Practice midterm, half length, from last year

330_midterm_w19/330_Midterm-student.ipynb

@@ -227,188 +227,6 @@
    "source": [
     "# Your code goes here.  There is no written answer for this question."
    ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Question 6 (Pivoting, 25 points max)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### Generate the following DataFrame from the movies data file:\n",
-    "![](assets/pivottable.png)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 13,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Your code goes here.  There is no written answer for this question "
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Question 7 (Natural Language Processing, 30 points max)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### List the adjectives that occur at least twice from the movie overviews for movies whose popularity values are greater than 100"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 14,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "# Your code goes here.  There is no written answer for this question."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "## Question 8 (Efficiency, 30 points max)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### 8a. Given the code and line profiler output below, identify where the majority of time is being spent. \n"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Your answer goes here"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 22,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "credits = pd.read_csv('data/tmdb_5000_credits_plus.csv')"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 23,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "import json\n",
-    "\n",
-    "# extract_genders will return a set of the genders found in whatever Series name is passed to it.\n",
-    "def extract_genders(x):\n",
-    "    genders = set() # Create an empty set to hold out results\n",
-    "    j = json.loads(x) # This loads the JSON string into a python data structure that consists of a list of dictionaries\n",
-    "    for character in j:\n",
-    "        str_gender = str(character['gender'])\n",
-    "        genders.add(str_gender)\n",
-    "    return \" \".join(genders)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 24,
-   "metadata": {},
-   "outputs": [
-    {
-     "name": "stdout",
-     "output_type": "stream",
-     "text": [
-      "208 ms ± 8.88 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)\n"
-     ]
-    }
-   ],
-   "source": [
-    "%timeit credits['cast'].apply(extract_genders)"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 25,
-   "metadata": {},
-   "outputs": [],
-   "source": [
-    "%load_ext line_profiler"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": 26,
-   "metadata": {},
-   "outputs": [
-    {
-     "data": {
-      "text/plain": [
-       "Timer unit: 1e-06 s\n",
-       "\n",
-       "Total time: 0.347512 s\n",
-       "File: <ipython-input-23-f09d4512344e>\n",
-       "Function: extract_genders at line 4\n",
-       "\n",
-       "Line #      Hits         Time  Per Hit   % Time  Line Contents\n",
-       "==============================================================\n",
-       "     4                                           def extract_genders(x):\n",
-       "     5      4803       2442.0      0.5      0.7      genders = set() # Create an empty set to hold out results\n",
-       "     6      4803     198111.0     41.2     57.0      j = json.loads(x) # This loads the JSON string into a python data structure that consists of a list of dictionaries\n",
-       "     7    111060      35616.0      0.3     10.2      for character in j:\n",
-       "     8    106257      63960.0      0.6     18.4          str_gender = str(character['gender'])\n",
-       "     9    106257      43705.0      0.4     12.6          genders.add(str_gender)\n",
-       "    10      4803       3678.0      0.8      1.1      return \" \".join(genders)"
-      ]
-     },
-     "metadata": {},
-     "output_type": "display_data"
-    }
-   ],
-   "source": [
-    "%lprun -f extract_genders credits['cast'].apply(extract_genders)"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "### 8b. Think carefully about what you might do to improve the performance and see if you can achieve an improvement.  Suggest some changes that you think might improve the performance."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "Insert your answer here."
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "metadata": {},
-   "source": [
-    "# END OF MID-TERM"
-   ]
-  },
-  {
-   "cell_type": "code",
-   "execution_count": null,
-   "metadata": {},
-   "outputs": [],
-   "source": []
   }
  ],
  "metadata": {