completed week 2 hw

homeworks/SA_week2_hw.ipynb

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+{
+ "cells": [
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "## ❌ DO NOT EDIT - MAKE A COPY\n",
+    "# Q1: 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": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['a', 'b', 'c']\n",
+      "['g', 'h', 'i']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve Alphabet slices here. \n",
+    "## Extra Credit:  Do this without 'hard coding' the alpahbet.\n",
+    "alpha = [\"a\", \"b\", \"c\", \"d\", \"e\", \"f\", \"g\", \"h\", \"i\", \"j\"]\n",
+    "print(alpha[0:3])\n",
+    "print(alpha[6:9])\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q2:  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": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "['Lil Wayne', 'Nicki Minaj', 'Drake']\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve rapper names here\n",
+    "rappers = ['lil wayne', 'nicki minaj', 'drake']\n",
+    "title_rappers = []\n",
+    "\n",
+    "for rapper in rappers:\n",
+    "    title_rappers.append(rapper.title())\n",
+    "print(title_rappers)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q3: Write a function that takes a number and returns:\n",
+    "* True if the input number is even.\n",
+    "* False if the input number is odd."
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve problem here\n",
+    "\n",
+    "def even_odd(lst):\n",
+    "   if lst % 2 == 0:\n",
+    "      return True\n",
+    "   return False\n",
+    "   # return something\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q4: 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": 11,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "353 35.3\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",
+    "list_average = list_sum / len(my_list)\n",
+    "\n",
+    "\n",
+    "# Keep this as your last line in this cell.\n",
+    "print(list_sum, list_average)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q5: \n",
+    "## Write a function that takes a list and returns a new list that has all the duplicates removed.\n",
+    "\n",
+    "Example input and expected output:\n",
+    "- input = `[\"Michele\", \"Robin\", \"Sara\", \"Michele\"]`\n",
+    "- expected output = `['Michele', 'Robin', 'Sara']`\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 15,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "{'Robin', 'Sara', 'Michele'}\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve problem here:\n",
+    "\n",
+    "names = [\"Michele\", \"Robin\", \"Sara\", \"Michele\"]\n",
+    "\n",
+    "def unique_names(names):\n",
+    "    unique = set(names)\n",
+    "    return unique\n",
+    "\n",
+    "print(unique_names(names))\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q6: Write a function that takes a list of numbers \n",
+    "(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",
+    "- input = `[5, 10, 15, 20, 25]`\n",
+    "- expected output = `[5, 25]`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 17,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[5, 25]\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve problem here:\n",
+    "input_list = [5, 10, 99, 20, 25]\n",
+    "\n",
+    "def first_and_last(input):\n",
+    "    new_list = []\n",
+    "    new_list.append(input[0])\n",
+    "    new_list.append(input[-1])\n",
+    "\n",
+    "    return new_list\n",
+    "\n",
+    "print(first_and_last(input_list))"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q7: \n",
+    "## Implement a function that takes as input three variables, and returns the largest of the three. \n",
+    "### Try doing this without using the `max()` function!\n",
+    "\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",
+    "- input: `your_function(1, 5, 10)`\n",
+    "- expected output:  `10`"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "8\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve Problem here:\n",
+    "\n",
+    "def my_max(a, b, c):\n",
+    "    # Fill in your code below and return max value of a, b, c\n",
+    "    if a >= b and a >= c:\n",
+    "        return a\n",
+    "    elif b >= a and b >= c:\n",
+    "        return b\n",
+    "    else:\n",
+    "        return c\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 22,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "10"
+      ]
+     },
+     "execution_count": 22,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "# Test to see if your function works properly.\n",
+    "my_max(1, 5, 10)"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "# Q8: Write a function that takes a number as input and 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": null,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "None\n"
+     ]
+    }
+   ],
+   "source": [
+    "# Solve Problem fizzbuzz here:\n",
+    "def fizz_buzz(input):\n",
+    "    if input % 3 == 0 and input % 5 == 0:\n",
+    "        return \"fizzbuzz\"\n",
+    "    elif input % 5 == 0:\n",
+    "        return \"buzz\"\n",
+    "    elif input % 3 == 0:\n",
+    "        return \"fizz\"\n",
+    "    else:\n",
+    "        return None\n",
+    "    "
+   ]
+  }
+ ],
+ "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": "3.12.1"
+  }
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+ "nbformat": 4,
+ "nbformat_minor": 4
+}