自己编程中遇到的Python错误和解决方法汇总整理

开个贴,用于记录平时经常碰到的Python的错误同时对导致错误的原因进行分析,并持续更新,方便以后查询,学习。
知识在于积累嘛!微笑
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
错误:


>>> def f(x,y): 
    print x,y 
>>> t = ('a','b') 
>>> f(t) 
 
Traceback (most recent call last): 
  File "<pyshell#65>",line 1,in <module> 
    f(t) 
TypeError: f() takes exactly 2 arguments (1 given) 

错误分析】不要误以为元祖里有两个参数,将元祖传进去就可以了,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.

>>> f(t,'var2') 
('a','b') var2 

更常用的用法: 在前面加*,代表引用元祖

>>> f(*t) 
'a','b' 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
错误:

>>> def func(y=2,x): 
    return x + y 
SyntaxError: non-default argument follows default argument 

错误分析】在C++,Python中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。

>>> def func(x,y=2): 
    return x + y 
>>> func(1) 


+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

错误:


>>> D1 = {'x':1,'y':2} 
>>> D1['x'] 

>>> D1['z'] 
 
Traceback (most recent call last): 
  File "<pyshell#185>",in <module> 
    D1['z'] 
KeyError: 'z' 

错误分析】这是Python中字典键错误提示,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.

>>> D1.get('z','Key Not Exist!') 
'Key Not Exist!' 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

错误:


>>> from math import sqrt 
>>> exec "sqrt = 1" 
>>> sqrt(4) 
 
Traceback (most recent call last): 
  File "<pyshell#22>",in <module> 
    sqrt(4) 
TypeError: 'int' object is not callable 

错误分析】exec语句最有用的地方在于动态地创建代码字符串,但里面存在的潜在的风险,它会执行其他地方的字符串,在CGI中更是如此!比如例子中的sqrt = 1,从而改变了当前的命名空间,从math模块中导入的sqrt不再和函数名绑定而是成为了一个整数。要避免这种情况,可以通过增加in <scope>,其中<scope>就是起到放置代码字符串命名空间的字典。

>>> from math import sqrt 
>>> scope = {} 
>>> exec "sqrt = 1" in scop
>>> sqrt(4) 
2.0 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
错误:

>>> seq = [1,2,3,4] 
>>> sep = '+' 
>>> sep.join(seq) 
 
Traceback (most recent call last): 
  File "<pyshell#25>",in <module> 
    sep.join(seq) 
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:


>>> seq = ['1','2','3','4'] 
>>> sep = '+' 
>>> sep.join(seq) 
'1+2+3+4' 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

错误:


>>> print r'C:\Program Files\foo\bar\' 
SyntaxError: EOL while scanning string litera

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:

>>> print r'C:\Program Files\foo\bar' "\\" 
C:\Program Files\foo\bar\ 
>>> print r'C:\Program Files\foo\bar' + "\\" 
C:\Program Files\foo\bar\ 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
代码:

bad = 'bad' 
 
try: 
    raise bad 
except bad: 
    print 'Got Bad!' 

错误:

>>>  
 
Traceback (most recent call last): 
  File "D:\Learn\Python\Learn.py",line 4,in <module> 
    raise bad 
TypeError: exceptions must be old-style classes or derived from BaseException,not str

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。


class Bad(Exception): 
    pass 
 
def raiseException(): 
    raise Bad() 
 
try: 
    raiseException() 
except Bad: 
    print 'Got Bad!' 
 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

class Super: 
    def method(self): 
        print "Super's method" 
 
class Sub(Super): 
    def method(self): 
        print "Sub's method" 
        Super.method() 
        print "Over..." 
 
S = Sub() 
S.method() 

执行上面一段代码错误如下:


>>>  
Sub's method 
 
Traceback (most recent call last): 
  File "D:\Learn\Python\test.py",line 12,in <module> 
    S.method() 
  File "D:\Learn\Python\test.py",line 8,in method 
    Super.method() 
TypeError: unbound method method() must be called with Super instance as first argument (got nothing instead) 

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。

class Super: 
    def method(self): 
        print "Super's method" 
 
class Sub(Super): 
    def method(self): 
        print "Sub's method" 
        Super.method(self) 
        print "Over..." 
 
S = Sub() 
S.method() 
 
 
#输出结果 
>>>  
Sub's method 
Super's method 
Over... 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> reload(sys) 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
NameError: name 'sys' is not defined 

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.

