mod10.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#  <font color=red> Module_10_Tuple</font>"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# P10-2\n",
    "## Tuple &rArr; 有序(sequence)、不可變(immutable)\n",
    ">### P3-3<img src='./img/py_data_type_3.png'>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup1 = ('bdse', 'aien', 2015, 2018)\n",
    "tup2 = (1, 2, 3, 4, 5 )\n",
    "tup3 = \"a\", \"b\", \"c\", \"d\"\n",
    "tup4 = () # empty tuple\n",
    "tup5 = (50,) # single value tuple\n",
    "print(tup1, tup2, tup3, tup4, tup5)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "t1=(50) # not tuple becasue of no ','\n",
    "t2=(50,) # ㇐個元素的 tuple 必須要多加㇐個逗點結尾\n",
    "print('type(t1) :', type(t1))\n",
    "print('type(t2) :', type(t2))\n",
    "print('len(t2) :',len(t2))\n",
    "print('len(t1) :',len(t1))"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-3 Tuple formation"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup6 = 50, # 此寫法是Tuple，只是省略掉小括號()\n",
    "print(type(tup6))\n",
    "int1 = (50)\n",
    "print(type(int1))"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup1 = ('aiot', 'python', 23.5, 2019)\n",
    "tup2 = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\n",
    "\n",
    "print(tup1[0])\n",
    "print(tup2[1:5])\n",
    "print(tup2[1:10:2])"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Tuple 內容是不可改變的(immutable)，亦即無法更新或刪除項目"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "t3=(0, 2, 4, 6, 8)\n",
    "print(t3)\n",
    "\n",
    "t3[1] = 5 # error,  immutable"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 可以利用已存在的 Tuple 來產生新的 Tuple"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = 4, 5, 6 ; tup1 = (4, 5, 6)\n",
    "print(id(tup),id(tup1))\n",
    "print(tup+tup1) # 與串列結合([1,2]+[1,2]=>[1,2,1,2])相似\n",
    "print(id(tup+tup1))"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-6"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Tuple裡的內容可以放入任何型態 &rArr; Tuple裡面就可以放入多層次的資料"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = (33, 'python', [11, 12, 13], 77)\n",
    "tup[2] "
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### <font color=\"red\">改動與否，視最後改動的對象為準。</font>\n",
    "<details>\n",
    "    <summary style=\"margin-left: 2em\">correct example</summary>\n",
    "    <img src='./img/tuple_hierarchy.png' style='height:450px'>\n",
    "</details>\n",
    "<details>\n",
    "    <summary style=\"margin-left: 2em\">incorrect example</summary>\n",
    "    <img src='./img/tuple_hierarchy_error.png' style='height:450px'>\n",
    "</details>"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "tup = ('foo', [1, 2], True)\n",
    "\n",
    "tup[1][0]=20\n",
    "print(tup)\n",
    "\n",
    "tup[1]=20  # error\n",
    "print(tup)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = ('foo', [1, 2], True)\n",
    "tup[0][0] = 'F'"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-7"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "len((10,20,30))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "('Hi',)*4"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "30 in (10,20,30)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "seq = (1, 2, 3, 4, 5, 6, 7, 8, 9) # iterable object\n",
    "for i in seq:\n",
    "    print(i)"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-8 Built-in function for Tuple"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "nested_tup = (4, 5, 6), (7, 8)\n",
    "print(len(nested_tup))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "tup = (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)\n",
    "print(max(tup))\n",
    "print(min(tup))\n",
    "print(sum(tup))\n",
    "print(sorted(tup))\n",
    "print(sorted(tup,reverse = True))"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "tags": []
   },
   "outputs": [],
   "source": [
    "t1 = tuple(range(0,10,2)) # lazy evaluation\n",
    "t2 = tuple([1,2,3,4])\n",
    "t3 = tuple('abcde')\n",
    "\n",
    "print(t1)\n",
    "print(t2)\n",
    "print(t3)"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-9 Tuple methods"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = (1, 2, 2, 2, 3, 4, 2)\n",
    "print(a.count(2))\n",
    "print(a.index(2))"
   ]
  },
  {
   "attachments": {},
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---\n",
    "# P10-11 list comprehensions"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "S = []\n",
    "for x in range(10) :\n",
    "    x = x**2\n",
    "    S.append(x)\n",
    "S"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "S = [x**2 for x in range(10)]\n",
    "print(S)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "V = [2**x for x in range(13)]\n",
    "print(V)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "M = []\n",
    "for x in S:\n",
    "    if x % 2 == 0:\n",
    "        M.append(x)\n",
    "print(M)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "M = [x for x in S if x % 2 == 0]\n",
    "print(M)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "%%timeit\n",
    "# V = (1,2,4,8,...2**10)\n",
    "V=[]\n",
    "for n in range(11):\n",
    "    V +=[2**n]\n",
    "V"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "%%timeit\n",
    "# V = (1,2,4,8,...2**10)\n",
    "V = [2**n for n in range(11)]\n",
    "V"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "# for loop start\n",
    "#公元年分除以4可整除但公元年分除以100不可整除」\n",
    "#或「公元年分除以400可整除」\n",
    "leap_year=[]\n",
    "for i in range(1000, 2020):\n",
    "    if( ( i%4==0 and  i%100!=0) or (i%400== 0) ):\n",
    "        leap_year.append(i)\n",
    "print(leap_year)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "leap_year1 = [i for i in range(1000,2020) if(( i%4==0 and  i%100!=0) or (i%400==0)) ]\n",
    "\n",
    "print(leap_year1)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Tuple沒有Comprehension的觀念"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup_a = (i for i in range(10))\n",
    "tup_a"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "---"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# &spades; 補充 "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### 在實務上，Tuple的速度比List快，在處理大數據時，使用Tuple會比較有效率。"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "%timeit list(range(5))\n",
    "%timeit tuple(range(5))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Unpacking tuples"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = (4, 5, 6)\n",
    "a, b, c = tup\n",
    "b"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = 4, 5, (6, 7)\n",
    "a, b, (c, d) = tup\n",
    "d"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "tup = 4, 5, (6, 7)\n",
    "a, b, c, d = tup\n",
    "d"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### P2-5 的變數建立方式就是利用unpacking tuple"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a, b, c = 1, 2., \"John\"\n",
    "print(a)\n",
    "print(b)\n",
    "print(c)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "seq = [(1, 2, 3), (4, 5, 6), (7, 8, 9)]\n",
    "for a, b, c in seq:\n",
    "    print(f'a={a}, b={b}, c={c}')"
   ]
  }
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