substring = 'z' result = any(substring.lower() in word.lower() for word in my_list) print(result) # 👉️ False if any(substring.lower() in word.lower() for word in my_list): # 👇️ this runs print('The substring is contained in at least one of the list items') else:...
例2，源字符串为：', string, ' 待查找字符串为：', sub_string)print('子字符串是否包含：', find_substring_regex(string, sub_string))在这个函数中，我们使用了re模块来编译一个正则表达式，然后使用search方法查找sub在s中的位置。如果sub出现了，返回True；否则返回False。执行结果为：总结 通过以上三种方...
found_substrings = [] for substring in substrings: if substring in text: found_substrings.append(substring) if found_substrings: print(f"The following substrings were found: {', '.join(found_substrings)}") else: print("No substrings were found.") 1. 2. 3. 4. 5. 6. 7. 8. 9...
# The string string = 'Hello World, this is a string' # The substring we are looking for substring = 'this' print(string.find(substring) # print:13, 13是字字符串的起始位置 使用正则表达式 re.search() 正则表达式是Python中非常强大的一个模块,我们用正则表达式同样可以查找子字符串(有点杀鸡用...
而S.find()在找不到substring时,不会报错,而会返回-1 list 和 tuples获取索引: list.index(),tuple.index() ist和tuple有一个索引方法来获取元素的位置 : alist = [10, 16, 26, 5, 2, 19, 105, 26] #返回元素为16的索引值 print(alist.index(16)) # 注意,这里是通过元素查找索引 print(alist...
substring=string[:3] print(substring) # Print"All", 停止索引不包含index(3)位置的字符 Udemy 2022 年完整 Python 开发课程:从零到精通 https://www.koudaizy.com/tutorials/complete-python-developer-zero-to-mastery/ 获取字符串的中间部分 同时设置起始索引和停止索引,你可以获得字符串的中间部分。例子: ...
Like S.find() but raise ValueError when the substring is not found. >>>str1="abcdefg">>>str1.index("a")0>>>str1.index("f")5 find S.find(sub[, start[, end]]) -> int #在字符串里查找指定的子串,未找到时返回-1,找到则返回子串在字符串中的索引值 ...
Return a copy of S with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced. 替换 1>>> a='1a2a3a4a5a'2>>> a.replace('a','b')3'1b2b3b4b5b4>>> a.replace('a','b',3)5'0b1b2b3a4a5a #替...
text ="Hello, world!"substring ="world"ifsubstringintext:print(f"{substring} is found in the text.")else:print(f"{substring} is not found in the text.") 使用循环和条件判断查找多个子字符串 如果要查找多个子字符串,可以使用循环和条件判断来实现。以下是一个示例: ...
count(sub[, start[, end]]) -> int 32 33 Return the number of non-overlapping occurrences of substring sub in 34 string S[start:end]. Optional arguments start and end are interpreted 35 as in slice notation. 36 """ 37 return 0 38 39 def decode(self, encoding=None, errors=None): ...