def find_substring_in(s, sub):""" 使用in关键字查找子字符串 """if sub in s:return Trueelse:return False# 定义一个字符串string = 'A New String Hello, World!'sub_string = "Hello"print('例1，源字符串为：', string, ' 待查找字符串为：', sub_string)print('子字符串是否包含：...
if any(substring in my_str for substring in my_list): # 👇️ this runs print('The string contains at least one element from the list') else: print('The string does NOT contain any of the elements in the list') 1. 2. 3. 4. 5. 6. 7. 8. 9. 如果需要检查列表中的任何元素...
使用in 运算符 如果字符串中存在子字符串,则 in 运算符将返回 True,否则返回 False。这是一个例子: # The string string = 'Hello World, this is a string' # substring substring = 'this' if substring in string: print('Found the substring!') else: print('Could not find the substring.') 使...
方法一:使用in运算符 Python中的in运算符用于检查一个字符串是否是另一个字符串的子字符串。可以使用in运算符来检查一个字符串是否包含多个子字符串。下面是一个简单的示例: string="Hello, world!"substrings=["Hello","world"]forsubstringinsubstrings:ifsubstringinstring:print(f"The string '{string}' con...
count(value) - value to search for in the string. count(value, start, end) - value to search for in the string, where search starts from start position till end position. 字符串数() txt = "hello world" print( txt.count("o") ) # 2 ...
substring2 =string[:-3] print(substring2) # Print"The quick brown " 获取字符串中的一个字符 这个很简单,如果切片中没有:字符,只包含数字,它将返回该索引处的字符。 例子: string='0123456789' substring =string[4] print(substring) # Print "4" ...
Note that find() function returns the index position of the substring if it’s found, otherwise it returns -1. count() functionto find the number of occurrences of a substring in the string. s = 'My Name is Pankaj' print('Substring count =', s.count('a')) ...
count(value)-valuetosearchforinthe string.count(value,start,end)-valuetosearchforinthe string,wheresearchstartsfromstartposition tillendposition. 字符串数() txt ="hello world"print( txt.count("o") )# 2print( txt.count("o", 4, 7) )# 1 ...
ascii_digits = string.digits# Output: 0123456789 forone_digitinascii_digits[:5]:# Loop through 01234 print(ord(one_digit)) Output: 48 49 50 51 52 在上面的代码片段中,我们遍历字符串 ABCDE 和 01234,并将每个字符转换为它们在 ASCII 表中的十进制表示。我们还可以使用 chr 函数执行反向操作,从而将...
count(sub[, start[, end]]) -> int Return the number of non-overlapping occurrences of substring sub in string S[start:end]. Optional arguments start and end are interpreted as in slice notation. """ return 0 def decode(self, encoding=None, errors=None): """ 解码 """ """ S....