text sublistas_jupyter_notebook.ipynb

Posted 2021-05-03

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    "Qual é a maneira mais Pythônica ou ter um código fácil de entender e que resolva:\n",
    "\n",
    "Tenho uma lista com sub-listas:\n",
    "\n",
    "```python\n",
    "items = [['a','b'], ['c','d'], ['e', None]]\n",
    "```\n",
    "\n",
    "\n",
    "Quero como resultado remover as sub-listas:\n",
    "\n",
    "```python\n",
    "resultado = ['a','b', 'c','d', 'e', None]\n",
    "```\n",
    "\n",
    "IMPORTANTE:\n",
    "- a lista de items será pequena, logo performance não é problema!\n",
    "- \"Readability counts\" The Zen of Python\n"
   ]
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   "cell_type": "code",
   "execution_count": 1,
   "metadata": {},
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   "source": [
    "\n",
    "items = [['a','b'], ['c','d'], ['e', None]]\n"
   ]
  },
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   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Tentativa 1\n",
    "=======\n",
    "Usar 'List Comprehensions' com 1 for dentro de outro é pythônico? É fácil de entender?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['a', 'b', 'c', 'd', 'e', None]\n"
     ]
    }
   ],
   "source": [
    "\n",
    "tentativa1 = [item for sublist in items for item in sublist]\n",
    "print(tentativa1)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Tentativa 2\n",
    "===========\n",
    "Usar itertools"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['a', 'b', 'c', 'd', 'e', None]\n",
      "['a', 'b', 'c', 'd', 'e', None]\n"
     ]
    }
   ],
   "source": [
    "\n",
    "from itertools import chain\n",
    "tentativa2 = list(chain(*items))\n",
    "print(tentativa2)\n",
    "\n",
    "# ou\n",
    "\n",
    "tentativa2 = list(chain.from_iterable(items))\n",
    "print(tentativa2)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Tentativa 3\n",
    "===========\n",
    "Usar a Built-in `sum(iterable [, start])` do Python. \n",
    "Parece simples, não precisa de import. Mas este [] como segundo parametro?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "scrolled": true
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['a', 'b', 'c', 'd', 'e', None]\n"
     ]
    }
   ],
   "source": [
    "\n",
    "tentativa3 = sum(items, [])\n",
    "print(tentativa3)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Tentativa 4\n",
    "===========\n",
    "Programação funcional?\n",
    "uma vez que `['a', 'b'] + ['c', 'd']` tem como resultado `['a', 'b', 'c', 'd']`\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "['a', 'b', 'c', 'd', 'e', None]\n"
     ]
    }
   ],
   "source": [
    "\n",
    "from functools import reduce\n",
    "tentativa4 = reduce(lambda x, y: x+y, items)\n",
    "print(tentativa4)\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "\n",
    "Tem mais formas? \n",
    "```\n",
    "\"There should be one-- and preferably only one --obvious way to do it.\"\n",
    "```\n"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": []
  }
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