0;31m    list_sum = ???\u001b[0m\n\u001b[0m               ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m invalid syntax\n"
     ]
    }
   ],
   "source": [
    "# Solve problem here:\n",
    "\n",
    "my_list = [1, 5, 10, 55, 88, 44, 42, 50, 20, 38]\n",
    "list_sum = sum(my_list)\n",
    "'''\n",
    "total = 0\n",
    "for i in my_list:\n",
    "    total += i\n",
    "'''\n",
    "list_avg = list_sum/10\n",
    "\n",
    "# Keep this as your last line in this cell."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem: Write a function that input is a number and output returns the following\n",
    "* If the input is divisible by three, return 'fizz'\n",
    "* If the input is divisible by five, return 'buzz'\n",
    "* If the input is divisible by three and by five, return 'fizzbuzz'\n",
    "* If the input is not divisible by three or five, return None."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve Problem fizzbuzz here:\n",
    "number = int(input())\n",
    "if number%3==0 and number%5==0: #if number is divisible by 3 & 5\n",
    "    print('fizzbuzz')\n",
    "elif number%5==0: #divisible by 5\n",
    "    print('buzz')\n",
    "elif number%3==0: #divisible by 3\n",
    "    print('fizz')\n",
    "else: #not divisible\n",
    "    print('None')"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem: \n",
    "Write a function that takes a list and returns a new list that contains all the elements of the first list minus all the duplicates.\n",
    "\n",
    "Example input and expected output\n",
    "\n",
    "input = [\"Michele\", \"Robin\", \"Sara\", \"Michele\"]\n",
    "\n",
    "expected output = ['Michele', 'Robin', 'Sara']\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve problem here:\n",
    "\n",
    "names = [\"Michele\", \"Robin\", \"Sara\", \"Michele\"]\n",
    "new_list=list(set(names))\n",
    "print(new_list) #check if it worked"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem: \n",
    "Write a function that takes a list of numbers (for example, a = [5, 10, 15, 20, 25]) and returns a new list of only the first and last elements of the given list.\n",
    "\n",
    "Example input and expected output:\n",
    "\n",
    "input = [5, 10, 15, 20, 25]\n",
    "\n",
    "expected output = [5, 25]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Solve problem here:\n",
    "input_list = [5, 10, 99, 20, 25]\n",
    "new_list = [input_list[0], input_list[-1]] #first & last element\n",
    "print(new_list) #check if it worked"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Problem:  Implement a function that takes as input three variables, and returns the largest of the three. \n",
    "Do this without using the Python max() function! \n",
    "_*Note all three input numbers will always be different, no need to account for a tie._\n",
    "\n",
    "Example input and expected output:\n",
    "\n",
    "input = your_function(1, 5, 10)\n",
    "\n",
    "expected output:  10"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {},
   "outputs": [
    {
     "ename": "SyntaxError",
     "evalue": "unexpected EOF while parsing (<ipython-input-10-ca9e2b626fbd>, line 5)",
     "output_type": "error",
     "traceback": [
      "\u001b[0;36m  File \u001b[0;32m\"<ipython-input-10-ca9e2b626fbd>\"\u001b[0;36m, line \u001b[0;32m5\u001b[0m\n\u001b[0;31m    \u001b[0m\n\u001b[0m    ^\u001b[0m\n\u001b[0;31mSyntaxError\u001b[0m\u001b[0;31m:\u001b[0m unexpected EOF while parsing\n"
     ]
    }
   ],
   "source": [
    "# Solve Problem here:\n",
    "\n",
    "def your_function(a, b, c):\n",
    "    # Fill in your code below and return max value of a, b, c\n",
    "    if a>b and a>c: #compares a with b & c\n",
    "        print(a)\n",
    "    elif b>a and b>c: #compares b with a & c\n",
    "        print(b)\n",
    "    else:\n",
    "        print(c)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# Test to see if your function works properly.\n",
    "your_function(1, 5, 10)"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.8.5"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 4
}