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   "source": [
    "# In Class Activity - Python - For loops"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 0: For-loop over a dictionary"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Here is a dictionary `colleges` that lists the city and zip (as values) for several schools in NYC (as keys)."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [],
   "source": [
    "colleges = {}\n",
    "colleges['NYU'] = 'New York, NY 10012'\n",
    "colleges['Columbia'] = 'New York, NY 10027'\n",
    "colleges['Fordham'] = 'Bronx, NY 10458'\n",
    "colleges['Brooklyn college'] = 'Brooklyn, NY 11210'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Write a for-loop to iterate over the keys and return the values, printing in this format:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "<img src=\"images-python-ica-2/colleges.png\" width=\"350\">"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "NYU : New York, NY 10012\n",
      "Columbia : New York, NY 10027\n",
      "Fordham : Bronx, NY 10458\n",
      "Brooklyn college : Brooklyn, NY 11210\n"
     ]
    }
   ],
   "source": [
    "# Your answer goes here\n",
    "for k in colleges.keys():\n",
    "    print(k, ':', colleges[k])"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Problem 1: Multiplication table"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Write a function `mult_table(n)` that produces the multiplication table below. Note that for each cell $c_{ij}$, its value is equal to the product of the row number $i$ and column number $j$, assuming they are numbered $1,\\dots,n$.\n",
    "\n",
    "Note, it's not necessary to get the spacing exactly as in the example, but kudos if you do!\n",
    "\n",
    "Here is what the output should look like when you run `mult_table(10)`\n",
    "<img src=\"images-python-ica-2/mult_table.png\" width=\"350\">"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "  1   2   3   4   5   6   7   8   9  10 \n",
      "  2   4   6   8  10  12  14  16  18  20 \n",
      "  3   6   9  12  15  18  21  24  27  30 \n",
      "  4   8  12  16  20  24  28  32  36  40 \n",
      "  5  10  15  20  25  30  35  40  45  50 \n",
      "  6  12  18  24  30  36  42  48  54  60 \n",
      "  7  14  21  28  35  42  49  56  63  70 \n",
      "  8  16  24  32  40  48  56  64  72  80 \n",
      "  9  18  27  36  45  54  63  72  81  90 \n",
      " 10  20  30  40  50  60  70  80  90 100 \n"
     ]
    }
   ],
   "source": [
    "# Your answer goes here\n",
    "def pad_num(x):\n",
    "    x = str(x)\n",
    "    npad = 3-len(x)\n",
    "    x = npad*' '+x\n",
    "    return x\n",
    "        \n",
    "def mult_table(n):\n",
    "    for i in range(1,n+1):\n",
    "        for j in range(1,n+1):\n",
    "            print(pad_num(i*j), end=' ')\n",
    "        print(\"\")\n",
    "        \n",
    "mult_table(10)"
   ]
  }
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