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{
"cells": [
{
"cell_type": "markdown",
"id": "2b789064",
"metadata": {},
"source": [
"# 实验9~实践12 参考答案"
]
},
{
"cell_type": "markdown",
"id": "e1fa4e6d",
"metadata": {},
"source": [
"## 实践9 元组和列表 \n"
]
},
{
"cell_type": "markdown",
"id": "d6b5ae4e",
"metadata": {},
"source": [
"### 小试身手1\n",
"(1)\t已知列表 L1 和L2，由 L1 和 L2 构造L3，并回答问题。 \n",
"```python\n",
" >>> L1=[1,2,3,4,5]\n",
" >>> L2=[\"one\",\"two\",\"three\",\"four\",\"five\"]\n",
" >>> L3=[[L1[1],L2[1]],[L1[2],L2[2]],[L1[3],L2[3]]]\n",
"L3 的值是__________\n",
"L3[1][1]的值是__________\n",
"执行 L4=L3.pop(2)后，列表L3的值是__________，L4的值是__________。\n",
"再执行L3.extend(L4)，列表L3的值是__________。\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": 8,
"id": "ee0db3a0",
"metadata": {},
"outputs": [],
"source": [
"L1=[1,2,3,4,5]\n",
"L2=[\"one\",\"two\",\"three\",\"four\",\"five\"]\n"
]
},
{
"cell_type": "code",
"execution_count": 11,
"id": "61ec7df3",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[2, 'two'], [3, 'three'], [4, 'four']]"
]
},
"execution_count": 11,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"L3=[[L1[1],L2[1]],[L1[2],L2[2]],[L1[3],L2[3]]]\n",
"L3"
]
},
{
"cell_type": "code",
"execution_count": 13,
"id": "873f2572",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"'three'"
]
},
"execution_count": 13,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"L3[1][1]"
]
},
{
"cell_type": "code",
"execution_count": 14,
"id": "d3ca6442",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[4, 'four']"
]
},
"execution_count": 14,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"L4=L3.pop(2)\n",
"L4"
]
},
{
"cell_type": "code",
"execution_count": 20,
"id": "7910a864",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[2, 'two'], [3, 'three'], 4, 'four']"
]
},
"execution_count": 20,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"L3"
]
},
{
"cell_type": "code",
"execution_count": 16,
"id": "f2a4aaa7",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[[2, 'two'], [3, 'three'], 4, 'four']"
]
},
"execution_count": 16,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"L3.extend(L4)\n",
"L3"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "a6e368db",
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "markdown",
"id": "de8197bc",
"metadata": {},
"source": [
"(2)假设执行了如下语句\n",
"```python\n",
">>> s1=[0,1,2,3,4,5,6]\n",
">>> s2=['SUN','MON','TUE','WED','THU','FRI','SAT']\n",
"利用 s1、s2 和列表操作，创建下列结果的序列对象。\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": 18,
"id": "d985454f",
"metadata": {},
"outputs": [],
"source": [
"s1=[0,1,2,3,4,5,6]\n",
"s2=['SUN','MON','TUE','WED','THU','FRI','SAT']"
]
},
{
"cell_type": "code",
"execution_count": 19,
"id": "a5a002be",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"'SUN|MON|TUE|WED|THU|FRI|SAT'"
]
},
"execution_count": 19,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#s3: 'SUN|MON|TUE|WED|THU|FRI|SAT'\n",
"s3=\"|\".join(s2)\n",
"s3"
]
},
{
"cell_type": "code",
"execution_count": 21,
"id": "f8bb4c02",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"[3, 4, 3, 4, 3, 4]"
]
},
"execution_count": 21,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"#s4:[3,4,3,4,3,4]\n",
"s4=s1[3:5]*3\n",
"s4"
]
},
{
"cell_type": "code",
"execution_count": 23,
"id": "bea9fa4d",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"[(0, 'SUN'), (1, 'MON'), (2, 'TUE'), (3, 'WED'), (4, 'THU'), (5, 'FRI'), (6, 'SAT')]\n"
]
}
],
"source": [
"#s5: [(0,'SUN'),(1,'MON'),(2,'TUE'),(3,'WED'),(4,'THU'),(5,'FRI'),(6,'SAT')]\n",
"s5=list(zip(s1,s2))\n",
"print(s5)"
]
},
{
"cell_type": "markdown",
"id": "b7e2a784",
"metadata": {},
"source": [
"## 小试身手2\n",
"### (1)合并列表 \n",
