python编程中遇到的错误和解决方法集锦

python scott 322℃ 0评论

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:

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>>> 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.

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>>> f(t, 'var2')  

('a', 'b') var2

更常用的用法: 在前面加*,代表引用元祖

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>>> f(*t)  

'a', 'b'

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错误:

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>>> def func(y=2, x):  

    return x + y  

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。

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>>> def func(x, y=2):  

    return x + y  

>>> func(1)  

3

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错误:

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>>> D1 = {'x':1, 'y':2}  

>>> D1['x']  

1  

>>> D1['z']  

 

Traceback (most recent call last):  

  File "<pyshell#185>", line 1, 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.

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>>> D1.get('z', 'Key Not Exist!')  

'Key Not Exist!'

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错误:

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>>> from math import sqrt  

>>> exec "sqrt = 1"  

>>> sqrt(4)  

 

Traceback (most recent call last):  

  File "<pyshell#22>", line 1, 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 ,其中就是起到放置代码字符串命名空间的字典。

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>>> from math import sqrt  

>>> scope = {}  

>>> exec "sqrt = 1" in scope  

>>> sqrt(4)  

2.0

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错误:

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>>> seq = [1, 2, 3, 4]  

>>> sep = '+'  

>>> sep.join(seq)  

 

Traceback (most recent call last):  

  File "<pyshell#25>", line 1, 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:

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>>> seq = ['1', '2', '3', '4']  

>>> sep = '+'  

>>> sep.join(seq)  

'1+2+3+4'

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错误:

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>>> print r'C:Program Filesfoobar'  

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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>>> print r'C:Program Filesfoobar' "\"  

C:Program Filesfoobar  

>>> print r'C:Program Filesfoobar' + "
\"  

C:Program Filesfoobar

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代码:

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bad = 'bad'  

 

try:  

    raise bad  

except bad:  

    print 'Got Bad!'

错误:

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>>>  

 

Traceback (most recent call last):  

  File "D:LearnPythonLearn.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。

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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()

执行上面一段代码,错误如下:

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>>>  

Sub's method  

 

Traceback (most recent call last):  

  File "D:LearnPythontest.py", line 12, in <module>  

    S.method()  

  File "D:LearnPythontest.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.

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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。

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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(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>", line 1, 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.

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>>> import sys  

>>> reload(sys)  

<module 'sys' (built-in)>

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>>> def f(x, y, z):  

    return x + y + z  

 

>>> args = (1,2,3)  

>>> print f(args)  

 

Traceback (most recent call last):  

  File "<pyshell#6>", line 1, 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

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>>> f(*args)  

6

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>>> def f(a,b,c,d):  

...   print a,b,c,d  

...  

>>> args = (1,2,3,4)  

>>> f(**args)  

Traceback (most recent call last):  

  File "<stdin>", line 1, 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:

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>>> args = {'a':1,'b':2,'c':3}  

>>> args['d'] = 4  

>>> f(**args)  

1 2 3 4

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python编程中遇到的错误和解决方法集锦

python编程中遇到的错误和解决方法集锦

【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

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>>> 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

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In [108]: type(T1)  

Out[108]: int  

 

In [109]: T1 = (1,)  

In [110]: T2 = (2,3)  

In [111]: T1 + T2  

Out[111]: (1, 2, 3)

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>>> 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

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>>> 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:

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>>> 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>", 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

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>>> L = [2,1,4,3]  

>>> L.reverse()  

>>> L.sort()  

>>> L  

[1, 2, 3, 4]

或者用下面的方法实现:

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

Out[103]: [1, 2, 3, 4]

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>>> 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

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>>> import and  

SyntaxError: invalid syntax

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>>> f = open('D:newtext.data','r')  

Traceback (most recent call last):  

  File "<stdin>", line 1, in <module>  

IOError: [Errno 22] invalid mode ('r') or filename: 'D:newtext.data'  

>>> f = open(r'D:newtext.data','r')  

>>> f.read()  

'Veryngoodnaaaaa'

【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:

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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)

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>>> [(x,y) for x in range(2) for y in range(3)]  

[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (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:

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>>>  

secretCount is: 1  

secretCount is: 2  

 

 

Traceback (most recent call last):  

  File "D:LearnPythonLearn.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:

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# 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>", 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

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>>> 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.

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>>> 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

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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 = (1,2,3,4)  

  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

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>>> f = file('1.txt')  

>>> f.readline()  

'AAAAAn'  

>>> f.readline()  

'BBBBBn'  

>>> f.next()  

'CCCCCn'

【错误分析】如果文件里面没有行了会报这种异常

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>>> 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.

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

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for line in open('test.txt','r'):  

    print line

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>>> 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

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>>> 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')

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

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>>> mydic={'a':1,'b':2}  

>>> mydic['a']  

1  

>>> mydic['c']  

Traceback (most recent call last):  

  File "<stdin>", line 1, in &#63;  

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

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>>> '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:

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>>> b = Bird()  

>>> b.eat()  

Ahaha...  

>>> b.eat()  

No, Thanks!

下面一个子类SingBird,

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class SingBird(Bird):  

    def __init__(self):  

        self.sound = 'squawk'  

    def sing(self):  

        print self.sound

输出结果:

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>>> s = SingBird()  

>>> s.sing()  

squawk

SingBird是Bird的子类,但如果调用Bird类的eat()方法时,

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>>> s.eat()  

Traceback (most recent call last):  

  File "<pyshell#5>", line 1, in <module>  

    s.eat()  

  File "D:LearnPythonPerson.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

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class SingBird(Bird):  

    def __init__(self):  

        self.sound = 'squawk'  

        self.hungry = Ture #加这么一句  

    def sing(self):  

        print self.sound

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

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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:LearnPythonPerson.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

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__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...

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

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>>> 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.

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>>> T = (1,2,3,4)  

>>> (22,) + T[1:]  

(22, 2, 3, 4)

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

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>>> 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 +=, *=

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>>> X += Y  

>>> X;Y  

3  

2

python编程中遇到的错误和解决方法就是这样,欢迎大家参考。。。。

转载请注明:osetc.com » python编程中遇到的错误和解决方法集锦

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