>>> import sys 
>>> reload(sys) 
<module 'sys' (built-in)> 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> def f(x,y,z): 
    return x + y + z 
 
>>> args = (1,3) 
>>> print f(args) 
 
Traceback (most recent call last): 
  File "<pyshell#6>",in <module> 
    print f(args) 
TypeError: f() takes exactly 3 arguments (1 given)

错误分析】args是一个元祖,如果是f(args),那么元祖是作为一个整体作为一个参数
*args,才是将元祖中的每个元素作为参数


>>> f(*args) 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> def f(a,b,c,d): 
...   print a,d 
... 
>>> args = (1,4) 
>>> f(**args) 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
TypeError: f() argument after ** must be a mapping,not tuple 

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:


>>> args = {'a':1,'b':2,'c':3} 
>>> args['d'] = 4 
>>> f(**args) 
1 2 3 4 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

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


>>> import __builtin__
>>> dir(__builtin__)
['ArithmeticError','AssertionError','AttributeError',..
  .........................................zip,filter,map]
 

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

In [105]: T1 = (1) 
In [106]: T2 = (2,3) 
In [107]: T1 + T2 
--------------------------------------------------------------------------- 
TypeError                                 Traceback (most recent call last) 
<ipython-input-107-b105c7b32d90> in <module>() 
----> 1 T1 + T2; 
 
TypeError: unsupported operand type(s) for +: 'int' and 'tuple' 

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

In [108]: type(T1) 
Out[108]: int 
 
In [109]: T1 = (1,) 
In [110]: T2 = (2,3) 
In [111]: T1 + T2 
Out[111]: (1,3) 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> hash(1,(2,[3,4])) 
 
Traceback (most recent call last): 
  File "<pyshell#95>",in <module> 
    hash((1,4]))) 
TypeError: unhashable type: 'list' 

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,元祖).
可用hash()判断某个对象是否可哈希


>>> hash('string') 
-1542666171 

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:

>>> D = {} 
>>> D[tuple([3,4])] = 5 
>>> D 
{(3,4): 5} 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> L = [2,1,4,3] 
>>> L.reverse().sort() 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
AttributeError: 'nonetype' object has no attribute 'sort' 
>>> L 
[3,2] 

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


>>> L = [2,3] 
>>> L.reverse() 
>>> L.sort() 
>>> L 
[1,4] 

或者用下面的方法实现:

In [103]: sorted(reversed([2,3])) 
Out[103]: [1,4] 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> class = 78 
SyntaxError: invalid Syntax 

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


>>> import and 
SyntaxError: invalid Syntax 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> f = open('D:\new\text.data','r') 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
IOError: [Errno 22] invalid mode ('r') or filename: 'D:\new\text.data' 
>>> f = open(r'D:\new\text.data','r') 
>>> f.read() 
'Very\ngood\naaaaa' 

错误分析】\n认为换行,\t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。
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


try: 
    print 1 / 0 
     
except ZeroDivisionError: 
    print 'integer division or modulo by zero' 
     
finally: 
    print 'Done' 
 
else:   
    print 'Continue Handle other part' 
报错如下: 
D:\>python Learn.py 
  File "Learn.py",line 11 
    else: 
       ^ 
SyntaxError: invalid Syntax 

错误分析】错误原因,else,finally执行位置;正确的程序应该如下:


try: 
    print 1 / 0 
     
except ZeroDivisionError: 
    print 'integer division or modulo by zero' 
 
 
else:   
    print 'Continue Handle other part' 
     
finally: 
    print 'Done' 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> [x,y for x in range(2) for y in range(3)] 
  File "<stdin>",line 1 
    [x,y for x in range(2) for y in range(3)] 
           ^ 
SyntaxError: invalid Syntax 

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)

>>> [(x,y) for x in range(2) for y in range(3)] 
[(0,0),(0,1),2),(1,2)] 
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
class JustCounter: 
    __secretCount = 0 
 
    def count(self): 
        self.__secretCount += 1 
        print 'secretCount is:',self.__secretCount 
 
count1 = JustCounter() 
 
count1.count() 
count1.count() 
 
count1.__secretCount 

报错如下:


>>>  
secretCount is: 1 
secretCount is: 2 
 
 
Traceback (most recent call last): 
  File "D:\Learn\Python\Learn.py",line 13,in <module> 
    count1.__secretCount 
AttributeError: JustCounter instance has no attribute '__secretCount' 

错误分析】双下划线的类属性__secretCount不可访问,所以会报无此属性错误.

解决办法如下:


# 1. 可以通过其内部成员方法访问 
# 2. 也可以通过访问 
ClassName._ClassName__Attr 
#或  
ClassInstance._ClassName__Attr 
#来访问,比如: 
print count1._JustCounter__secretCount 
print JustCounter._JustCounter__secretCount  

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> print x 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
NameError: name 'x' is not defined 
>>> x = 1 
>>> print x 

错误分析】Python不允许使用未赋值变量
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> t = (1,2) 
>>> t.append(3) 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
AttributeError: 'tuple' object has no attribute 'append' 
>>> t.remove(2) 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
AttributeError: 'tuple' object has no attribute 'remove' 
>>> t.pop() 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
AttributeError: 'tuple' object has no attribute 'pop' 

错误分析】属性错误,归根到底在于元祖是不可变类型,所以没有这几种方法.
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> t = () 
>>> t[0] 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
IndexError: tuple index out of range 
>>> l = [] 
>>> l[0] 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
IndexError: list index out of range 

错误分析】空元祖和空列表,没有索引为0的项
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> if X>Y: 
...  X,Y = 3,4 
...   print X,Y 
  File "<stdin>",line 3 
    print X,Y 
    ^ 
IndentationError: unexpected indent 
 