"使用随机函数randint()，建立两个长度为5的二维列表A和B。包含的数据如下：‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"A列表：包含子列表 [学号，p成绩，m成绩]，学号为从1开始递增的正整数，成绩为0-100之间的随机整数。‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"B列表：包含子列表 [学号，q成绩]，学号为从1开始递增的正整数，成绩为0-100之间的随机整数‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"先需要将A,B两个列表数据进行合并，即将B列表中的‘q成绩’添加到A列表中的相同学号的子列表中。‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"输入一个整数k，将B列表中的'q成绩'插入到A列表相同学号的子列表的k位置上。‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"合并完成后，输出合并后的A列表。‪‬‪‬‪‬‪‬‪‬‮‬‫‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‭‬‫‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‭‬\n",
"\n",
"注意：为保证生成固定序列，本题需要使用同一个循环结构生成两个随机列表A和B，类似：\n",
"```python\n",
"for #######：\n",
" #A列表添加 学号，p成绩，m成绩\n",
" #B列表添加 学号，q成绩\n",
"运行示例：\n",
"A: [[1, 94, 58], [2, 6, 84], [3, 26, 42], [4, 39, 98], [5, 22, 18]]\n",
"B: [[1, 33], [2, 82], [3, 29], [4, 26], [5, 24]]\n",
"3\n",
"[[1, 94, 58, 33], [2, 6, 84, 82], [3, 26, 42, 29], [4, 39, 98, 26], [5, 22, 18, 24]]\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "c4adadb0",
"metadata": {},
"outputs": [],
"source": [
"import random\n",
"a=[]\n",
"b=[]\n",
"\n",
"for i in range(1,6): \n",
" a.append([i,random.randint(0,100),random.randint(0,100)]) \n",
" b.append([i,random.randint(0,100)])\n",
"print(\"A:\",a)\n",
"print(\"B:\",b)\n",
"p=int(input()) \n",
"for i in range(len(a)):\n",
" a[i].insert(p,b[i][1]) \n",
"print(a) "
]
},
{
"cell_type": "markdown",
"id": "f9dd4a7a",
"metadata": {},
"source": [
"### (2)成绩统计\n",
"输入用空格分隔的两个正整数n和m，表示后面要接收m个人的n门课的成绩的输入。\n",
"随后的m行，每行输入用空格分隔的n个成绩。题目保证m和n都在(0-100]之间。\n",
"请统计\n",
"（1）每门课的平均成绩。\n",
"（2）每个学生的平均成绩\n",
"（3）每门课的最高分"
]
},
{
"cell_type": "code",
"execution_count": 31,
"id": "23a32ac5",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"n=3\n",
"m=5\n",
"63 87 76\n",
"98 92 90\n",
"86 83 82\n",
"69 98 71\n",
"77 74 92\n",
"[[63, 87, 76], [98, 92, 90], [86, 83, 82], [69, 98, 71], [77, 74, 92]]\n",
"每个学生的平均成绩: [75.33, 93.33, 83.67, 79.33, 81.0]\n",
"每门课的平均成绩: [78.6, 86.8, 82.2]\n",
"每门课的最高分: [98, 98, 92]\n"
]
}
],
"source": [
"n,m = int(input(\"n=\")),int(input(\"m=\"))\n",
"Scores = []\n",
"for i in range(m):\n",
" Scores.append([int(x) for x in input().split()])\n",
"print(Scores)\n",
"avgs=[]#每个学生的平均成绩\n",
"for stu in Scores:\n",
" avgs.append(round(sum(stu)/len(stu),2))\n",
"print(\"每个学生的平均成绩:\",avgs)\n",
"\n",
"ksums=[0]*n #每门课的平均成绩\n",
"kmaxs=[0]*n #每门课的最高分\n",
"for stu in Scores:\n",
" for i in range(n):\n",
" ksums[i]+=stu[i]\n",
" if stu[i]>kmaxs[i]:\n",
" kmaxs[i] = stu[i]\n",
"kavgs = [ round(x/m,2) for x in ksums]\n",
"print(\"每门课的平均成绩:\",kavgs)\n",
"print(\"每门课的最高分:\",kmaxs)"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "84af1780",
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "markdown",
"id": "5262941a",
"metadata": {},
"source": [
"## 实践10 集合和字典\n",
"小试身手1 （略）"
]
},
{
"cell_type": "markdown",
"id": "471a3f13",
"metadata": {},
"source": [
"### 小试身手2\n",
"输入两个整数，在这两个整数组成的闭区间范围内生成100个随机整数，并统计出现数据的次数，按照生成随机数从小到大的顺序，每行输出一个生成的整数以及其出现的次数，以空格间隔。 \n",
"```python\n",
"例如：\n",
"输入：\n",
"3 5\n",
"输出：\n",
"3 36\n",
"4 39\n",
"5 25\n",
"```"
]
},
{
"cell_type": "code",
"execution_count": 32,
"id": "c82977a4",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"3 5\n",
"3 36\n",
"4 39\n",
"5 25\n"
]
}
],
"source": [
"import random\n",
"dic = {}\n",
"m,n=map(int,input().split())\n",
"random.seed(10)\n",
"\n",
"for i in range (100):\n",
" key = random.randint(m,n)\n",
" if key in dic:\n",
" dic[key] += 1\n",
" else:\n",
" dic[key] = 1\n",
"for i in sorted(dic.keys()):\n",
" print('{} {}'.format(i,dic[i]))"
]
},
{
"cell_type": "markdown",
"id": "9d5b06fc",
"metadata": {},
"source": [
"## 实践11 模块化程序设计 \n",
"1.完整程序，程序功能实现统计1000以内每100个整数中素数的个数。"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "dcba82ab",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
" 1- 100 有26个素数\n",
"101- 200 有21个素数\n",
"201- 300 有16个素数\n",
"301- 400 有16个素数\n",
"401- 500 有17个素数\n",
"501- 600 有14个素数\n",
"601- 700 有16个素数\n",
"701- 800 有14个素数\n",
"801- 900 有15个素数\n",
"901-1000 有14个素数\n"
]
}
],
"source": [
"from math import sqrt\n",
"def isPrime(n):\n",
" m=int(sqrt(n))\n",
