Added copy of lab 1

Exams/Mid-term/Exam.ipynb

@@ -20,7 +20,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 1,
    "metadata": {},
    "outputs": [],
    "source": [
@@ -43,9 +43,20 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 2,
    "metadata": {},
-   "outputs": [],
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "False\n",
+      "True\n",
+      "False\n",
+      "True\n"
+     ]
+    }
+   ],
    "source": [
     "print(\"A\">\"B\")\n",
     "print(\"B\">\"A\")\n",
@@ -60,20 +71,148 @@
     "Make sure your implementation isn't case sensitive. Do not use python's built-in `min`, `max`, `sort` or any other sort function you find."
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "Alabama\n"
+     ]
+    }
+   ],
+   "source": [
+    "def first_alphabetically(lst):\n",
+    "    # return min(lst)\n",
+    "    first = lst[0]\n",
+    "    for state in lst:\n",
+    "        if first.upper() > state.upper():\n",
+    "            first = state\n",
+    "    return first\n",
+    "\n",
+    "\n",
+    "print(first_alphabetically(states))"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "2. Write a function `arg_first_alphabetically(lst)`, which does the same thing as in exercise 1 but returns the index of the first string alphabetically."
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 13,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "23\n"
+     ]
+    }
+   ],
+   "source": [
+    "def arg_first_alphabetically(lst):\n",
+    "    # return lst.index(min(lst))\n",
+    "     \n",
+    "    first = lst[0]\n",
+    "    first_index = 0\n",
+    "    for i in range(1, len(lst)):\n",
+    "        if first.upper() > lst[i].upper():\n",
+    "            first = lst[i]\n",
+    "            first_index = i\n",
+    "    return first_index\n",
+    "\n",
+    "\n",
+    "print(arg_first_alphabetically(states))"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "3. Use your result in question 2 to implement a function `arg_sort_alphabetically(lst)` that returns a list that is alphabetically sorted. Sorting can be accomplished by successively applying the function in question 1 and removing the first element alphabetically. You can remove an element from a list using `pop()`. Make sure your implementation isn't case sensitive. Do not use python's built-in `min`, `max`, `sort` or any other sort function you find."
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 19,
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/plain": [
+       "['Alabama',\n",
+       " 'Alaska',\n",
+       " 'Arizona',\n",
+       " 'Arkansas',\n",
+       " 'California',\n",
+       " 'Colorado',\n",
+       " 'Connecticut',\n",
+       " 'Delaware',\n",
+       " 'Florida',\n",
+       " 'Georgia',\n",
+       " 'Hawaii',\n",
+       " 'Idaho',\n",
+       " 'Illinois',\n",
+       " 'Indiana',\n",
+       " 'Iowa',\n",
+       " 'Kansas',\n",
+       " 'Kentucky',\n",
+       " 'Louisiana',\n",
+       " 'Maine',\n",
+       " 'Maryland',\n",
+       " 'Massachusetts',\n",
+       " 'Michigan',\n",
+       " 'Minnesota',\n",
+       " 'Mississippi',\n",
+       " 'Missouri',\n",
+       " 'Montana',\n",
+       " 'Nebraska',\n",
+       " 'Nevada',\n",
+       " 'New Hampshire',\n",
+       " 'New Jersey',\n",
+       " 'New Mexico',\n",
+       " 'New York',\n",
+       " 'North Carolina',\n",
+       " 'North Dakota',\n",
+       " 'Ohio',\n",
+       " 'Oklahoma',\n",
+       " 'Oregon',\n",
+       " 'Pennsylvania',\n",
+       " 'Rhode Island',\n",
+       " 'South Carolina',\n",
+       " 'South Dakota',\n",
+       " 'Tennessee',\n",
+       " 'Texas',\n",
+       " 'Utah',\n",
+       " 'Vermont',\n",
+       " 'Virginia',\n",
+       " 'Washington',\n",
+       " 'West Virginia',\n",
+       " 'Wisconsin',\n",
+       " 'Wyoming']"
+      ]
+     },
+     "execution_count": 19,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "def arg_sort_alphabetically(lst):\n",
+    "    return [lst.pop(arg_first_alphabetically(lst)) for i in range(len(lst))]\n",
+    "\n",
+    "\n",
+    "arg_sort_alphabetically(states)"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
@@ -93,13 +232,71 @@
     "In other words the elements of matrix C which is the outer product of A and B are $c_{ij} = a_i b_j$."
