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  • 字符串知识 ---len ---count --- startswith --- endswith ---split ---strip ---replace ---isalnum ---isalpha ---isdecimal ---find ---index ---captalize,swapcase,title ---conter ---eval ---整数加法计算器

    # len   长度 迭代
    s1 = "nihaoma"
    print(len(s1)) #   list s1  的长度
    #  len 的运用
    i = 0
    while i < len(s1):
        print(s1[i])
        i+=1
    
        #s1.count          总  数    查看出现的次数
        #s1.startswith    判断是否以...开头
        #s1.endswith      判断是否以...结尾
    s1 = "nihaonihaonihao"
    re = s1.startswith("n",1,6)
    print(re)
    # False
    re = s1.count("n",1,6)
    print(re)
    # 1
    re = s1.count("n",1,5)
    print(re)
    # 0
    re = s1.count("ni")
    print(re)
    # 3
    
        # split      返还的列表  分割
    s1 = "nihaonihaonihao"
    re = s1.split("a")
    print(re)
    # ['nih', 'onih', 'onih', 'o']
    
        # strip   传出去掉空格
    s1 = " nihao "
    re = s1.strip()
    print(s1)
    print(re)
    #  nihao
    # nihao
    
         # replace   替换
    s1 = " *nihao* "
    re = s1.replace('*','#')
    print(re)
    # nihao#
    
       # isalnum()) #字符串由字母或数字组成
       # isalpha()) #字符串只由字母组成
       # isdecimal()) #字符串只由十进制组成
       # 判断是否有数字组成
    s1 = "nihao359k"
    re = s1.isalnum()   # True
    print(re)
    
    #    # find 查找       返回的找到的元素的索引位置,如果找不到返回-1
    s1 = "nihao359k"
    re = s1.find("a")  #在第三位 返 3
    re = s1.find("l")  #  不存在 返回 -1
    print(re)
    
    #    # index   返回的找到的元素的索引位置,找不到报错
    s1 = "nihao359k"
    re = s1.index('3')    #找到在第 5 位   5
    # re = s1.index('8')    #找不到 报错
    re = s1.index('3',2,55) # no ou of range  超出范围不会出错
    re = s1.index('3')
    print(re)
    
    
         #captalize,swapcase,title
    s2 = "nihao359k"
    print(s2.capitalize()) #首字母大写   ------   Nihao359k
    print(s2.swapcase()) #大小写翻转    -------  NIHAO359K
    s3 = "hi,i,m,a  rapo"
    print(s3.title())#每个单词的首字母大写   -----  Hi,I,M,A  Rapo
    
    
    #     # conter内同居中,总长度,空白处填充
    s2 = "nihao359k"
    re =s2.center(20,"3") #   s2.center(20,"3")  " "  只能填一个字符
    print(re)#    33333nihao359k333333
    print(len(re)) # 验证长度  20
    
         #Eval(str):官方解释为:将字符串str当成标准的表达式求值并返回计算结果
    print(eval('5+9'))#  14
    
    #      # 整数加法计算器
    content = input('>>>>').strip()
    i = content.split('+')
    k = 0
    for l in i :
        k+=int(l)
    print(k)
    
    
    # 使用while 循环和for循环分别打印字符串s ='miaoge'中每个元素
    s ='miaoge'
    
    for i in s:
         print(i)
    
    i = 0
    while i<=5:
        print(s[i])
        i+=1
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  • 原文地址:https://www.cnblogs.com/lxcai213/p/13281173.html
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