" for i in range(2,m+1):\n",
" if n%i ==0:\n",
" return False\n",
" return True\n",
"def countPrimes(start,end):\n",
" c=0\n",
" for n in range(start,end+1):\n",
" if isPrime(n):\n",
" c=c+1\n",
" return c \n",
"\n",
"#主程序\n",
"for start in range(1,1000,100):\n",
" print(f\"{start:3d}-{start+99:4d} 有{countPrimes(start,start+99)}个素数\")\n",
" "
]
},
{
"cell_type": "markdown",
"id": "e33847ab",
"metadata": {},
"source": [
" ## 3.斐波那契数列\n",
" ### (1) 单变量实现"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "efa91d0c",
"metadata": {},
"outputs": [],
"source": [
"def fib(n):\n",
" if n<=2:\n",
" return 1\n",
" f1=f2=1\n",
" while n>=3:\n",
" f3=f1+f2\n",
" f1,f2=f2,f3\n",
" n=n-1\n",
" return f3\n",
"\n",
"\n",
"n=int(input(\"n=\"))\n",
"print(fib(n))\n"
]
},
{
"cell_type": "markdown",
"id": "4d0b3b92",
"metadata": {},
"source": [
"## (2) 列表实现"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "48aa767e",
"metadata": {},
"outputs": [],
"source": [
"def fib( n ):\n",
" if n<=2:\n",
" return 1\n",
" FL=[1,1]\n",
" i=2\n",
" while i<n:\n",
" FL.append(FL[i-1]+FL[i-2])\n",
" i=i+1\n",
" return FL[-1] \n",
" \n",
"n=int(input(\"n=\"))\n",
"print(fib(n))\n"
]
},
{
"cell_type": "markdown",
"id": "5dee8615",
"metadata": {},
"source": [
"## (3) 字典实现"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "4cde2e34",
"metadata": {},
"outputs": [],
"source": [
"def fib(n): \n",
" a = 0\n",
" b = 1\n",
" dic = {} ## {} or dict() \n",
" for i in range(n): ## range \n",
" dic[i]=a ## a \n",
" a,b=b,a+b\n",
" \n",
" return dic ## dic \n",
" \n",
"#调用函数生成斐波那契数列中的前20个斐波那契数\n",
"fibonac = fib(20) ## fib(20) 调用函数 \n",
" print(fibonac[key], end=\",\")\n"
]
},
{
"cell_type": "markdown",
"id": "7494690d",
"metadata": {},
"source": [
"## (4)递归函数"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "5da061f6",
"metadata": {},
"outputs": [],
"source": [
"def Fib(n):\n",
" if n==1 or n==2 :\n",
" return 1\n",
" return Fib(n-1)+Fib(n-2)\n",
"\n",
"n=int(input(\"n=\"))\n",
"print(Fib(n))\n",
"\n",
"\n"
]
},
{
"cell_type": "markdown",
"id": "fdcc945d",
"metadata": {},
"source": [
"## (5) 优化的递归函数"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "46f27dea",
"metadata": {},
"outputs": [],
"source": [
"d_fib={1:1,2:1}\n",
"def Fib(n):\n",
" if n in d_fib :\n",
" return d_fib[n]\n",
" m=Fib(n-1)+Fib(n-2)\n",
" d_fib[n]= m\n",
" return m\n",
"\n",
"n=int(input(\"n=\"))\n",
"print(Fib(n))"
]
},
{
"cell_type": "markdown",
"id": "eecb1ba6",
"metadata": {},
"source": [
"## 实践12 再谈Python函数 \n",
"### 小试身手(1) \n",
"阅读可变参数的vfunc函数，仿写milti函数实现返回所有参数的乘积"
]
},
{
"cell_type": "code",
"execution_count": 33,
"id": "b4245d1d",
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"720"
]
},
"execution_count": 33,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"def milti(*num):\n",
" p=1\n",
" for x in num:\n",
" p=p*x\n",
" return p\n",
"milti(1,2,3,4,5,6)"
]
},
{
"cell_type": "markdown",
"id": "e2c1ee22",
"metadata": {},
"source": [
"### 小试身手(2) \n",
"1.判断奇数 \n",
"判断奇数的函数lambda函数的定义和使用 \n",
"（1）编写一个函数isOdd，能够判断一个整数是否是奇数，返回True或False。 \n"
]
},
{
"cell_type": "code",
"execution_count": 35,
"id": "af90b317",
"metadata": {},
"outputs": [],
"source": [
"\n",
"isodd = lambda x: x%2 == 1"
]
},
{
"cell_type": "markdown",
"id": "519d41a6",
"metadata": {},
"source": [
"（2）请写出输入一个数x，调用isOdd函数判断x是否是奇数的程序段"
]
},
{
"cell_type": "code",
"execution_count": 36,
"id": "b75e2c05",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"x=8\n",
"False\n"
]
}
],
"source": [
"x = int(input(\"x=\"))\n",
"print(isodd(x))"
]
},
{
"cell_type": "markdown",
"id": "71637790",
"metadata": {},
"source": [
"（3）请写出求50-100之间所有奇数的和的程序段,（使用isOdd实现奇数判断）。"
]
},
{
"cell_type": "code",
"execution_count": 43,
"id": "22543b83",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"s= 1875\n"
]
}
],
"source": [
"s=0\n",
"for x in range(50,100):\n",
" if isodd(x):\n",
" # print(x)\n",
" s+=x\n",
"print(\"s=\",s)\n"
]
},
{
"cell_type": "markdown",
"id": "3ad734be",
