Ten common mistakes Python programmers make-Python Tutorial-php.cn

No matter in the process of studying or working, people will make mistakes. Although Python's syntax is simple and flexible, there are still some big pitfalls. If you are not careful, both beginners and experienced Python programmers may stumble. This article shares 10 common mistakes with you. Friends in need can refer to them

Common mistake 1: Wrongly using an expression as the default parameter of a function

In Python, we can set a default value for a certain parameter of the function. Make this parameter optional. While this is a nice language feature, it can also lead to some confusing situations when the default value is a mutable type. Let’s take a look at the following Python function definition:

>>> def foo(bar=[]):    # bar是可选参数，如果没有提供bar的值，则默认为[]，
...  bar.append("baz")  # 但是稍后我们会看到这行代码会出现问题。
...  return bar

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A common mistake Python programmers make is to take it for granted that every time a function is called, if no value is passed in for the optional parameter, then the optional parameter will will be set to the specified default value. In the above code, you may think that repeatedly calling the foo() function should always return 'baz', because by default, every time the foo() function is executed (the value of the bar variable is not specified), the bar variable is set to [ ] (that is, a new empty list).

However, the actual running result is like this:

>>> foo()
["baz"]
>>> foo()
["baz", "baz"]
>>> foo()
["baz", "baz", "baz"]

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Isn’t it strange? Why is the default value "baz" added to the existing list every time the foo() function is called, instead of creating a new empty list?

The answer is that the setting of the default value of the optional parameter will only be executed once in Python, that is, when the function is defined. Therefore, only when the foo() function is defined, the bar parameter will be initialized to the default value (that is, an empty list), but every time the foo() function is called thereafter, the bar parameter will continue to be initialized with the original value. of that list.

Of course, a common solution is:

>>> def foo(bar=None):
...  if bar is None:  # or if not bar:
...    bar = []
...  bar.append("baz")
...  return bar
...
>>> foo()
["baz"]
>>> foo()
["baz"]
>>> foo()
["baz"]

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FAQ 2: Incorrect use of class variables

Let’s look at the following example:

>>> class A(object):
...   x = 1
...
>>> class B(A):
...   pass
...
>>> class C(A):
...   pass
...
>>> print A.x, B.x, C.x
1 1 1

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This result is normal.

>>> B.x = 2
>>> print A.x, B.x, C.x
1 2 1

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Well, the result is as expected.

>>> A.x = 3
>>> print A.x, B.x, C.x
3 2 3

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In the Python language, class variables are processed in the form of dictionaries and follow the method resolution order (MRO). Therefore, in the above code, since there is no attribute x in class C, the interpreter will look for its base class (although Python supports multiple inheritance, in this example, the base class of C is only A) . In other words, C does not have its own x attribute that is independent of A and truly belongs to it. Therefore, referencing C.x actually refers to A.x. If the relationship here is not handled properly, it will lead to the problem in the example.

Common mistake 3: Incorrectly specifying the parameters of the exception block (exception block)

Please look at the following code:

>>> try:
...   l = ["a", "b"]
...   int(l[2])
... except ValueError, IndexError: # To catch both exceptions, right?
...   pass
...
Traceback (most recent call last):
 File "<stdin>"</stdin>, line 3, in <module>
IndexError: list index out of range

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The problem with this code is that the except statement does not support specifying exceptions in this way. In Python 2.x, you need to use the variable e to bind the exception to the optional second parameter to further view the exception. Therefore, in the above code, the except statement does not catch the IndexError exception; instead, it binds the exception that occurs to a parameter named IndexError.

To correctly catch multiple exceptions in an except statement, you should specify the first parameter as a tuple, and then write the exception type you want to catch in the tuple. Also, to improve portability, use the as keyword, which is supported in both Python 2 and Python 3.

