if any(substring.lower() in my_str.lower() 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. 在检查每个...
def f1(list): s ="" for substring in list: s += substring return s # ✅ pythonic 的方法 @timeshow def f2(list): s = "".join(list) return s l = ["I", "Love", "Python"] * 1000 # 为了看到差异,我们把这个列表放大了 f1(l) f2(l) 运行输出: f1 : 0.000227 sec f2 : 0.00...
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...
参考资料:https://www.geeksforgeeks.org/python-finding-strings-with-given-substring-in-list/ 方法1: In [1]: data=["北京市","福建省","河南省","杭州市"] In [2]: word="福建" In [3]: [iforiindataifwordini] Out[3]: ['福建省'] 方法2: In [4]: data=["北京市","福建省","...
In [36]: import re ...: ...: # Create a Regex pattern to match the substring ...: regexPattern = re.compile("sample") ...: ...: # Get a list of strings that matches the given pattern i.e. substring ...: listOfMatches = regexPattern.findall(mainStr) ...: ...: print(...
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. >>>str1="abcdefg">>>str2="ababab">>>str1.find("bc")1>>>str2.find("b")1>>...
# 定义一个检查函数defcontains_substring(s,substring):returnsubstringins# 使用filter()函数查找包含特定字符串的元素result=list(filter(contains_substring,str_list,[search_str]))print(result)# 输出: ['strawberry', 'blueberry'] 1. 2. 3.
'replace', 'rfind', 'rindex', 'rjust', 'rpartition', 'rsplit', 'rstrip', 'split', 'splitlines', 'startswith', 'strip', 'swapcase', 'title', 'translate', 'upper', 'zfill'] Help on class str in module __builtin__: class str(basestring) | str(object='') -> string | | ...
argv 命令行参数list,第一个是程序本身的路径 path 返回模块的搜索路径 modules.keys() 返回已经导入的所有模块的列表 exit(0) 退出程序 a in s or b in s or c in s简写 采用any方式:all() 对于任何可迭代对象为空都会返回True # 方法一 Truein[iins...
print('Substring count =', s.count('Th')) Output: Find all indexes of substring There is no built-in function to get the list of all the indexes for the substring. However, we can easily define one using find() function. def find_all_indexes(input_str, substring): ...