码迷,mamicode.com
首页 > 编程语言 > 详细

Python错误汇总

时间:2014-12-20 11:46:07      阅读:341      评论:0      收藏:0      [点我收藏+]

标签:python

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!技术分享

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

>>> hash(1,(2,[3,4]))

Traceback (most recent call last):
  File "<pyshell#95>", line 1, in <module>
    hash((1,2,(2,[3,4])))
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>", line 1, in <module>
AttributeError: 'NoneType' object has no attribute 'sort'
>>> L
[3, 4, 1, 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 = [2,1,4,3]
>>> L.reverse()
>>> L.sort()
>>> L
[1, 2, 3, 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>", line 1, 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。
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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), (0, 2), (1, 0), (1, 1), (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,所以会报无此属性的错误. 

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>", line 1, in <module>
NameError: name 'x' is not defined
>>> x = 1
>>> print x
1
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>", line 1, in <module>
AttributeError: 'tuple' object has no attribute 'append'
>>> t.remove(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'tuple' object has no attribute 'remove'
>>> t.pop()
Traceback (most recent call last):
  File "<stdin>", line 1, 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>", line 1, in <module>
IndexError: tuple index out of range
>>> l = []
>>> l[0]
Traceback (most recent call last):
  File "<stdin>", line 1, 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 = (1,2,3,4)
  File "<stdin>", line 1
    t = (1,2,3,4)
    ^
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>", line 1, 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.

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

>>> string = 'SPAM'
>>> a,b,c = string
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: too many values to unpack
【错误分析】接受的变量少了,应该是
>>> a,b,c,d = string
>>> a,d
('S', 'M')
#除非用切片的方式
>>> a,b,c = string[0],string[1],string[2:]
>>> a,b,c
('S', 'P', 'AM')
或者
>>> a,b,c = list(string[:2]) + [string[2:]]
>>> a,b,c
('S', 'P', 'AM')
或者
>>> (a,b),c = string[:2],string[2:]
>>> a,b,c
('S', 'P', 'AM')
或者
>>> ((a,b),c) = ('SP','AM')
>>> a,b,c
('S', 'P', 'AM')

简单点就是:
>>> a,b = string[:2]
>>> c   = string[2:]
>>> a,b,c
('S', 'P', 'AM')
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
>>> mydic={'a':1,'b':2}
>>> mydic['a']
1
>>> mydic['c']
Traceback (most recent call last):
  File "<stdin>", line 1, 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
    ^
IndentationError: 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!"
该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>", line 1, 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>", line 1, 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, Thanks!"

class SingBird(Bird):
    def __init__(self):
        super(SingBird,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, 2, 3, 4)
>>> T[0] = 22 
Traceback (most recent call last):
  File "<pyshell#129>", line 1, 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 = (1,2,3,4)
>>> (22,) + T[1:]
(22, 2, 3, 4)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

Python错误汇总

标签:python

原文地址:http://blog.csdn.net/jerry_1126/article/details/39395899

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!