05-advanced-python/05.09-iterators.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# 迭代器"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 简介"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "迭代器对象可以在 `for` 循环中使用："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2\n",
      "4\n",
      "6\n"
     ]
    }
   ],
   "source": [
    "x = [2, 4, 6]\n",
    "\n",
    "for n in x:\n",
    "    print n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "其好处是不需要对下标进行迭代，但是有些情况下，我们既希望获得下标，也希望获得对应的值，那么可以将迭代器传给 `enumerate` 函数，这样每次迭代都会返回一组 `(index, value)` 组成的元组："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "pos 0 is 2\n",
      "pos 1 is 4\n",
      "pos 2 is 6\n"
     ]
    }
   ],
   "source": [
    "x = [2, 4, 6]\n",
    "\n",
    "for i, n in enumerate(x):\n",
    "    print 'pos', i, 'is', n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "迭代器对象必须实现 `__iter__` 方法："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<listiterator object at 0x0000000003CAE630>\n"
     ]
    }
   ],
   "source": [
    "x = [2, 4, 6]\n",
    "i = x.__iter__()\n",
    "print i"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "`__iter__()` 返回的对象支持 `next` 方法，返回迭代器中的下一个元素："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "2\n"
     ]
    }
   ],
   "source": [
    "print i.next()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "当下一个元素不存在时，会 `raise` 一个 `StopIteration` 错误："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "4\n",
      "6\n"
     ]
    }
   ],
   "source": [
    "print i.next()\n",
    "print i.next()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "ename": "StopIteration",
     "evalue": "",
     "output_type": "error",
     "traceback": [
      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
      "\u001b[1;31mStopIteration\u001b[0m                             Traceback (most recent call last)",
      "\u001b[1;32m<ipython-input-6-e590fe0d22f8>\u001b[0m in \u001b[0;36m<module>\u001b[1;34m()\u001b[0m\n\u001b[1;32m----> 1\u001b[1;33m \u001b[0mi\u001b[0m\u001b[1;33m.\u001b[0m\u001b[0mnext\u001b[0m\u001b[1;33m(\u001b[0m\u001b[1;33m)\u001b[0m\u001b[1;33m\u001b[0m\u001b[0m\n\u001b[0m",
      "\u001b[1;31mStopIteration\u001b[0m: "
     ]
    }
   ],
   "source": [
    "i.next()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "很多标准库函数返回的是迭代器："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<listreverseiterator object at 0x0000000003D615F8>\n"
     ]
    }
   ],
   "source": [
    "r = reversed(x)\n",
    "print r"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "调用它的 `next()` 方法："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "6\n",
      "4\n",
      "2\n"
     ]
    }
   ],
   "source": [
    "print r.next()\n",
    "print r.next()\n",
    "print r.next()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "字典对象的 `iterkeys, itervalues, iteritems` 方法返回的都是迭代器："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 9,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<dictionary-itemiterator object at 0x0000000003D51B88>\n"
     ]
    }
   ],
   "source": [
    "x = {'a':1, 'b':2, 'c':3}\n",
    "i = x.iteritems()\n",
    "print i"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "迭代器的 `__iter__` 方法返回它本身："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 10,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "<dictionary-itemiterator object at 0x0000000003D51B88>\n"
     ]
    }
   ],
   "source": [
    "print i.__iter__()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 11,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "('a', 1)\n"
     ]
    }
   ],
   "source": [
    "print i.next()"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## 自定义迭代器"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "自定义一个 list 的取反迭代器："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 12,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "class ReverseListIterator(object):\n",
    "    \n",
    "    def __init__(self, list):\n",
    "        self.list = list\n",
    "        self.index = len(list)\n",
    "        \n",
    "    def __iter__(self):\n",
    "        return self\n",
    "    \n",
    "    def next(self):\n",
    "        self.index -= 1\n",
    "        if self.index >= 0:\n",
    "            return self.list[self.index]\n",
    "        else:\n",
    "            raise StopIteration"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 13,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "9 8 7 6 5 4 3 2 1 0\n"
     ]
    }
   ],
   "source": [
    "x = range(10)\n",
    "for i in ReverseListIterator(x):\n",
    "    print i,"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "只要我们定义了这三个方法，我们可以返回任意迭代值："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 14,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "class Collatz(object):\n",
    "    \n",
    "    def __init__(self, start):\n",
    "        self.value = start\n",
    "        \n",
    "    def __iter__(self):\n",
    "        return self\n",
    "    \n",
    "    def next(self):\n",
    "        if self.value == 1:\n",
    "            raise StopIteration\n",
    "        elif self.value % 2 == 0:\n",
    "            self.value = self.value / 2\n",
    "        else:\n",
    "            self.value = 3 * self.value + 1\n",
    "        return self.value"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "这里我们实现 [Collatz 猜想](http://baike.baidu.com/view/736196.htm)：\n",
    "\n",
    "- 奇数 n：返回 3n + 1\n",
    "- 偶数 n：返回 n / 2\n",
    "\n",
    "直到 n 为 1 为止："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 15,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1\n"
     ]
    }
   ],
   "source": [
    "for x in Collatz(7):\n",
    "    print x,"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "不过迭代器对象存在状态，会出现这样的问题："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 16,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "22 11\n",
      "34 17\n",
      "52 26\n",
      "13 40\n",
      "20 10\n",
      "5 16\n",
      "8 4\n",
      "2 1\n"
     ]
    }
   ],
   "source": [
    "i = Collatz(7)\n",
    "for x, y in zip(i, i):\n",
    "    print x, y"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "一个比较好的解决方法是将迭代器和可迭代对象分开处理，这里提供了一个二分树的中序遍历实现："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 17,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "class BinaryTree(object):\n",
    "    def __init__(self, value, left=None, right=None):\n",
    "        self.value = value\n",
    "        self.left = left\n",
    "        self.right = right\n",
    "\n",
    "    def __iter__(self):\n",
    "        return InorderIterator(self)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 18,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "class InorderIterator(object):\n",
    "    \n",
    "    def __init__(self, node):\n",
    "        self.node = node\n",
    "        self.stack = []\n",
    "    \n",
    "    def next(self):\n",
    "        if len(self.stack) > 0 or self.node is not None:\n",
    "            while self.node is not None:\n",
    "                self.stack.append(self.node)\n",
    "                self.node = self.node.left\n",
    "            node = self.stack.pop()\n",
    "            self.node = node.right\n",
    "            return node.value\n",
    "        else:\n",
    "            raise StopIteration()"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 19,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": [
    "tree = BinaryTree(\n",
    "    left=BinaryTree(\n",
    "        left=BinaryTree(1),\n",
    "        value=2,\n",
    "        right=BinaryTree(\n",
    "            left=BinaryTree(3),\n",
    "            value=4,\n",
    "            right=BinaryTree(5)\n",
    "        ),\n",
    "    ),\n",
    "    value=6,\n",
    "    right=BinaryTree(\n",
    "        value=7,\n",
    "        right=BinaryTree(8)\n",
    "    )\n",
    ")"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 20,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 2 3 4 5 6 7 8\n"
     ]
    }
   ],
   "source": [
    "for value in tree:\n",
    "    print value,"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "不会出现之前的问题："
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 21,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 1\n",
      "2 2\n",
      "3 3\n",
      "4 4\n",
      "5 5\n",
      "6 6\n",
      "7 7\n",
      "8 8\n"
     ]
    }
   ],
   "source": [
    "for x,y in zip(tree, tree):\n",
    "    print x, y"
   ]
  }
 ],
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