>>> try:
...   l = ["a", "b"]
...   int(l[2])
... except (ValueError, IndexError) as e: 
...   pass
...
>>>

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Common Mistake 4: Misunderstanding variable name resolution in Python
Variable name resolution in Python follows the so-called LEGB principle, that is, "L: local scope; E: local scope of def or lambda in the upper layer structure ;G: Global scope; B: Built-in scope" (Local, Enclosing, Global, Builtin), search in order. Doesn’t it look simple? However, there are actually some special features in the way this principle takes effect. Speaking of this, we have to mention the following common Python programming errors. Please look at the code below:

>>> x = 10
>>> def foo():
...   x += 1
...   print x
...
>>> foo()
Traceback (most recent call last):
 File "<stdin>"</stdin>, line 1, in <module>
 File "<stdin>"</stdin>, line 2, in foo
UnboundLocalError: local variable &#39;x&#39; referenced before assignment

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What’s wrong?

The above error occurs because when you assign a value to a variable in a certain scope, the variable is automatically regarded as a local variable of the scope by the Python interpreter and will replace any variable with the same name in the upper scope. variable.

It is precisely because of this that the code that started out well appears, but after adding an assignment statement inside a function, an UnboundLocalError occurs. No wonder it surprises many people.

Python programmers are especially likely to fall into this trap when working with lists.

Please take a look at this code example:

>>> lst = [1, 2, 3]
>>> def foo1():
...   lst.append(5)  # 这里没问题
...
>>> foo1()
>>> lst
[1, 2, 3, 5]

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>>> lst = [1, 2, 3]
>>> def foo2():
...   lst += [5]   # ... 但这里就不对了!
...
>>> foo2()
Traceback (most recent call last):
 File "<stdin>"</stdin>, line 1, in <module>
 File "<stdin>"</stdin>, line 2, in foo
UnboundLocalError: local variable &#39;lst&#39; referenced before assignment

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Huh? Why does function foo1 run normally but error occurs in foo2?

答案与上一个示例相同，但是却更难捉摸清楚。foo1函数并没有为lst变量进行赋值，但是foo2却有赋值。我们知道，lst += [5]只是lst = lst + [5]的简写，从中我们就可以看出，foo2函数在尝试为lst赋值（因此，被Python解释器认为是函数本地作用域的变量）。但是，我们希望为lst赋的值却又是基于lst变量本身（这时，也被认为是函数本地作用域内的变量），也就是说该变量还没有被定义。这才出现了错误。

常见错误5：在遍历列表时更改列表

下面这段代码的问题应该算是十分明显：

>>> odd = lambda x : bool(x % 2)
>>> numbers = [n for n in range(10)]
>>> for i in range(len(numbers)):
...   if odd(numbers[i]):
...     del numbers[i] # BAD: Deleting item from a list while iterating over it
...
Traceback (most recent call last):
   File "<stdin>"</stdin>, line 2, in <module>
IndexError: list index out of range

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在遍历列表或数组的同时从中删除元素，是任何经验丰富的Python开发人员都会注意的问题。但是尽管上面的示例十分明显，资深开发人员在编写更为复杂代码的时候，也很可能会无意之下犯同样的错误。

幸运的是，Python语言融合了许多优雅的编程范式，如果使用得当，可以极大地简化代码。简化代码还有一个好处，就是不容易出现在遍历列表时删除元素这个错误。能够做到这点的一个编程范式就是列表解析式。而且，列表解析式在避免这个问题方面尤其有用，下面用列表解析式重新实现上面代码的功能：

>>> odd = lambda x : bool(x % 2)
>>> numbers = [n for n in range(10)]
>>> numbers[:] = [n for n in numbers if not odd(n)] # ahh, the beauty of it all
>>> numbers
[0, 2, 4, 6, 8]

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常见错误6：不理解Python在闭包中如何绑定变量

请看下面这段代码：

>>> def create_multipliers():
...   return [lambda x : i * x for i in range(5)]
>>> for multiplier in create_multipliers():
...   print multiplier(2)
...