"metadata": {},
"source": [
"（4）请写出求50-100之间所有偶数的和的程序段（使用isOdd实现偶数判断）。"
]
},
{
"cell_type": "code",
"execution_count": 41,
"id": "23d97c6a",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"s= 1950\n"
]
}
],
"source": [
"s=0\n",
"for x in range(50,101):\n",
" if not isodd(x):\n",
" # print(x)\n",
" s+=x\n",
"print(\"s=\",s)\n"
]
},
{
"cell_type": "markdown",
"id": "71f64423",
"metadata": {},
"source": [
"2.构造整数:求由任意个整数上的个位数构造的新整数并输出。 \n",
"具体要求 \n",
"（1）编写lambda函数getLastBit，该函数返回正整数number的个位数，例如正整数1234，则返回4。 \n",
"（2）编写可变参数的函数makeNumber，求任意个整数的个位数构成的一个新整数。 \n",
"（3）输出（45、81、673、938）4个数的个位数构成的新整数，得到的新的整数为：5138。请填空完整程序。 \n"
]
},
{
"cell_type": "code",
"execution_count": 45,
"id": "2b89059b",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"新整数是5138\n"
]
}
],
"source": [
"#定义lambda函数getLastBit返回一个数的个位数\n",
"getLastBit=lambda number:number%10\n",
"\n",
"#定义一个可变参数的函数makeNumber，求任意个整数的个位数构成的一个新整数\n",
"def makeNumber(*b):\n",
" s=0\n",
" for i in b:\n",
" s=s*10+getLastBit(i)\n",
" return s\n",
"\n",
"#输出（45、81、673、938）4个数的个位数构成的新整数。\n",
"\n",
"print(f\"新整数是{makeNumber(45,81,673,938)}\")\n"
]
},
{
"cell_type": "markdown",
"id": "cf9617cc",
"metadata": {},
"source": [
"## 实践13 递归函数"
]
},
{
"cell_type": "markdown",
"id": "b26582d3",
"metadata": {},
"source": [
"### 小试身手"
]
},
{
"cell_type": "markdown",
"id": "0bff8c63",
"metadata": {},
"source": [
"1.使用递归求最大公约数"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "0bd1eb79",
"metadata": {},
"outputs": [],
"source": [
"def gcd(m,n):\n",
" if m%n == 0:\n",
" return n\n",
" else:\n",
" return gcd(n,m%n)\n",
"print(gcd(15,25))"
]
},
{
"cell_type": "markdown",
"id": "d8006798",
"metadata": {},
"source": [
"2.使用递归方法逆序输出一个整数中的数字"
]
},
{
"cell_type": "code",
"execution_count": 3,
"id": "099d4634",
"metadata": {},
"outputs": [],
"source": [
" def reverseDisplay(value):\n",
" if value>0:\n",
" print(value%10,end=\"\")\n",
" reverseDisplay(value//10)\n",
" return"
]
},
{
"cell_type": "code",
"execution_count": 4,
"id": "cc25a415",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"12345"
]
}
],
"source": [
"reverseDisplay(54321)"
]
},
{
"cell_type": "markdown",
"id": "0e80ecad",
"metadata": {},
"source": [
"3.使用递归方法逆序输出一个字符串中的字母 "
]
},
{
"cell_type": "code",
"execution_count": 6,
"id": "a2ff79ad",
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"edcba"
]
}
],
"source": [
"def reverseDisplay(s):\n",
" if len(s)!=1:\n",
" reverseDisplay(s[1:])\n",
" print(s[0],end=\"\")\n",
"reverseDisplay(\"abcde\")"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "fcc70358",
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.7.10"
}
},
"nbformat": 4,
"nbformat_minor": 5
}

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{
"cells": [
{
"cell_type": "markdown",
"metadata": {},
"source": [
"# 实践12 再谈python函数\n",
"\n",
"1. 理解函数定义四要素：函数名、参数表、函数体和返回值，本章对每一个部分都进行了更深入的说明，尤其是一些特殊的用法；\n",
"2. 理解 函数定义内外是两个不同的“**作用域**（*scope*）”，区分出全局变量和局部变量，需要充分理解其运作原理；\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们已经见过不少函数，也自己写过一些函数。我们已经理解函数的概念来自代数：从**输入参数**出发，**计算**出函数的**返回值**；我们也知道可以用 `def foo():` 来定义函数。其实函数的定义非常复杂，我们不太能够在第一次介绍时就讲清楚，所以之前我们就采取“先引入用起来”的方法，这也是一种知识上的“提前引用”。\n",
"\n",
"本章将围绕函数几个重要的要素深入看看。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 1. 为函数命名"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"哪怕一个函数内部什么都不干，它也得有个名字，然后名字后面要加上圆括号 `()`，以明示它是个函数，而不是某个变量。"
]
},
{
"cell_type": "code",
"execution_count": 1,
"metadata": {},
"outputs": [],
"source": [
"def do_nothing():\n",
" pass\n",
"\n",
"do_nothing()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"这就是个“什么也不干”的函数，关键字 `pass` 就是什么也不干的意思。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"给函数命名（给变量命名也一样）需要遵循的一些规则如下：\n",
"* 首先，名称不能以数字开头，能用在名称开头的只有大小写字母和下划线 `_`；\n",
"* 其次，名称中不能有空格，如果一个名字里有好几个词汇，可以用下划线来分割（`do_nothing`），也可以用所谓 *Camel Case* 风格（*doNothing*），习惯上更推荐使用下划线；\n",
"* 最后，绝对不能与 Python 语言的**关键字**（*keyword*）重复。\n",
