而且我也试过这样做: l1 = l.strip().split(';') 但是Python 给我一个错误: AttributeError: 'list' object has no attribute 'strip' 因此,如果“list”对象没有属性“strip”或“split”,我该如何拆分列表? 谢谢 你想要做的是 - strtemp = ";".join(l) 第一行在;的末尾添加---MySpace,这样在拆分时,它不会给出MySpaceA...
AttributeError: 'int' object has no attribute 'strip' 翻译成中文意思是说，属性错误：整数对象没有...
a ='sashiaynshihai'print(a.strip('shi')) 结果是:ashiaynshiha 》 3.注意: >>>*** 复制代码 a ='shi an'c= a.split('')print(c)print(c.strip())#AttributeError: 'list' object has no attribute 'strip'报错的原因是:strip()不能作用在list上,应该是str类型 》4.例子: 复制代码 a ='ed...
} info_list.append(info) 运行时,由于html中cell_44、cell_45中是空值,程序返回以下错误: AttributeError: 'NoneType' object has no attribute 'strip' 请各位大佬帮助以上问题该如何解决?请回复详细代码,谢谢。
>>> f = open('ip.txt') >>> for ip in f.readlines().strip():... if ip.startswith('172.16'): ... print ip ... Traceback (most recent call last): File "<stdin>", line 1, in <module>AttributeError: 'list' object has no attribute 'strip' 验证了下改为 ...
Error:'list'object has no attribute'rmove'25>>>mylist26[1, 2,'abc', 4,'word','zizi','qwse', 1234, 123, 123]27#这个也是错误的28>>> mylist.remove(['abc',2])29Traceback (most recent call last):30File"<stdin>", line 1,in<module>31ValueError: list.remove(x): xnotinlist...
ob1 = "abc"ob2 = iter("abc")ob3 = iter("abc")# ob1它遍历for i in ob1:print(i, end = " ") # a b cfor i in ob1: print(i, end = " ") # a b c# ob1自遍历ob1.__next__() # 报错： 'str' object has no attribute '__next__'# ob2它遍历for i in ...
对于元素None,如果运行方法strip(),程序将报错,'NoneType' object has no attribute 'strip'。所幸的是None本身返回False。因此,对于元素None, s.strip()并不会运行,从而也不会报错。 另外,对于带两个空格的变量' ',该变量返回的布尔值是True,采用s.strip()方法后,得到的是空字符串(布尔值为False),所以,...
units = units.strip() AttributeError: 'numpy.float64' object has no attribute 'strip' During handling of the above exception, another exception occurred: Traceback (most recent call last): File "process.py", line 3, in f = cf.read("GEBCO_2019_SID.nc") ...
4 v3 = num.isnumeric() # AttributeError 'bytes' object has no attribute 'isnumeric' 5 print(v1,v2,v3) 1. 2. 3. 4. 5. 1 num = "②" # 非普通数字字符② 2 v1 = num.isdigit() # True 3 v2 = num.isdecimal() # False ...