Python常见的错误汇总
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错误:
TypeError: 'NoneType' object is not callable
【错误分析】我是在别的文件中写了一个函数,然后在python console调用使用该函数,出现了上述错误。原因是因为没有对别的文件中函数进行编译,只要打开该调用函数的文件,然后点击运行,之后再在python console中调用使用该函数就不会报错。
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错误:
【错误分析】第二个参数必须为类,否则会报TypeError,所以正确的应该是这样的:
但如果第二个参数是类型对象,则不会报上面的错误,是允许的,比如说:
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错误:
【错误分析】这个涉及到调用顺序问题,即解析方法的MRO调用顺序,在Python2.7版本之后,这样调用会报错,
必须是子类先放前面,然后才是父类.如下所示,方不会报错.
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错误:
【错误分析】foo()未带参数self,也未带cls参数,属于类的静态方法,类的静态方法调用,实例不能直接调用,需要再声明一个静态方法
或者通过@staticmethod来调用
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错误:
【错误分析】__dict__是实例的特殊属性,但在内建属性中,不存在__dict__属性,一般的情况是:
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错误:
【错误分析】如果定义了构造器,它不应当返回任何对象,因为实例对象是自动在实例化调用后返回的。相应地,__init__()就不应当返回任何对象(应当为None);否则就可能出现冲突,因为只能返回实例。试着返回非None的任何其他对象都会导致TypeError异常
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错误:
1 >>> def f(x, y): 2 print x, y 3 >>> t = ('a', 'b') 4 >>> f(t) 5 6 Traceback (most recent call last): 7 File "<pyshell#65>", line 1, in <module> 8 f(t) 9 TypeError: f() takes exactly 2 arguments (1 given)
【错误分析】不要误以为元组里有两个参数,将元组传进去就可以了,实际上元祖作为一个整体只是一个参数,
实际需要两个参数,所以报错。必需再传一个参数方可.
1 >>> f(t, 'var2') 2 ('a', 'b') var2
更常用的用法: 在前面加*,代表引用元组
1 >>> f(*t) 2 'a', 'b'
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错误:
1 >>> def func(y=2, x): 2 return x + y 3 SyntaxError: non-default argument follows default argument
【错误分析】在C++,Python中默认参数从左往右防止,而不是相反。这可能跟参数进栈顺序有关。
1 >>> def func(x, y=2): 2 return x + y 3 >>> func(1) 4 3
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错误:
1 >>> D1 = {'x':1, 'y':2} 2 >>> D1['x'] 3 1 4 >>> D1['z'] 5 6 Traceback (most recent call last): 7 File "<pyshell#185>", line 1, in <module> 8 D1['z'] 9 KeyError: 'z'
【错误分析】这是Python中字典键错误的提示,如果想让程序继续运行,可以用字典中的get方法,如果键存在,则获取该键对应的值,不存在的,返回None,也可打印提示信息.
1 >>> D1.get('z', 'Key Not Exist!') 2 'Key Not Exist!'
错误:
1 >>> from math import sqrt 2 >>> exec "sqrt = 1" 3 >>> sqrt(4) 4 5 Traceback (most recent call last): 6 File "<pyshell#22>", line 1, in <module> 7 sqrt(4) 8 TypeError: 'int' object is not callable
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错误:
1 >>> seq = [1, 2, 3, 4] 2 >>> sep = '+' 3 >>> sep.join(seq) 4 5 Traceback (most recent call last): 6 File "<pyshell#25>", line 1, in <module> 7 sep.join(seq) 8 TypeError: sequence item 0: expected string, int found
【错误分析】join是split的逆方法,是非常重要的字符串方法,但不能用来连接整数型列表,所以需要改成:
1 >>> seq = ['1', '2', '3', '4'] 2 >>> sep = '+' 3 >>> sep.join(seq) 4 '1+2+3+4'
错误:
1 >>> print r'C:\Program Files\foo\bar\' 2 SyntaxError: EOL while scanning string literal
【错误分析】Python中原始字符串以r开头,里面可以放置任意原始字符,包括\,包含在字符中的\不做转义。
但是,不能放在末尾!也就是说,最后一个字符不能是\,如果真 需要的话,可以这样写:
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代码:
1 bad = 'bad' 2 3 try: 4 raise bad 5 except bad: 6 print 'Got Bad!'