"\n"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"在程序里给变量、函数命名是个挺重要的事情，影响到程序的可读性，就像小说的语言，最好能有一种流畅清晰、又始终一致的**风格**（*style*）。为了让全世界的 Python 程序员都有相对一致的风格，Python 社区有专门的一套建议规范，放在专门维护 Python 语言特性的社区 [PEP](https://www.python.org/dev/peps/) 上：\n",
"\n",
"* [PEP 8 -- Style Guide for Python Code: Naming Conventions](https://www.python.org/dev/peps/pep-0008/#naming-conventions)\n",
"\n",
"> PEP，是 *Python enhancement proposal* 的缩写，每当有重要的语言特性新需求新想法，就放在这里，经过广大 Python 用户和开发者的讨论完善，在某个版本放进 Python 中。很多 PEP 早已从 *proposal* 毕业变成官方特性，但也还在这里保留着。PEP 8 就是一个古老的 *proposal*，现在已为大多数 Python 用户采纳。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 2. 没有、一个和多个参数"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"函数可以没有参数，也可以有一个或者多个参数。\n",
"\n",
"没有参数就意味着，这个函数执行不依赖于输入，比如我们定义一个函数来在程序结束时打印一句退出提示："
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Program exits. Bye.\n"
]
}
],
"source": [
"def exit_info():\n",
" print('Program exits. Bye.')\n",
" \n",
"exit_info()"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"注意即使没有参数，无论定义还是调用时，函数名后面的括号都是不可省略的，这是函数身份的标志。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"函数也可以有多个参数，调用时输入参数的值是严格按照参数的顺序去匹配的。比如我们写一个函数输出某年到某年之间的所有闰年："
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2000\n",
"2004\n",
"2008\n",
"2012\n",
"2016\n"
]
}
],
"source": [
"def leap_years(begin, end):\n",
" year = begin\n",
" while year < end:\n",
" if (year % 4 == 0 and year % 100 != 0) or year % 400 == 0:\n",
" print(year)\n",
" year += 1\n",
" \n",
"leap_years(2000, 2020)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"当我们调用 `leap_years(2000, 2020)` 时，输入两个参数值 2000 和 2020，按照顺序匹配函数定义 `leap_years(begin, end)`，于是 `begin = 2000` `end = 2020`。所以参数的顺序是不能搞错的，有些函数参数很多，要是开发过程中还调整过顺序的话，那简直就是灾难，所以一般情况下还是保持函数参数不要乱动为好。\n",
"\n",
"顺便说一句，判断闰年的算法虽然不难，但要写的简洁也不容易。建议你可以先自己思考和实现一遍，然后尝试搞清楚为啥上面代码里的那行 `if` 是对的。实际上闰年的判断有很多正确的写法，你应该尝试写出自己的版本并确认它的正确性。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 3. 没有、一个和多个返回值"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"和参数一样，Python 的函数可以没有返回值，也可以有一个或者多个返回值。\n",
"\n",
"上面的 `exit_info` 和 `leap_year` 也是没有返回值的例子，它们的效果都通过 `print` 函数来体现。实际上没有返回语句的函数，等价于在其最后有一句 `return None`，表示函数返回了一个空值 `None`，`None` 在 Python 中是一个合法的值，表示什么都没有，它在逻辑上等价于 `False`："
]
},
{
"cell_type": "code",
"execution_count": 5,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"False"
]
},
"execution_count": 5,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"bool(None)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"所以即使没有返回值的函数，也可以用在 `if` 后面做逻辑表达式，不过我们并不推荐这么做，因为可读性很差。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"大部分情况下函数是有返回值的，因为绝大部分情况下函数的作用都是做“数据处理”，从输入出发得到输出。\n",
"\n",
"一般情况下函数都只有一个返回值，我们已经见过不少例子；但 Python 也允许多返回值，比如我们想用一个函数来计算两个整数相除的商和余数，可以这么写："
]
},
{
"cell_type": "code",
"execution_count": 6,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"8 2\n"
]
}
],
"source": [
"def idiv(a, b):\n",
" quotient = a // b\n",
" remainder = a % b\n",
" return quotient, remainder\n",
"\n",
"q, r = idiv(50, 6)\n",
"print(q, r)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"和多参数的情况类似，多返回值的情况下，赋值也是按照顺序匹配的，上面的代码中赋值语句左边的 `q` 匹配到第一个返回值，`r` 匹配第二个。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 4. 函数内与函数外：变量的作用域"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"下面的代码经常会把人搞晕："
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"2\n",
"1\n"
]
}
],
"source": [
"def increase_one(n):\n",
" n += 1\n",
" return n\n",
"\n",
"n = 1\n",
"print(increase_one(n))\n",
"print(n)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"请你思考一下，为什么这段代码里的两个 `print` 函数输出分别是 2 和 1。\n",
"\n",
"这个问题就涉及到变量的**作用域**（*scope*）问题，也就是说在不同地方出现的同名变量和函数，可能是完全不同的两个东西：\n",
"* 函数定义体中的变量的作用域是该函数内，程序的其他部分不知道其存在，这种变量叫**局部变量**（*local variable*）；函数的输入参数也是局部变量，也只在函数定义体中有效；\n",
"* 不在任何函数、类定义体中的变量的作用域是全局的，在任何地方都可以访问，这种变量称为**全局变量**（*global variable*）；\n",
"* 如果局部变量和全局变量同名，函数定义体内会优先局部变量，不会把它当做全局变量。\n",
"\n",
"这样我们就能理解上面代码输出的 2 和 1 了：\n",
"* 第一个 `print()` 打印的是函数调用 `increase_one(n)` 的返回值，这个语句不在任何函数定义体中，所以它里面用到的变量都是全局变量：\n",