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你可能觉得输出结果应该是这样的：

但是，实际的输出结果却是：

吓了一跳吧！

这个结果的出现，主要是因为Python中的迟绑定（late binding ）机制，即闭包中变量的值只有在内部函数被调用时才会进行查询。因此，在上面的代码中，每次create_multipliers()所返回的函数被调用时，都会在附近的作用域中查询变量i的值（而到那时，循环已经结束，所以变量i最后被赋予的值为4）。

要解决这个常见Python问题的方法中，需要使用一些hack技巧：

>>> def create_multipliers():
...   return [lambda x, i=i : i * x for i in range(5)]
...
>>> for multiplier in create_multipliers():
...   print multiplier(2)
...
0
2
4
6
8

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请注意！我们在这里利用了默认参数来实现这个lambda匿名函数。有人可能认为这样做很优雅，有人会觉得很巧妙，还有人会嗤之以鼻。但是，如果你是一名Python程序员，不管怎样你都应该要了解这种解决方法。

常见错误7：模块之间出现循环依赖（circular dependencies）

假设你有两个文件，分别是a.py和b.py，二者相互引用，如下所示：

a.py文件中的代码:

import b
def f():
  return b.x
print f()

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b.py文件中的代码：

import a
x = 1
def g():
  print a.f()

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首先，我们尝试导入a.py模块：

代码运行正常。也许这出乎了你的意料。毕竟，我们这里存在循环引用这个问题，想必应该是会出现问题的，难道不是吗？

答案是，仅仅存在循环引用的情况本身并不会导致问题。如果一个模块已经被引用了，Python可以做到不再次进行引用。但是如果每个模块试图访问其他模块定义的函数或变量的时机不对，那么你就很可能陷入困境。

那么回到我们的示例，当我们导入a.py模块时，它在引用b.py模块时是不会出现问题的，因为b.py模块在被引用时，并不需要访问在a.py模块中定义的任何变量或函数。b.py模块中对a模块唯一的引用，就是调用了a模块的foo()函数。但是那个函数调用发生在g()函数当中，而a.py或b.py模块中都没有调用g()函数。所以，不会出现问题。

但是，如果我们试着导入b.py模块呢（即之前没有引用a.py模块的前提下）：

>>> import b
Traceback (most recent call last):
   File "<stdin>"</stdin>, line 1, in <module>
   File "b.py", line 1, in <module>
  import a
   File "a.py", line 6, in <module>
 print f()
   File "a.py", line 4, in f
 return b.x
AttributeError: &#39;module&#39; object has no attribute &#39;x&#39;

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糟糕。情况不太妙！这里的问题是，在导入b.py的过程中，它试图引用a.py模块，而a.py模块接着又要调用foo()函数，这个foo()函数接着又试图去访问b.x变量。但是这个时候，b.x变量还没有被定义，所以才出现了AttributeError异常。

解决这个问题有一种非常简单的方法，就是简单地修改下b.py模块，在g()函数内部才引用a.py：

x = 1
def g():
  import a # This will be evaluated only when g() is called
  print a.f()

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现在我们再导入b.py模块的话，就不会出现任何问题了：

>>> import b
>>> b.g()
1 # Printed a first time since module &#39;a&#39; calls &#39;print f()&#39; at the end
1 # Printed a second time, this one is our call to &#39;g&#39;

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常见错误8：模块命名与Python标准库模块名冲突

Python语言的一大优势，就是其本身自带的强大标准库。但是，正因为如此，如果你不去刻意注意的话，你也是有可能为自己的模块取一个和Python自带标准库模块相同的名字（例如，如果你的代码中有一个模块叫email.py，那么这就会与Python标准库中同名的模块相冲突。）

这很可能会给你带来难缠的问题。举个例子，在导入模块A的时候，假如该模块A试图引用Python标准库中的模块B，但却因为你已经有了一个同名模块B，模块A会错误地引用你自己代码中的模块B，而不是Python标准库中的模块B。这也是导致一些严重错误的原因。

