下面是将以上四个步骤整合在一起的完整代码示例: defindex_of_substring(string,substring):string_list=list(string)forindex,charinenumerate(string_list):ifchar==substring[0]:# 判断当前字符是否与子字符串的第一个字符相同ifstring[index:index+len(substring)]==substring:# 判断当前字符后续字符是否与子字符...
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The index method can’t return a number because the substring isn’t there, so we get a value error instead: In order to avoid thisTraceback Error, we can use the keywordinto check if a substring is contained in a string. In the case of Loops, it was used for iteration, whereas in...
我们可以先看一个简单的例子: s ="Duan Yixuan"x=len(s)print('The Length of %s is %d'%(s,x))#输出结果:The Length of Duan Yixuan is 11 分析: 'The length of %s is %d'这部分叫做:格式控制符 (s,x)这部分叫做:转换说明符 %字符,表示标记转换说明符的开始,类似于C语言中的用的逗号 接下来...
# Function to find index of substring def find_Index(str1, pos): # Check if pos longer than str1 if len(pos) > len(str1): return 'Not found' # Iterate through str1 for i in range(len(str1)): # Iterate through pos for j in range(len(pos)): ...
2. Substring or slicing 通过使用slice语法,我们可以获得一系列字符。索引从零开始。 str[m:n] 从位置2(包括)到5(不包括)返回字符串。 从索引2到5的子字符串 str = 'hello world' print(str[2:5]) # llo 负切片将从末尾返回子字符串。 子串从索引-5到-2 ...
C:\Python27\python.exe D:/git/Python/FullStack/Study/index.py ['__add__','__class__','__contains__','__delattr__','__doc__','__eq__','__format__','__ge__','__getattribute__','__getitem__','__getnewargs__','__getslice__','__gt__','__hash__','__init...
ValueError: substring not found 解决办法: 对于不确定子字符串是否在被搜索字符串中存在,可以用find函数,如果不存在,它会返回-1. s = "hello, world" print(s.find("world!")) 35. ValueError: subsection not found 对于字节串,使用index来搜索子串,容易出现这个错误。
C:\Python27\python.exe D:/git/Python/FullStack/Study/index.py['__add__', '__class__', '__contains__', '__delattr__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__getitem__', '__getnewargs__', '__getslice__', '__gt__', '__...
) S.find(sub [,start [,end]]) -> int Return the lowest index in S where substring sub is found, such that sub is contained within S[start:end]. Optional arguments start and end are interpreted as in slice notation. Return -1 on failure. In [102]: s1.find("i") #元素第一次...