ValueError: substring not found >>> str.index("n") #同find类似,返回第一次匹配的索引值 4 >>> str.rindex("n") #返回最后一次匹配的索引值 11 >>> str.count('a') #字符串中匹配的次数 0 >>> str.count('n') #同上 2 >>> str.replace('EAR','ear') #匹配替换 'string learn' >...
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(...
Return the lowest index in the string where substring sub is found, such that sub is contained in the slice s[start:end]. Optional arguments start and end are interpreted as in slice notation. Return -1 if sub is not found.Note: The find() method should be used only if you need to ...
参考:如何使用Python将字符串复制到Windows上的剪贴板上?实现代码如下:from Tkinter import Tk r = Tk() r.withdraw() r.clipboard_clear() r.clipboard_append('复制到剪贴板的内容') r.update() r.destroy()写成函数的形式:def addToClipboard( string ): ...
a nice string representation of the object. | If the argument is a string, the return value is the same object. | | Method resolution order: | str | basestring | object | | Methods defined here: | | __add__(...) | x.__add__(y) <==> x+y | | __contains__(...) | x...
substring = string1[0:5] # 结果为 'Hello' (6)长度(Length) 使用len()函数可以获取字符串的长度。 length_of_string = len(combined_string) # 结果为 13 (7)大小写转换 使用字符串的方法可以将字符串转换为大写或小写。 uppercase_string = combined_string.upper() # 结果为 'HELLO, WORLD!' ...
Return a copy of the string with all occurrences of substring old replaced by new. If the optional argument count is given, only the first count occurrences are replaced. >>> b='abc123efe12abcde1212' >>> b.replace('12','AB') 'abcAB3efeABabcdeABAB' >>> b.replace('12','AB',1)...
If the separator is not found, returns a 3-tuple containing the original string and two empty strings. """ pass def replace(self, *args, **kwargs): # real signature unknown """ Return a copy with all occurrences of substring old replaced by new. ...
) 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...
return[False,True][len(strString)>6] importtimeit print(timeit.timeit('isLen1("5fsdfsdfsaf")',setup="from __main__ import isLen1")) print(timeit.timeit('isLen("5fsdfsdfsaf")',setup="from __main__ import isLen")) contextlib ...