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
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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'
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错误:
【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。
>>> def func(x,y=2):
return x + y
>>> func(1)
3
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
错误:
>>> D1 = {'x':1,'y':2}
>>> D1['x']
1
>>> 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.
>>> D1.get('z','Key Not Exist!')
'Key Not Exist!'
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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。
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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'
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错误:
【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\
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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!'
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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()
>>>
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...
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>>> 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)>
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>>> 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)
6
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>>> 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
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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
>>> import __builtin__
>>> dir(__builtin__)
['ArithmeticError','AssertionError','AttributeError',..
.........................................zip,filter,map]
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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)
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>>> 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}
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>>> 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
或者用下面的方法实现:
In [103]: sorted(reversed([2,3]))
Out[103]: [1,4]
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【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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>>> 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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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'
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>>> [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)]
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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
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>>> print x
Traceback (most recent call last):
File "<stdin>",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
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>>> 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.
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>>> 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
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>>> 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
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>>> f = file('1.txt')
>>> f.readline()
'AAAAA\n'
>>> f.readline()
'BBBBB\n'
>>> f.next()
'CCCCC\n'
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
for line in open('test.txt','r'):
print line
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>>> 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')
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>>> mydic={'a':1,'b':2}
>>> mydic['a']
1
>>> 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!'
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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
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>>>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,
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...
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>>> 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)
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
【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
3
2
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