因此，Python程序员要格外注意，避免使用与Python标准库模块相同的名称。毕竟，修改自己模块的名称比提出PEP提议修改上游模块名称且让提议通过，要来得容易的多。

常见错误9：未能解决Python 2与Python 3之间的差异

假设有下面这段代码：

import sys
def bar(i):
  if i == 1:
    raise KeyError(1)
  if i == 2:
    raise ValueError(2)
def bad():
  e = None
  try:
    bar(int(sys.argv[1]))
  except KeyError as e:
    print(&#39;key error&#39;)
  except ValueError as e:
    print(&#39;value error&#39;)
  print(e)
bad()

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如果是Python 2，那么代码运行正常：

$ python foo.py 1
key error
1
$ python foo.py 2
value error
2

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但是现在，我们换成Python 3再运行一遍：

   
$ python3 foo.py 1
key error
Traceback (most recent call last):
 File "foo.py", line 19, in <module>
  bad()
 File "foo.py", line 17, in bad
  print(e)
UnboundLocalError: local variable &#39;e&#39; referenced before assignment

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这到底是怎么回事？这里的“问题”是，在Python 3中，异常对象在except代码块作用域之外是无法访问的。（这么设计的原因在于，如果不这样的话，堆栈帧中就会一直保留它的引用循环，直到垃圾回收器运行，将引用从内存中清除。）

避免这个问题的一种方法，就是在except代码块的作用域之外，维持一个对异常对象的引用（reference），这样异常对象就可以访问了。下面这段代码就使用了这种方法，因此在Python 2和Python 3中的输出结果是一致的：

import sys
def bar(i):
  if i == 1:
    raise KeyError(1)
  if i == 2:
    raise ValueError(2)
def good():
  exception = None
  try:
    bar(int(sys.argv[1]))
  except KeyError as e:
    exception = e
    print(&#39;key error&#39;)
  except ValueError as e:
    exception = e
    print(&#39;value error&#39;)
  print(exception)
good()

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在Python 3下运行代码：

$ python3 foo.py 1
key error
1
$ python3 foo.py 2
value error
2

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太棒了！

常见错误10：错误使用del方法

假设你在mod.py的文件中编写了下面的代码：

import foo
class Bar(object):
    ...
  def __del__(self):
    foo.cleanup(self.myhandle)

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之后，你在another_mod.py文件中进行如下操作：

import mod
mybar = mod.Bar()

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如果你运行another_mod.py模块的话，将会出现AttributeError异常。

为什么？因为当解释器结束运行的时候，该模块的全局变量都会被设置为None。因此，在上述示例中，当__del__方法被调用之前，foo已经被设置成了None。

要想解决这个有点棘手的Python编程问题，其中一个办法就是使用atexit.register()方法。这样的话，当你的程序执行完成之后（即正常退出程序的情况下），你所指定的处理程序就会在解释器关闭之前运行。

应用了上面这种方法，修改后的mod.py文件可能会是这样子的：

import foo
import atexit
def cleanup(handle):
  foo.cleanup(handle)
class Bar(object):
  def __init__(self):
    ...
    atexit.register(cleanup, self.myhandle)

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这种实现支持在程序正常终止时干净利落地调用任何必要的清理功能。很明显，上述示例中将会由foo.cleanup函数来决定如何处理self.myhandle所绑定的对象。

综述

Python是一门强大而又灵活的编程语言，提供的许多编程机制和范式可以极大地提高工作效率。但是与任何软件工具或语言一样，如果对该语言的能力理解有限或无法欣赏，那么有时候自己反而会被阻碍，而不是受益了。正如一句谚语所说，“自以为知道够多，但实则会给自己或别人带来危险。

不断地熟悉Python语言的一些细微之处，尤其是本文中提到的10大常见错误，将会帮助你有效地使用这门语言，同时也能避免犯一些比较常见的错误。