1.循环
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# 1.for...in循环,依次把list或tuple中的每个元素迭代出来 studentNames = [ "Willard" , "ChenJD" , "ChenBao" , "LinWenYu" ] for studentName in studentNames: print (studentName) print ( "------------------------------------------------" ) # 计算1-10的累加和 sumOfNums = 0 for num in [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 ]: sumOfNums = sumOfNums + num print ( "1-10的累加和是:" ,sumOfNums) print ( "------------------------------------------------" ) # 使用range()函数进行整数序列生成,range()函数为左闭右开 # 计算1-1000的累加和 sumOfNums = 0 for num in range ( 1001 ): sumOfNums + = num print ( "1-1000的累加和是:" ,sumOfNums) |
# 结果输出:
Willard
ChenJD
ChenBao
LinWenYu
------------------------------------------------
1-10的累加和是: 55
------------------------------------------------
1-1000的累加和是: 500500
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# 2.while循环,只要条件满足,就不断循环,条件不满足时退出循环 # 计算100以内的奇数和 sumOfNums = 0 n = 99 while n > 0 : sumOfNums + = n n = n - 2 print ( "100以内的奇数累加和为:" ,sumOfNums) |
# 结果输出:
100以内的奇数累加和为: 2500
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# 3.使用break语句提前退出循环 n = 1 while n < 20 : if n > 10 : break print ( "n的值为:" ,n) n + = 1 print ( "The End." ) |
n的值为: 1
n的值为: 2
n的值为: 3
n的值为: 4
n的值为: 5
n的值为: 6
n的值为: 7
n的值为: 8
n的值为: 9
n的值为: 10
The End.
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# 4.continue语句,跳过当前循环,直接开始下一次循环 n = 10 while n < 20 : n + = 1 if n = = 15 : continue print ( "n的值为:" ,n) print ( "The End." ) |
# 结果输出:
n的值为: 11
n的值为: 12
n的值为: 13
n的值为: 14
n的值为: 16
n的值为: 17
n的值为: 18
n的值为: 19
n的值为: 20
The End.
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# 5.登录实例 totalFrequency = 3 inputFrequency = 0 userName = input ( "请输入您的账号:" ) password = input ( "请输入您的密码:" ) while inputFrequency < totalFrequency: if ((userName = = "Willard" ) and (password = = "JD584520" )): print ( "账号密码正确,登录成功!" ) break else : print ( "账号或密码输入错误,登录失败!" ) if (totalFrequency - inputFrequency - 1 ) = = 0 : break print ( "请重新登录!您还有%d次输入账号密码的机会!" % (totalFrequency - inputFrequency - 1 )) print ( "----------------------------------------" ) inputFrequency + = 1 userName = input ( "请重新输入您的账号:" ) password = input ( "请重新输入您的密码:" ) |
# 结果输出:
# 输出1:
请输入您的账号:Willard
请输入您的密码:JD584520
账号密码正确,登录成功!
---------------------------# 输出2:
请输入您的账号:Willard
请输入您的密码:jd584520
账号或密码输入错误,登录失败!
请重新登录!您还有2次输入账号密码的机会!
----------------------------------------
请重新输入您的账号:Willard
请重新输入您的密码:JD584520
账号密码正确,登录成功!--------------------------
# 输出3:
请输入您的账号:willard
请输入您的密码:JD584520
账号或密码输入错误,登录失败!
请重新登录!您还有2次输入账号密码的机会!
----------------------------------------
请重新输入您的账号:Willard
请重新输入您的密码:jd584520
账号或密码输入错误,登录失败!
请重新登录!您还有1次输入账号密码的机会!
----------------------------------------
请重新输入您的账号:willard
请重新输入您的密码:jd584520
账号或密码输入错误,登录失败!