 
>>>   t = (1,4) 
  File "<stdin>",line 1 
    t = (1,4) 
    ^ 
IndentationError: unexpected indent 

错误分析】一般出在代码缩进的问题
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> f = file('1.txt') 
>>> f.readline() 
'AAAAA\n' 
>>> f.readline() 
'BBBBB\n' 
>>> f.next() 
'CCCCC\n' 

错误分析】如果文件里面没有行了会报这种异常


>>> f.next() # 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
stopiteration

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

for line in open('test.txt','r'): 
    print line 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>> string = 'SPAM' 
>>> a,c = string 
Traceback (most recent call last): 
  File "<stdin>",in <module> 
ValueError: too many values to unpack 

错误分析】接受的变量少了,应该是

>>> a,d = string 
>>> a,d 
('S','M') 
#除非用切片的方式 
>>> a,c = string[0],string[1],string[2:] 
>>> a,c 
('S','P','AM') 
或者 
>>> a,c = list(string[:2]) + [string[2:]] 
>>> a,'AM') 
或者 
>>> (a,b),c = string[:2],'AM') 
或者 
>>> ((a,c) = ('SP','AM') 
>>> a,'AM') 
 
简单点就是: 
>>> a,b = string[:2] 
>>> c   = string[2:] 
>>> a,'AM') 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> mydic={'a':1,'b':2} 
>>> mydic['a'] 

>>> mydic['c'] 
Traceback (most recent call last): 
  File "<stdin>",in ? 
KeyError: 'c' 

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

>>> 'a' in mydic.keys() 
True 
>>> 'c' in mydic.keys()              #用in做成员归属测试 
False 
>>> 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 
'"c" is not exist!' 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

File "study.py",line 3 
  return None 
  ^ 
dentationError: unexpected indent 

错误分析】一般是代码缩进问题,TAB键或空格键不一致导致

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


>>>def A(): 
return A() 
>>>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   
File "<pyshell#2>",line 2,in A return A()RuntimeError: maximum recursion depth exceeded 
class Bird: 
    def __init__(self): 
        self.hungry = True 
    def eat(self): 
        if self.hungry: 
            print "Ahaha..." 
            self.hungry = False 
        else: 
            print "No,Thanks!" 

类定义鸟的基本功能吃,吃饱了就不再吃 
输出结果: 

>>> b = Bird() 
>>> b.eat() 
Ahaha... 
>>> b.eat() 
No,Thanks! 

下面一个子类SingBird, 

class SingBird(Bird): 
    def __init__(self): 
        self.sound = 'squawk' 
    def sing(self): 
        print self.sound
  
输出结果: 

>>> s = SingBird() 
>>> s.sing() 
squawk 

SingBird是Bird的子类,但如果调用Bird类的eat()方法时, 


>>> s.eat() 
Traceback (most recent call last): 
  File "<pyshell#5>",in <module> 
    s.eat() 
  File "D:\Learn\Python\Person.py",line 42,in eat 
    if self.hungry: 
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


class SingBird(Bird): 
    def __init__(self): 
        self.sound = 'squawk' 
        self.hungry = Ture #加这么一句 
    def sing(self): 
        print self.sound 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++


class Bird: 
    def __init__(self): 
        self.hungry = True 
    def eat(self): 
        if self.hungry: 
            print "Ahaha..." 
            self.hungry = False 
        else: 
            print "No,Thanks!" 
 
class SingBird(Bird): 
    def __init__(self): 
        super(SingBird,self).__init__() 
        self.sound = 'squawk' 
    def sing(self): 
        print self.sound 
>>> sb = SingBird() 
Traceback (most recent call last): 
  File "<pyshell#5>",in <module> 
    sb = SingBird() 
  File "D:\Learn\Python\Person.py",line 51,in __init__ 
    super(SingBird,self).__init__() 
TypeError: must be type,not classobj 

错误分析】在模块首行里面加上__Metaclass__=type,具体还没搞清楚为什么要加

__Metaclass__=type 
class Bird: 
    def __init__(self): 
        self.hungry = True 
    def eat(self): 
        if self.hungry: 
            print "Ahaha..." 
            self.hungry = False 
        else: 
            print "No,self).__init__() 
        self.sound = 'squawk' 
    def sing(self): 
        print self.sound 
>>> S = SingBird() 
>>> S. 
SyntaxError: invalid Syntax 
>>> S. 
SyntaxError: invalid Syntax 
>>> S.eat() 
Ahaha... 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> T 
(1,4) 
>>> T[0] = 22  
Traceback (most recent call last): 
  File "<pyshell#129>",in <module> 
    T[0] = 22 
TypeError: 'tuple' object does not support item assignment

错误分析】元祖不可变,所以不可以更改;可以用切片或合并的方式达到目的.

>>> T = (1,4) 
>>> (22,) + T[1:] 
(22,4) 

+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

>>> X = 1; 
>>> Y = 2; 
>>> X + = Y 
  File "<stdin>",line 1 
    X + = Y 
        ^ 
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 +=,*=

>>> X += Y 
>>> X;Y 


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