" * 在调用 `increase_one()` 时参数 `n`，按照作用域原理，是全局变量 `n` 当时的值，也就是 1；\n",
" * 在 `increase_one()` 函数定义内，参数 `n` 是输入参数即局部变量，带着传进来的值 1，经过加一之后返回，返回值是 2；\n",
" * `print` 打印这个返回值，输出 2；\n",
" * 这个过程中处理的都是局部变量，完全不影响全局变量 `n` 的值；\n",
"* 第二个 `print()` 打印的是全局变量 `n` 的值，输出出 1。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"以上的文字，可能需要反复阅读若干遍；几遍下来，消除了疑惑，以后就彻底没问题了；若是这个疑惑并未消除，或者关键点并未消化，以后则会反复被这个疑惑所坑害，浪费无数时间。\n",
"\n",
"顺便说一句，上面这个例子用来说明作用域的概念很有用，但是平时写程序最好别这么写，减少重名的变量可以提升代码的清晰度和可读性。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"与此相关的，我们在介绍列表等数据容器时，会为上面的规则作出重要的补充，这里先留一个伏笔。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 5. 带缺省值的参数"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们其实已经见过带缺省值的参数（*argument with default value*），这里我们更细致的看看这个特性。\n",
"\n",
"在函数定义中可以在某个参数后面用等号 `=` 给它一个缺省值，调用时可以省略传入这个参数的值，直接采用缺省值；当然也可以在调用时传入这个参数的值来覆盖掉缺省值。这种特性相当于给了这个函数两个版本，一个带某个参数，一个不带，不带的版本就当该参数是某个缺省值。看看下面的例子："
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [],
"source": [
"def greeting(name, msg='Hi'):\n",
" print(f'{msg}, {name}!')"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo!\n"
]
}
],
"source": [
"greeting('Neo')"
]
},
{
"cell_type": "code",
"execution_count": 10,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Good morning, Neo!\n"
]
}
],
"source": [
"greeting('Neo', 'Good morning')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"一个函数可以有多个带缺省值的参数，但有一个限制：所有这些带缺省值的参数只能堆在参数表的最后，也就是说你定义的参数表里，出现一个带缺省值的参数，则它后面的都必须带缺省值。如果把上面的 `greeting()` 函数的两个参数调换一下，会扔出一个 `SyntaxError: non-default argument follows default argument` 的异常。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 6. 指定参数名来调用函数"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们前面说过，调用函数时传入的参数值会严格按照顺序去匹配参数变量，第一个输入值赋给第一个参数变量，第二个值赋给第二个参数变量，依此类推。因为有了上面说的带缺省值参数，这个规则出现了变通的可能。\n",
"\n",
"如果一个函数有多个带缺省值的参数，我们想忽略掉某几个参数（就用其缺省值），但指定后面某一个参数的值（覆盖缺省值），例如下面这个函数："
]
},
{
"cell_type": "code",
"execution_count": 11,
"metadata": {},
"outputs": [],
"source": [
"def greeting(name, msg='Hi', punc='!'):\n",
" print(f'{msg}, {name}{punc}')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"在这个版本的 `greeting()` 函数中，包含一个普通参数 `name` 和两个带缺省值的参数 `msg` `punc`，如果我们想跳过 `msg` 只传入 `name`（这个是必须的，因为没有缺省值）和 `punc` 的值，那么就可用下面的语法："
]
},
{
"cell_type": "code",
"execution_count": 12,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo.\n"
]
}
],
"source": [
"greeting('Neo', punc='.')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"这里第一个值按照顺序位置匹配到参数变量 `name`，这叫 *positional argument*（即“按照位置顺序匹配的参数”），而按照位置下一个是 `msg`，是我们想跳过的，所以要注明参数变量名，说明下一个传入的值 `'.'` 是给 `punc` 参数变量的，这叫 *keyword argument*（即“按照参数名匹配的参数”）。\n",
"\n",
"由于所有带缺省值的参数都在普通参数的后面，所以我们只要记住：\n",
"* 调用函数时先传入所有不带缺省值的参数的值，严格按照函数定义的位置顺序（*positional*）；\n",
"* 然后想指定哪些带缺省值参数的值，就用 `变量名=值` 这样的格式在后面列出（*keyword*），未列出的就还用缺省值了。\n",
"\n",
"在后半部分，顺序就无所谓了，可以和定义时不一样，反正是用名字指定的（*keyword*），比如我们完全可以这么干："
]
},
{
"cell_type": "code",
"execution_count": 13,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Good nite, Neo.\n"
]
}
],
"source": [
"greeting('Neo', punc='.', msg='Good nite')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"### 7. 变长参数"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"到目前为止，Python 的函数定义还是很简单清晰的，无论参数还是返回值，都没什么难懂的。下面开始就要进入比较混沌的领域了。\n",
"\n",
"所谓变长参数就是函数定义时名字前面带个星号 `*` 的参数变量，这表示这个变量其实是一组值，多少个都可以。我们先来看个简单的例子："
]
},
{
"cell_type": "code",
"execution_count": 14,
"metadata": {},
"outputs": [],
"source": [
"def say_hi(*names):\n",
" for name in names:\n",
" print('Hi,', name)"
]
},
{
"cell_type": "code",
"execution_count": 15,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo\n"
]
}
],
"source": [
"say_hi('Neo')"
]
},
{
"cell_type": "code",
"execution_count": 16,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo\n",
"Hi, Trinity\n"
]
}
],
"source": [
"say_hi('Neo', 'Trinity')"
]
},
{
"cell_type": "code",
"execution_count": 17,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo\n",
"Hi, Trinity\n",
"Hi, Morpheus\n"
]
}
],
"source": [
