deffind_all_occurrences(string,sub_string):positions=[]start=0whileTrue:position=string.find(sub_string,start)ifposition==-1:breakpositions.append(position)start=position+1returnpositions# 示例text="Hello, hello,
deffind_all_occurrences(string,sub_string):occurrences=[]start=0whileTrue:index=string.find(sub_string,start)ifindex==-1:breakoccurrences.append(index)start=index+1returnoccurrences# 示例用法string="abracadabra"sub_string="a"occurrences=find_all_occurrences(string,sub_string)print(occurrences)# 输出 ...
如果找不到，则返回-1，实例代码如下：>>>string='笨鸟工具，x1y1z1.com'>>>string.find('笨鸟')...
replace(old, new[, count]) -> string Return a copy of string S with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced. """ return "" def rfind(self, sub, start=None, end=None): """ S.rfind(...
print("all occurrences of number:", re.findall("d+", s)) # 输出:all occurrences of number: ['123'] 使用第三方库fuzzywuzzy进行模糊匹配 from fuzzywuzzy import fuzz s1 = "Hello world" s2 = "hello world" print(fuzz.ratio(s1, s2)) # 输出:90 ...
计算一串字符串中每个字符出现的次数 import java.util.HashMap; import java.util.Scanner; public class demo { public static void main(String[] args) { //1、使用Scanner获取用户输入的字符串 Scanner scanner = new Scanner(System.in); System.out.println("请输入字符串："
>>> str='string learn' >>> str.startswith('str') #判断字符串以'str'开头 True >>> str.endswith('arn') #判读字符串以'arn'结尾 True 字符串搜索定位与替换 >>> str='string lEARn' >>> >>> str.find('a') #查找字符串,没有则返回-1,有则返回查到到第一个匹配的索引 ...
>>> symbols[-2] # Negative indices are from end of string ? >>> 在Python 语言中,字符串是只读的。尝试通过将 symbols 字符串的第一个字符变为小写字母 ‘a’ 来验证这一点。>>> symbols[0] = 'a' Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: '...
concatenate ordinary characters, so last matches the string 'last'. The special characters are: "." Matches any character except a newline. "^" Matches the start of the string. "$" Matches the end of the string or just before the newline at ...
sub) Help on function sub in module re: sub(pattern, repl, string, count=0, flags=0) Return the string obtained by replacing the leftmost non-overlapping occurrences of the pattern in string by the replacement repl. repl can be either a string or a callable; if a string, backslash ...