Course 3 Zip.ipynb

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    "# Zip\n",
    "\n",
    "One more common pattern with lists, besides accumulation, is to step through a pair of lists (or several lists), doing something with all of the first items, then something with all of the second items, and so on. For example, given two lists of numbers, you might like to add them up pairwise, taking [3, 4, 5] and [1, 2, 3] to yield [4, 6, 8]"
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     "text": [
      "[4, 6, 8]\n"
     ]
    }
   ],
   "source": [
    "L1 = [3, 4, 5]\n",
    "L2 = [1, 2, 3]\n",
    "L3 = []\n",
    "\n",
    "for i in range(len(L1)):\n",
    "    L3.append(L1[i] + L2[i])\n",
    "\n",
    "print(L3)\n"
   ]
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   "execution_count": 3,
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   "outputs": [
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     "name": "stdout",
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     "text": [
      "[(3, 1), (4, 2), (5, 3)]\n"
     ]
    }
   ],
   "source": [
    "L1 = [3, 4, 5]\n",
    "L2 = [1, 2, 3]\n",
    "L4 = list(zip(L1, L2))\n",
    "print(L4)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[4, 6, 8]\n"
     ]
    }
   ],
   "source": [
    "L1 = [3, 4, 5]\n",
    "L2 = [1, 2, 3]\n",
    "L3 = []\n",
    "L4 = list(zip(L1, L2))\n",
    "\n",
    "for (x1, x2) in L4:\n",
    "    L3.append(x1+x2)\n",
    "\n",
    "print(L3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[9, 12, 14]\n"
     ]
    }
   ],
   "source": [
    "L1 = [3,4,5]\n",
    "L2 = [6,8,9]\n",
    "L3 = [x1 + x2 for (x1, x2) in list(zip(L1, L2))]\n",
    "print(L3)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {},
   "outputs": [
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     "name": "stdout",
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     "text": [
      "<map object at 0x7fcca4142490>\n"
     ]
    }
   ],
   "source": [
    "L1 = [3, 4, 5]\n",
    "L2 = [1, 2, 3]\n",
    "L3 = map(lambda x: x[0] + x[1], zip(L1, L2))\n",
    "print(L3)"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Consider a function called possible, which determines whether a word is still possible to play in a game of hangman, given the guesses that have been made and the current state of the blanked word."
   ]
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   "cell_type": "code",
   "execution_count": 11,
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     "text": [
      "False\n",
      "True\n"
     ]
    }
   ],
   "source": [
    "def possible(word, blanked, guesses_made):\n",
    "    if len(word) != len(blanked):\n",
    "        return False\n",
    "    for i in range(len(word)):\n",
    "        bc = blanked[i]\n",
    "        wc = word[i]\n",
    "        if bc == '_' and wc in guesses_made:\n",
    "            return False\n",
    "        elif bc != '_' and bc != wc:\n",
    "            return False\n",
    "    return True\n",
    "\n",
    "print(possible(\"wonderwall\", \"_on__r__ll\", \"weotnqurl\"))\n",
    "print(possible(\"wonderwall\", \"_on__r__ll\", \"otnqurl\"))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "However, we can rewrite that using zip, to be a little more comprehensible."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 29,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "False\n",
      "False\n"
     ]
    }
   ],
   "source": [
    "def compatible_char(bc, wc, guesses_made):\n",
    "    if bc == '_' and wc in guesses_made:\n",
    "        return False\n",
    "    elif bc != '_' and bc != wc:\n",
    "        return False\n",
    "    return True\n",
    "\n",
    "\n",
    "def possible(word, blanked, guesses_made):\n",
    "    if len(word) != len(blanked):\n",
    "        return False\n",
    "    for (bc, wc) in zip(blanked, word):\n",
    "        if not compatible_char(bc, wc, guesses_made):\n",
    "            return False\n",
    "    return True\n",
    "\n",
    "print(possible(\"wonderwall\", \"_on__r__ll\", \"otnqurl\"))\n",
    "print(possible(\"wonderwall\", \"_on__r__ll\", \"wotnqurl\"))"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "1. Below we have provided two lists of numbers, L1 and L2. Using zip and list comprehension, create a new list, L3, that sums the two numbers if the number from L1 is greater than 10 and the number from L2 is less than 5. This can be accomplished in one line of code."
   ]
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   "cell_type": "code",
   "execution_count": 16,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "[18, 57, 103]\n"
     ]
    }
   ],
   "source": [
    "L1 = [1, 5, 2, 16, 32, 3, 54, 8, 100]\n",
    "L2 = [1, 3, 10, 2, 42, 2, 3, 4, 3]\n",
    "L3 = [x1 + x2 for (x1, x2) in list(zip(L1, L2)) if x1 > 10 and x2 < 5]\n",
    "print(L3)"
   ]
  }
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