"say_hi('Neo', 'Trinity', 'Morpheus')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"在这个例子里，`*names` 是一个变长参数（*arbitrary argument*），调用时可以传入一个或者多个值，函数会把这些值看做一个列表，赋给局部变量 `names`——后面我们会知道，其实不是**列表**（*list*），而是一个**元组**（*tuple*）——然后我们在函数体中可以用 `for...in` 来对这个 `names` 做循环。\n",
"\n",
"> 有些中文书籍把 *arbitrary arguments* 翻译成“可变参数”或者“任意参数”。事实上，在这样的地方，无论怎样的中文翻译都是很难准确表达原意的。这还算好的，甚至还见过翻译成“武断的参数”的——这样的翻译肯定会使读者产生说不明道不白的疑惑。\n",
">\n",
"> 所以，**入门之后就尽量只用英文**是个好策略。虽然刚开始有点吃力，但后面会很省心，很长寿——是呀，少浪费时间、少浪费生命，其实就相当于更长寿了呀！"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"在使用 *arbitrary argument* 的场合，有几点需要注意：\n",
"* 参数变量名最好用复数单词，一看就知道是一组数据；这个变量在函数里通常都会被 `for...in` 循环处理，用复数名词在写类似 `for name in names` 的循环语句时会很舒服、很地道（*idiomatic*），是的，写程序和学外语一样，不写则已，写就要尽量写得“地道”；\n",
"* 这种参数变量只能有一个，因为从它开始后面的输入值都会被当做它的一部分，多了就不知道怎么分了，显然，如果有这种参数，必须放在参数表的最后。\n",
"\n",
"上面的第二点，有一个不太常见的例外，那就是一个函数既有 *arbitrary arguments* 又有 *arguments with default values* 的情况，那么可以有两个 *arbitrary arguments*，其中第二个必须带缺省值，然后参数表排列成这样：\n",
"\n",
"`def monstrosity(*normal arguments*, *normal arbitrary argument*, *arguments with defaults*, *arbitrary argument with default*)`\n",
"\n",
"这样是完全符合语法要求的，调用时传入参数值还是按照前面讲的规则，先按照位置顺序匹配前两部分，多出来的都归 *normal arbitrary argument*；然后按照参数变量名指定对应值，没指定的都用缺省值。不过这实在是太麻烦了，不知道什么情况下才必须用这么可怕的函数，还是祈祷我们不会碰到这样的场景吧！\n",
"\n",
"当然，只有上面列出的前三个部分的情况还是有的，比如下面的例子："
]
},
{
"cell_type": "code",
"execution_count": 18,
"metadata": {},
"outputs": [],
"source": [
"def say_hi(*names, msg='Hi', punc='!'):\n",
" for name in names:\n",
" print(f'{msg}, {name}{punc}')"
]
},
{
"cell_type": "code",
"execution_count": 19,
"metadata": {},
"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Hi, Neo.\n",
"Hi, Trinity.\n"
]
}
],
"source": [
"say_hi('Neo', 'Trinity', punc='.')"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 小试身手(1)\n",
"阅读可变参数的vfunc函数，仿写milti函数实现返回所有参数的乘积。"
]
},
{
"cell_type": "code",
"execution_count": 7,
"metadata": {},
"outputs": [],
"source": [
"def vfunc(*b):\n",
" a=0\n",
" for n in b:\n",
" a+=n\n",
" return a\n"
]
},
{
"cell_type": "code",
"execution_count": 8,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"15"
]
},
"execution_count": 8,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"vfunc(1,2,3,4,5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 函数别名"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"在 Python 中，所有函数也是对象，证据就是它们都有对象 id。Python 会为创建的每一个对象（不管基本数据类型，还是某个 *class* 的实例）指定一个唯一的 id，可以用内置函数 `id()` 来查看，比如："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"n = 42\n",
"id(n)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"函数也有这个 id，比如："
]
},
{
"cell_type": "code",
"execution_count": 3,
"metadata": {},
"outputs": [],
"source": [
"def _is_leap(year):\n",
" return (year % 4 == 0 and year % 100 != 0) or year % 400 == 0"
]
},
{
"cell_type": "code",
"execution_count": 4,
"metadata": {},
"outputs": [
{
"data": {
"text/plain": [
"2989669354808"
]
},
"execution_count": 4,
"metadata": {},
"output_type": "execute_result"
}
],
"source": [
"id(_is_leap)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"既然函数有 id，是个对象，那是什么类型的对象呢？可以用内置函数 `type` 来看："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"type(_is_leap)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"所以函数是个 `function` 类型的对象。\n",
"\n",
"既然是个对象，我们就可以用赋值语句来创建函数的**别名**（*alias*）："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"is_leap = _is_leap"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"id(is_leap)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"可以看到，这两个函数的 id 完全一样，是同一个对象的两个名字而已。我们可以用这两个名字来调用这个函数，完全没区别："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"_is_leap(2018)"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"is_leap(2018)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"那么，我们为什么需要给函数取别名呢？\n",
"\n",
"很多时候是为了提供更好的代码可读性，比如在特定上下文让某个函数的作用更显而易见，比如以前的例子里，我们曾经在 `Cat` 类里给父类 `Animal` 的 `voice()` 方法定义别名叫 `meow()`。\n",
"\n",
