{"cells":[{"cell_type":"markdown","metadata":{"id":"4e2hspYloycG"},"source":["# Control Flow\n","## Practice Problems"]},{"cell_type":"markdown","metadata":{"id":"W6-SVI7SqNWb"},"source":["### 1. Write a conditional that prints different messages if a bank account balance is below `$`3,000, between `$`3,000 and `$`10,000, or over `$`10,000.\n","\n","HINT: You can check if a value `x` is between two other values with a condition like `5 <= x <= 10`, but you can also solve this problem by strategically ordering conditions."]},{"cell_type":"code","execution_count":null,"metadata":{"colab":{"base_uri":"https://localhost:8080/"},"executionInfo":{"elapsed":199,"status":"ok","timestamp":1668023835926,"user":{"displayName":"Kaylie Lau","userId":"01284785813595846851"},"user_tz":300},"id":"x92XbNTc9eYN","outputId":"5fd6a60f-13fa-45b9-f231-7aa111d1de7b"},"outputs":[],"source":["# Your code goes here"]},{"cell_type":"markdown","metadata":{},"source":["<details>\n"," <summary>Answer</summary>\n","\n"," ```python\n"," balance = 3000\n","\n"," if balance < 3000:\n"," print('Your balance is under $3,000')\n"," elif balance <= 10000:\n"," print('Your balance is between $3,000 and $10,000')\n"," else:\n"," print('Your balance is over $10,000.')\n","\n"," # This would also be a valid solution:\n"," \n"," if balance < 3000:\n"," print('Your balance is under $3,000')\n"," elif balance >= 3000 and balance <= 10000:\n"," print('Your balance is between $3,000 and $10,000')\n"," else:\n"," print('Your balance is over $10,000.')\n","\n"," # And so would this:\n","\n"," if balance < 3000:\n"," print('Your balance is under $3,000')\n"," elif 3000 <= balance <= 10000:\n"," print('Your balance is between $3,000 and $10,000')\n"," else:\n"," print('Your balance is over $10,000.')\n"," ```\n","</details>"]},{"cell_type":"markdown","metadata":{"id":"-eLr_oCiqRiH"},"source":["### 2. Create a list, `books`, containing the following items: `'War and Peace', 'Pride and Prejudice', 'Mockingjay', 'Three Musketeers', 'The Adventures of Robinson Crusoe', 'Yevgeniy Onegin'`. Then:\n","\n","- Using slicing or indexing, create the following:\n"," - An empty list\n"," - The last item of `books`\n"," - List of three items: 'Three Musketeers', 'The Adventures of Robinson Crusoe', 'Yevgeniy Onegin'.\n"," \n","- Using list methods:\n"," - Remove 'Pride and Prejudice' from the list.\n"," - Insert 'Harry Potter and the Chamber of Secrets' after 'Mockingjay'.\n","\n","HINT: Try using the same number as the starting index and ending index of a slice. Useful list methods include `remove()` and `insert()`."]},{"cell_type":"code","execution_count":null,"metadata":{"id":"7DuSnDQ9-Cs0"},"outputs":[],"source":["# Your code goes here"]},{"cell_type":"markdown","metadata":{},"source":["<details>\n"," <summary>Answer</summary>\n","\n"," ```python\n"," books = ['War and Peace', 'Pride and Prejudice', 'Mockingjay', 'Three Musketeers', 'The Adventures of Robinson Crusoe', 'Yevgeniy Onegin']\n","\n"," # an empty list\n"," books[0:0]\n","\n"," # the last item\n"," books[-1]\n","\n"," # the last three items\n"," books[-3:]\n","\n"," # remove 'Pride and Prejudice'\n"," books.remove('Pride and Prejudice')\n","\n"," # Insert Harry Potter and the Chamber of Secrets after Mockingjay\n"," # If you did not remove Pride and Prejudice, the index would be 3, not 2\n"," books.insert(2, 'Harry Potter and the Chamber of Secrets')\n","\n"," books\n"," ```\n","</details>"]},{"cell_type":"markdown","metadata":{"id":"1P3cHY0MqWP_"},"source":["### 3. Given the list `people`, sort it by people's first name, last name and age. Store the sorted lists as `by_first_name`, `by_last_name`, and `by_age`, respectively. \n","`people = [('Mark', 'Harrison', 56), ('Ken', 'Wolseley', 23), ('Emily', 'Robinson', 77)]`\n","\n","HINT 1: We want our sorting function to return a new list, rather than modifying in place.