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 40,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[6, 7, 8]\n",
+      "[12, 14, 16]\n",
+      "[18, 21, 24]\n",
+      "[24, 28, 32]\n",
+      "[30, 35, 40]\n"
+     ]
+    }
+   ],
+   "source": [
+    "def outer_product(lst1, lst2):\n",
+    "    lst3 = [[0 for i in range(len(lst2))] for j in range(len(lst1))]\n",
+    "    for i in range(len(lst1)):\n",
+    "        for j in range(len(lst2)):\n",
+    "            lst3[i][j] = lst1[i]*lst2[j]\n",
+    "    return lst3\n",
+    "\n",
+    "tl1 = [1, 2, 3, 4, 5]\n",
+    "tl2 = [6, 7, 8]\n",
+    "op = outer_product(tl1, tl2)\n",
+    "for l in op:\n",
+    "    print(l)"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
    "source": [
     "5. Implement a function `cumulative_sum(lst)` that takes a list of numbers and returns a list of same size where the element `i` is the sum of the elements `0` to `i` of the input list. For example given `[1,2,3]`, you should return [1,3,6]."
    ]
   },
+  {
+   "cell_type": "code",
+   "execution_count": 57,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "[1, 2, 3, 4, 5]\n",
+      "[1, 3, 6, 10, 15]\n"
+     ]
+    }
+   ],
+   "source": [
+    "def cumulative_sum(lst):\n",
+    "    for i in range(1, len(lst)):\n",
+    "        lst[i] += lst[i-1]\n",
+    "    return lst\n",
+    "\n",
+    "\n",
+    "tl = list(range(1,6))\n",
+    "print(tl)\n",
+    "print(cumulative_sum(tl))"
+   ]
+  },
   {
    "cell_type": "markdown",
    "metadata": {},
@@ -118,10 +315,29 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 68,
    "metadata": {},
    "outputs": [],
    "source": [
+<<<<<<< HEAD
+    "import random\n",
+    "import math\n",
+    "\n",
+    "\n",
+    "# all code in this block from lectures #\n",
+    "\n",
+    "\n",
+    "def arange(x_min, x_max, steps=10):\n",
+    "    step_size = (x_max - x_min) /steps\n",
+    "    x = x_min\n",
+    "    out = list()\n",
+    "    for i in range(steps):\n",
+    "        out.append(x)\n",
+    "        x += step_size\n",
+    "    return out\n",
+    "\n",
+    "\n",
+=======
     "import math,random\n",
     "\n",
     "def arange(x_min,x_max,steps=10):\n",
@@ -133,6 +349,7 @@
     "        x+=step_size\n",
     "    return out\n",
     "\n",
+>>>>>>> dd964aa33bbeea4d02fcfdd92e806792198daa05
     "def generate_normal(N,m=0,s=1):\n",
     "    out = list() \n",
     "    \n",
@@ -169,10 +386,38 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 130,
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "1.3119894260482794\n",
+      "      x_90: 1.311989\n",
+      "under x_90: 90.0%\n"
+     ]
+    }
+   ],
+   "source": [
+    "def find_x_90(x):\n",
+    "    h = histogram(x, len(x))\n",
+    "    N = float(len(x))\n",
+    "    x_90 = h[1][1] # second to lowest edge by default\n",
+    "    for i in range(1, len(h[0])):\n",
+    "        if cumulative_sum(h[0][:i])[-1] < 0.9*N:\n",
+    "            x_90 = h[1][i+1]\n",
+    "        else:\n",
+    "            break\n",
+    "    return x_90\n",
+    "\n",
+    "x = generate_normal(1000)\n",
+    "x_90 = find_x_90(x)\n",
+    "hist = histogram(x,len(x))\n",
+    "print(hist[1][hist[1].index(x_90)])\n",
+    "print(\"      x_90: {:f}\".format(x_90))\n",
+    "print(\"under x_90: {:3.1f}%\".format(100 * cumulative_sum(hist[0][:hist[1].index(x_90)])[-1]/len(x)))\n"
+   ]
   }
  ],
  "metadata": {
@@ -191,7 +436,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.8.1"
+   "version": "3.7.2"
   }
  },
  "nbformat": 4,