错误:
【错误分析】因所用的Python版本2.7,比较高的版本,raise触发的异常,只能是自定义类异常,而不能是字符串。所以会报错,字符串改为自定义类,就可以了。
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错误:
1 class Super: 2 def method(self): 3 print "Super's method" 4 5 class Sub(Super): 6 def method(self): 7 print "Sub's method" 8 Super.method() 9 print "Over..." 10 11 S = Sub() 12 S.method()
执行上面一段代码,错误如下:
【错误分析】Python中调用类的方法,必须与实例绑定,或者调用自身.
ClassName.method(x, 'Parm')
ClassName.method(self)
所以上面代码,要调用Super类的话,只需要加个self参数即可。
1 class Super: 2 def method(self): 3 print "Super's method" 4 5 class Sub(Super): 6 def method(self): 7 print "Sub's method" 8 Super.method(self) 9 print "Over..." 10 11 S = Sub() 12 S.method() 13 14 15 #输出结果 16 >>> 17 Sub's method 18 Super's method 19 Over...
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【错误分析】reload期望得到的是对象,所以该模块必须成功导入。在没导入模块前,不能重载.
1 >>> import sys 2 >>> reload(sys) 3 <module 'sys' (built-in)>
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错误
1 >>> def f(x, y, z): 2 return x + y + z 3 4 >>> args = (1,2,3) 5 >>> print f(args) 6 7 Traceback (most recent call last): 8 File "<pyshell#6>", line 1, in <module> 9 print f(args) 10 TypeError: f() takes exactly 3 arguments (1 given)
*args,才是将元组中的每个元素作为参数
1 >>> f(*args) 2 6
错误:
【错误分析】错误原因**匹配并收集在字典中所有包含位置的参数,但传递进去的却是个元祖。
所以修改传递参数如下:
1 >>> args = {'a':1,'b':2,'c':3} 2 >>> args['d'] = 4 3 >>> f(**args) 4 1 2 3 4
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【错误分析】在函数hider()内使用了内置变量open,但根据Python作用域规则LEGB的优先级:
先是查找本地变量==》模块内的其他函数==》全局变量==》内置变量,查到了即停止查找。
所以open在这里只是个字符串,不能作为打开文件来使用,所以报错,更改变量名即可。
可以导入__builtin__模块看到所有内置变量:异常错误、和内置方法
1 import __builtin__ 2 >>> dir(__builtin__) 3 ['ArithmeticError', 'AssertionError', 'AttributeError',...
........................................zip,filter,map]
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错误:
【错误分析】(1)的类型是整数,所以不能与另一个元祖做合并操作,如果只有一个元素的元祖,应该用(1,)来表示
1 In [108]: type(T1) 2 Out[108]: int 3 4 In [109]: T1 = (1,) 5 In [110]: T2 = (2,3) 6 In [111]: T1 + T2 7 Out[111]: (1, 2, 3)
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错误:
【错误分析】字典中的键必须是不可变对象,如(整数,浮点数,字符串,元组).
可用hash()判断某个对象是否可哈希
但列表中元素是可变对象,所以是不可哈希的,所以会报上面的错误.
如果要用列表作为字典中的键,最简单的办法是:
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错误:
【错误分析】列表属于可变对象,其append(),sort(),reverse()会在原处修改对象,不会有返回值,
或者说返回值为空,所以要实现反转并排序,不能并行操作,要分开来写。
或者用下面的方法实现:
1 In [103]: sorted(reversed([2,1,4,3])) 2 Out[103]: [1, 2, 3, 4]
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错误:
【错误分析】class是Python保留字,Python保留字不能做变量名,可以用Class,或klass
同样,保留字不能作为模块名来导入,比如说,有个and.py,但不能将其作为模块导入
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错误:
【错误分析】\n默认为换行,\t默认为TAB键.
所以在D:\目录下找不到ew目录下的ext.data文件,将其改为raw方式输入即可。
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错误:
【错误分析】错误原因,else, finally执行位置;正确的程序应该如下:
1 try: 2 print 1 / 0 3 4 except ZeroDivisionError: 5 print 'integer division or modulo by zero' 6 7 8 else: 9 print 'Continue Handle other part' 10 11 finally: 12 print 'Done'
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错误:
【错误分析】错误原因,列表解析中,x,y必须以数组的方式列出(x,y)
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报错如下:
1 >>> 2 secretCount is: 1 3 secretCount is: 2 4 5 6 Traceback (most recent call last): 7 File "D:\Learn\Python\Learn.py", line 13, in <module> 8 count1.__secretCount 9 AttributeError: JustCounter instance has no attribute '__secretCount'
【错误分析】双下划线的类属性__secretCount不可访问,所以会报无此属性的错误.