2.字典
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# dict:字典,使用键-值对(key-value)存储; # 实例: studentScore = { "Willard" :{ "Math" : 100 , "Chinese" : 98 , "Eng" : 90 }, "ChenJD" :{ "Math" : 99 , "Chinese" : 100 , "Eng" : 98 }, "ChenBao" :{ "Math" : 100 , "Chinese" : 99 , "Eng" : 96 }} print ( "Willard的成绩是:" ,studentScore[ "Willard" ]) print ( "ChenJD的成绩是:" ,studentScore[ "ChenJD" ]) print ( "ChenBao的成绩是:" ,studentScore[ "ChenBao" ]) print ( "---------------------------------------------" ) # 修改元素的值 print ( "Willard原来的成绩是:" ,studentScore[ "Willard" ]) studentScore[ "Willard" ] = { "Math" : 100 , "Chinese" : 100 , "Eng" : 100 } print ( "修改后Willard的成绩是:" ,studentScore[ "Willard" ]) print ( "---------------------------------------------" ) # 判断key是否存在 # 1.通过in判断key是否存在 print ( "判断是否存在'ChenXiaoBao'这个key.\n" ) if "ChenXiaoBao" in studentScore: print ( "存在'ChenXiaoBao'这个key." ) else : print ( "不存在'ChenXiaoBao'这个key." ) print ( "---------------------------------------------" ) # 2.通过get()方法,如果key不存在,返回None,或指定的value print (studentScore.get( "Willard" )) print (studentScore.get( "Willard" ), - 1 ) print ( "---------------------------------------------" ) # 删除一个key,使用pop(key) print ( "删除前的字典:\n" ,studentScore, "\n" ) studentScore.pop( "ChenBao" ) print ( "删除后的字典:" ,studentScore) # Tips: # 字典的key必须是不可变对象,如:字符串、整数等,list是可变的; |
# 结果输出:
Willard的成绩是: {'Math': 100, 'Chinese': 98, 'Eng': 90}
ChenJD的成绩是: {'Math': 99, 'Chinese': 100, 'Eng': 98}
ChenBao的成绩是: {'Math': 100, 'Chinese': 99, 'Eng': 96}
---------------------------------------------
Willard原来的成绩是: {'Math': 100, 'Chinese': 98, 'Eng': 90}
修改后Willard的成绩是: {'Math': 100, 'Chinese': 100, 'Eng': 100}
---------------------------------------------
判断是否存在'ChenXiaoBao'这个key.不存在'ChenXiaoBao'这个key.
---------------------------------------------
{'Math': 100, 'Chinese': 100, 'Eng': 100}
{'Math': 100, 'Chinese': 100, 'Eng': 100} -1
---------------------------------------------
删除前的字典:
{'Willard': {'Math': 100, 'Chinese': 100, 'Eng': 100}, 'ChenJD': {'Math': 99, 'Chinese': 100, 'Eng': 98}, 'ChenBao': {'Math': 100, 'Chinese': 99, 'Eng': 96}}删除后的字典: {'Willard': {'Math': 100, 'Chinese': 100, 'Eng': 100}, 'ChenJD': {'Math': 99, 'Chinese': 100, 'Eng': 98}}
3.集合
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# 集合:set;是一组key的集合,但不存储value,且key不能重复,具有唯一性; # 1.创建一个set,提供一个list作为输入集合 setEg = set ([ 1 , 2 , 3 ]) print ( "集合setEg的内容:" ,setEg) print ( "------------------------------" ) # 2.集合的元素唯一性 setEg2 = set ([ 1 , 1 , 1 , 2 , 3 , 4 , 5 , 3 , 2 ]) print ( "集合setEg2的内容:" ,setEg2) print ( "------------------------------" ) # 3.添加元素 setEg = set ([ 1 , 2 , 3 ]) print ( "添加元素前集合的内容:" ,setEg) setEg.add( 5 ) print ( "添加元素后集合的内容:" ,setEg) print ( "------------------------------" ) # 4.删除元素 setEg = set ([ 1 , 2 , 3 ]) print ( "删除元素前集合的内容:" ,setEg) setEg.remove( 1 ) print ( "删除元素后集合的内容:" ,setEg) print ( "------------------------------" ) # 5.交集、并集 setOne = set ([ 1 , 2 , 3 , 4 , 5 ]) setTwo = set ([ 1 , 2 , 4 , 6 ]) print ( "setOne集合的内容:" ,setOne) print ( "setTwo集合的内容:" ,setTwo) print ( "setOne和setTwo的交集:" ,(setOne & setTwo)) print ( "setOne和setTwo的并集:" ,(setOne | setTwo)) |
# 结果输出:
集合setEg的内容: {1, 2, 3}
------------------------------
集合setEg2的内容: {1, 2, 3, 4, 5}
------------------------------
添加元素前集合的内容: {1, 2, 3}
添加元素后集合的内容: {1, 2, 3, 5}
------------------------------
删除元素前集合的内容: {1, 2, 3}
删除元素后集合的内容: {2, 3}
------------------------------
setOne集合的内容: {1, 2, 3, 4, 5}
setTwo集合的内容: {1, 2, 4, 6}
setOne和setTwo的交集: {1, 2, 4}
setOne和setTwo的并集: {1, 2, 3, 4, 5, 6}
总结
本篇文章就到这里了,希望能够给你带来帮助,也希望您能够多多关注服务器之家的更多内容!
原文链接:https://fuxi-willard.blog.csdn.net/article/details/122610808