\n","\n","HINT 2: `sorted()` accepts a `key` argument. This argument is the name of a function to use when sorting list elements."]},{"cell_type":"code","execution_count":null,"metadata":{"id":"ReUtuKtk_CH6"},"outputs":[],"source":["# Your code goes here"]},{"cell_type":"markdown","metadata":{},"source":["<details>\n"," <summary>Answer</summary>\n","\n"," ```python\n"," #Here, we have to write functions to get the first, second, and third items in a list.\n"," #Then, we can use those functions as the keys for sorted().\n","\n"," people = [('Mark', 'Harrison', 56), ('Ken', 'Wolseley', 23), ('Emily', 'Robinson', 77)]\n","\n"," def get_first_item(lst):\n"," return lst[0]\n","\n"," def get_second_item(lst):\n"," return lst[1]\n","\n"," def get_third_item(lst):\n"," return lst[2]\n","\n"," by_first_name = sorted(people, key=get_first_item)\n"," by_last_name = sorted(people, key=get_second_item)\n"," by_age = sorted(people, key=get_third_item)\n"," ```\n","</details>"]},{"cell_type":"markdown","metadata":{"id":"UpMWlbwrqZDz"},"source":["### 4. Write a function called `dict_intersect` that takes two dictionaries, `d1` and `d2`, as arguments and returns a set that contains only the keys found in both of the original dictionaries.\n","\n","HINT: Let's break this down into steps. We need to:\n","* Get the keys in `d1`\n","* Get the keys in `d2`\n","* Convert them both to sets\n","* Find their intersection\n","\n","HINT: Some useful functions and methods are `keys()`, `set()`, and `intersection()`."]},{"cell_type":"code","execution_count":null,"metadata":{"id":"TdBwT7dYACOY"},"outputs":[],"source":["# Your code goes here"]},{"cell_type":"markdown","metadata":{},"source":["<details>\n"," <summary>Answer</summary>\n","\n"," ```python\n"," def dict_intersect(d1, d2):\n"," '''Return the set of keys found in both d1 and d2.\n"," >>> dict_intersect({'a': 'A', 'b': 'B', 'c': 'C'}, {'a': 'alpha', 'b': 'beta'})\n"," {'a', 'b'}\n"," >>> dict_intersect({'a': 1, 'b': 2}, {'c': 3, 'd': 2})\n"," set()\n"," '''\n"," d1_keys = set(d1.keys())\n"," d2_keys = set(d2.keys())\n"," return d1_keys.intersection(d2_keys)\n"," ```\n","</details>"]},{"cell_type":"markdown","metadata":{"id":"hW3Q-AHFqbZL"},"source":["### 5. Write a loop that iterates over the two lists below simultaneously. For each pair of values, print the first number divided by the second. When the program encounters a zero divisor, it should skip the pair without printing anything. \n","\n","HINT: We can exit a loop early with `break`.\n","\n","HINT 2: Remember that we can bundle two lists pairwise with `zip()`."]},{"cell_type":"code","execution_count":null,"metadata":{"colab":{"base_uri":"https://localhost:8080/"},"executionInfo":{"elapsed":7,"status":"ok","timestamp":1668024643760,"user":{"displayName":"Kaylie Lau","userId":"01284785813595846851"},"user_tz":300},"id":"JgcenI9Mngvz","outputId":"630f1104-533a-4210-ebd1-575a9b69760e"},"outputs":[],"source":["# Your code goes here"]},{"cell_type":"markdown","metadata":{},"source":["<details>\n"," <summary>Answer</summary>\n","\n"," ```python\n"," dividends = [100, 37.5, -12]\n"," divisors = [8, 0, -3]\n","\n"," for x, y in zip(dividends, divisors):\n"," if y == 0:\n"," continue\n"," else:\n"," print(x/y)\n"," ```\n","</details>"]}],"metadata":{"colab":{"authorship_tag":"ABX9TyOqN1klgwmxsG8wtOHzdOL5","collapsed_sections":[],"provenance":[]},"kernelspec":{"display_name":"Python 3","name":"python3"},"language_info":{"name":"python"}},"nbformat":4,"nbformat_minor":0}