解决办法如下:
1 # 1. 可以通过其内部成员方法访问 2 # 2. 也可以通过访问 3 ClassName._ClassName__Attr 4 #或 5 ClassInstance._ClassName__Attr 6 #来访问,比如: 7 print count1._JustCounter__secretCount 8 print JustCounter._JustCounter__secretCount
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【错误分析】属性错误,归根到底在于元组是不可变类型,所以没有这几种方法.
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【错误分析】空元组和空列表,没有索引为0的项
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【错误分析】一般出在代码缩进的问题
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【错误分析】如果文件里面没有行了会报这种异常
有可迭代的对象的next方法,会前进到下一个结果,而在一系列结果的末尾时,会引发StopIteration的异常.
next()方法属于Python的魔法方法,这种方法的效果就是:逐行读取文本文件的最佳方式就是根本不要去读取。
取而代之的用for循环去遍历文件,自动调用next()去调用每一行,且不会报错
1 for line in open('test.txt','r'): 2 print line
1 >>> string = 'SPAM' 2 >>> a,b,c = string 3 Traceback (most recent call last): 4 File "<stdin>", line 1, in <module> 5 ValueError: too many values to unpack
【错误分析】接受的变量少了,应该是
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【错误分析】当映射到字典中的键不存在时候,就会触发此类异常, 或者可以,这样测试
1 >>> 'a' in mydic.keys() 2 True 3 >>> 'c' in mydic.keys() #用in做成员归属测试 4 False 5 >>> D.get('c','"c" is not exist!') #用get或获取键,如不存在,会打印后面给出的错误信息 6 '"c" is not exist!'
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【错误分析】一般是代码缩进问题,TAB键或空格键不一致导致
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1 >>>def A(): 2 return A() 3 >>>A() #无限循环,等消耗掉所有内存资源后,报最大递归深度的错误 4 File "<pyshell#2>", line 2, in A return A()RuntimeError: maximum recursion depth exceeded 5 class Bird: 6 def __init__(self): 7 self.hungry = True 8 def eat(self): 9 if self.hungry: 10 print "Ahaha..." 11 self.hungry = False 12 else: 13 print "No, Thanks!" 14 该类定义鸟的基本功能吃,吃饱了就不再吃 15 输出结果: 16 >>> b = Bird() 17 >>> b.eat() 18 Ahaha... 19 >>> b.eat() 20 No, Thanks! 21 下面一个子类SingBird, 22 class SingBird(Bird): 23 def __init__(self): 24 self.sound = 'squawk' 25 def sing(self): 26 print self.sound 27 输出结果: 28 >>> s = SingBird() 29 >>> s.sing() 30 squawk 31 SingBird是Bird的子类,但如果调用Bird类的eat()方法时, 32 >>> s.eat() 33 Traceback (most recent call last): 34 File "<pyshell#5>", line 1, in <module> 35 s.eat() 36 File "D:\Learn\Python\Person.py", line 42, in eat 37 if self.hungry: 38 AttributeError: SingBird instance has no attribute 'hungry'
【错误分析】代码错误很清晰,SingBird中初始化代码被重写,但没有任何初始化hungry的代码
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1 __metaclass__=type 2 class Bird: 3 def __init__(self): 4 self.hungry = True 5 def eat(self): 6 if self.hungry: 7 print "Ahaha..." 8 self.hungry = False 9 else: 10 print "No, Thanks!" 11 12 class SingBird(Bird): 13 def __init__(self): 14 super(SingBird,self).__init__() 15 self.sound = 'squawk' 16 def sing(self): 17 print self.sound 18 >>> S = SingBird() 19 >>> S. 20 SyntaxError: invalid syntax 21 >>> S. 22 SyntaxError: invalid syntax 23 >>> S.eat() 24 Ahaha...
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【错误】
1 >>> T 2 (1, 2, 3, 4) 3 >>> T[0] = 22 4 Traceback (most recent call last): 5 File "<pyshell#129>", line 1, in <module> 6 T[0] = 22 7 TypeError: 'tuple' object does not support item assignment
【错误分析】元组不可变,所以不可以更改;可以用切片或合并的方式达到目的.
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1 >>> X = 1; 2 >>> Y = 2; 3 >>> X + = Y 4 File "<stdin>", line 1 5 X + = Y 6 ^ 7 SyntaxError: invalid syntax
【错误分析】增强行赋值不能分开来写,必须连着写比如说 +=, *=
1 >>> X += Y 2 >>> X;Y 3 3 4 2
参考:http://blog.csdn.net/jerry_1126/article/details/39395899