"还有一种情况是一个函数需要在运行时动态指向不同的实现版本。这里只简单描述一个典型场景：假定我们要渲染一段视频，如果系统里有兼容的显卡（GPU），就调用显卡来渲染，会更快更省电，如果没有则用 CPU 来渲染，会慢一点和更耗电一点，于是我们把用 GPU 渲染的算法写成函数 `_render_by_gpu()`，用 CPU 渲染的算法写成函数 `_render_by_cpu()`，而检测是否存在可用 GPU 的算法写成函数 `is_gpu_available()`，然后可以用下面的方法来定义一个函数 `render`：\n",
"\n",
"```python\n",
"if is_gpu_available():\n",
" render = _render_by_gpu\n",
"else:\n",
" render = _render_by_cpu\n",
"```\n",
"\n",
"这样 `render()` 就成为一个当前系统中最优化的渲染函数，在程序的其他地方就不用管细节，直接用这个函数就好。这就是动态函数别名的价值。\n",
"\n",
"顺便说一句，在任何一个工程里，为函数或者变量取名都是**很简单却不容易**的事情——因为可能会重名（虽然已经尽量用变量的作用域隔离了），可能会因取名含混而令后来者费解，所以，仅仅为了少敲几下键盘而给一个函数取个更短的别名，实际上并不是好主意，更不是好习惯。尤其现在的编辑器都支持自动补全和多光标编辑的功能，变量名长点不是什么大问题。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 匿名函数"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"有的函数需要两个甚至更多名字，有的函数却一个也不要，人生就是这么丰富多彩啊！\n",
"\n",
"所谓匿名函数，就是有时候我们需要一个函数，但就在一个地方，用完就扔，再也不会用了，Python 对这种情况提供了一个方便的语法，不需要 `def` 那套严肃完整的语法，一行就可以写完一个函数，这个语法使用关键字 `lambda`。`lambda` 是希腊字母 `λ` 的英语音译，在计算机领域是个来头不小的词儿，代表了一系列高深的理论，[和阿伦佐·丘奇（Alonzo Church）的理论有关](https://en.wikipedia.org/wiki/Lambda_calculus)，有兴趣的话可以自行研究。\n",
"\n",
"不过目前我们不需要管那么多，只要了解怎么快速创建“用过即扔”的小函数就好了。"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"def add(x, y):\n",
" return x + y\n",
"add(3, 5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"我们可以用 lambda 来改写："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"add = lambda x, y: x + y\n",
"add(3, 5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"甚至更简单一点，名字也不要了："
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"(lambda x, y: x + y)(3, 5)"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"最后这种形式，就是典型的匿名函数了。简单地说，`lambda` 可以生成一个函数对象，出现在所有需要一个函数的地方，可以将其赋给一个变量（如上面的 `add`），这个变量就称为函数变量（别名），可以当函数用；也可以直接把 `lambda` 语句用括号括起来当一个函数用（上面后一种形式）。\n",
"\n",
"在 Python 官方文档中，`lambda` 语句的语法定义是这样的：\n",
"\n",
"`lambda_expr ::= \"lambda\" [parameter_list] \":\" expression`\n",
"\n",
"这个语法定义采用的是 [巴克斯范式（BNF）](https://en.wikipedia.org/wiki/Backus%E2%80%93Naur_form)标注，现在不明白没关系（虽然对照上面的例子也能猜出个大概吧），以后我们会专门介绍。\n",
"\n",
"其实也很简单，就是这个样子：\n",
"\n",
"```python\n",
"lambda x, y: x + y\n",
"```\n",
"\n",
"先写上 `lambda` 关键字，其后分为两个部分，`:` 之前是参数表，之后是表达式，这个表达式的值，就是这个函数的返回值。**注意**：`lambda` 语句中，`:` 之后有且只能有一个表达式，所以它搞不出很复杂的函数，比较适合一句话的函数。\n",
"\n",
"而这个函数呢，没有名字，所以被称为 “匿名函数”。\n",
"\n",
"`add = lambda x, y: x + y`\n",
"\n",
"就相当于是给一个没有名字的函数取了个名字。"
]
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"## 小试身手（2）"
]
},
{
"cell_type": "raw",
"metadata": {},
"source": [
"1.判断奇数 \n",
"判断奇数的函数lambda函数的定义和使用 \n",
"（1）使用匿名函数编写一个函数isOdd，能够判断一个整数是否是奇数，返回True或False。\n",
"isodd = lambda x: x%2 == 1\n",
"（2）请写出输入一个数x，调用isOdd函数判断x是否是奇数的程序段。 \n",
"（3）请写出求50-100之间所有奇数的和的程序段，（使用isOdd实现奇数判断）。 \n",
"（4）请写出求50-100之间所有偶数的和的程序段（使用isOdd实现偶数判断）。\n"
]
},
{
"cell_type": "code",
"execution_count": 9,
"metadata": {},
"outputs": [],
"source": [
"isodd = lambda x: x%2 == 1"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
},
{
"cell_type": "markdown",
"metadata": {},
"source": [
"2.构造整数\n",
"问题描述\n",
"求由任意个整数上的个位数构造的新整数并输出。\n",
"具体要求\n",
"（1）编写lambda函数getLastBit，该函数返回正整数number的个位数，例如正整数1234，则返回4。\n",
"（2）编写可变参数的函数makeNumber，求任意个整数的个位数构成的一个新整数。\n",
"（3）输出（45、81、673、938）4个数的个位数构成的新整数，得到的新的整数为：5138。请填空完整程序。\n",
"\n",
"新整数是5138"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": [
"#定义lambda函数getLastBit返回一个数的个位数\n",
"getLastBit= \n",
"\n",
"#定义一个可变参数的函数makeNumber，求任意个整数的个位数构成的一个新整数\n",
"def makeNumber(a,*b):\n",
" s=getLastBit(a)\n",
" \n",
"\n",
"\n",
" return s\n",
"\n",
"#输出（45、81、673、938）4个数的个位数构成的新整数。\n",
"print(f\"新整数是{makeNumber(45,81,673,938)}\")\n",
"\n"
]
},
{
"cell_type": "code",
"execution_count": null,
"metadata": {},
"outputs": [],
"source": []
}
],
"metadata": {
"kernelspec": {
"display_name": "Python 3",
"language": "python",
"name": "python3"
},
"language_info": {
"codemirror_mode": {
"name": "ipython",
"version": 3
},
"file_extension": ".py",
"mimetype": "text/x-python",
"name": "python",
"nbconvert_exporter": "python",
"pygments_lexer": "ipython3",
"version": "3.7.10"
}
},
"nbformat": 4,
"nbformat_minor": 4
}

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