http://ilian.i-n-i.org/python-interview-question-and-answers/
http://www.geekinterview.com/Interview-Questions/Programming/Python
http://www.reddit.com/r/Python/comments/1knw7z/python_interview_questions
http://www.toptal.com/python/interview-questions
http://careerride.com/python-interview-questions.aspx
http://www.oschina.net/translate/top-10-mistakes-that-python-programmers-make
http://www.toptal.com/python#hiring-guide
- Singleton
1.
class Singleton(type):
def __init__(cls, name, bases, dict):
super(Singleton, cls).__init__(name, bases, dict);
cls.instance = None
def __call__(cls, *arg, **kw):
if cls.instance is None:
cls.instance = super(Singleton, cls).__call__(*arg, **kw);
return cls.instance;
class MyClass1(Singleton):
__metaclass__ = Singleton;
2.
def signleton(cls):
instance = [];
def getinstance():
if cls not in instance:
instance[ cls ] = cls();
return instnace[ cls ]
- What will be the output of the code below?
list = ['a', 'b', 'c', 'd', 'e']
print list[10:]
>>>[]
- What will be the output of the code below in Python 2? Explain your answer.
def div1(x,y):
print "%s/%s = %s" % (x, y, x/y)
def div2(x,y):
print "%s//%s = %s" % (x, y, x//y)
div1(5,2)
div1(5.,2)
div2(5,2)
div2(5.,2.)
Also, how would the answer differ in Python 3 (assuming, of course, that the above print statements were converted to Python 3 syntax)?
5/2 = 2
5.0/2 = 2.5
5//2 = 2
5.0//2.0 = 2.0-What will be the output of the code below? Explain your answer.
def extendList(val, list=[]):
list.append(val)
return list
list1 = extendList(10)
list2 = extendList(123,[])
list3 = extendList('a')
print "list1 = %s" % list1
print "list2 = %s" % list2
print "list3 = %s" % list3
How would you modify the definition of extendList to produce the presumably desired behavior?
-What will be the output of the code below? Explain your answer.
class Parent(object):
x = 1
class Child1(Parent):
pass
class Child2(Parent):
pass
print Parent.x, Child1.x, Child2.x
Child1.x = 2
print Parent.x, Child1.x, Child2.x
Parent.x = 3
print Parent.x, Child1.x, Child2.x
1 1 1
1 2 1
3 2 3
- What will be the output of the code below? Explain your answer.
def multipliers():
return [lambda x : i * x for i in range(4)]
print [m(2) for m in multipliers()]
How would you modify the definition of multipliers to produce the presumably desired behavior?
[6, 6, 6, 6]
def multipliers():
return [lambda x, i=i : i * x for i in range(4)]- 去掉 list 中的重复元素
def unique(L):
return list(set(L))
ll = [1, 3, 3, 4, 5, 5, 5]
print unique(ll)
循环的数据结构会导致循环
尽管这在实际情况中很少见，但是如果一个对象的集合包含了到它自己的引用，这被称为循环对象（cyclic object）。如果在一个对象中发现一个循环，Python会输出一个[…]，以避免在无限循环中卡住：
>>> L = ['grail'] # 在 L中又引用L自身会>>> L.append(L) # 在对象中创造一个循环>>> L['grail', [...]]赋值语句不会创建对象的副本，仅仅创建引用
>>> L = [1, 2, 3] # 共用的列表对象>>> M = ['X', L, 'Y'] # 嵌入一个到L的引用>>> M['X', [1, 2, 3], 'Y']>>> L[1] = 0 # 也改变了M>>> M['X', [1, 0, 3], 'Y']>>> L = [1, 2, 3]>>> M = ['X', L[:], 'Y'] # 嵌入一个L的副本>>> L[1] = 0 # 仅仅改变了L，但是不影响M>>> L[1, 0, 3]>>> M['X', [1, 2, 3], 'Y']在执行def语句时，默认参数的值只被解析并保存一次，而不是每次在调用函数的时候。这通常是你想要的那样，但是因为默认值需要在每次调用时都保持同样对象，
>>> def saver(x=[]): # 保存一个列表对象
... x.append(1) # 并每次调用的时候
... print x # 改变它的值
...
>>> saver([2]) # 未使用默认值
[2, 1]
>>> saver() # 使用默认值
[1]
>>> saver() # 每次调用都会增加！
[1, 1]
>>> saver()
[1, 1, 1]
有的人将这个视作Python的一个特点——因为可变的默认参数在每次函数调用时保持了它们的状态，它们能提供像C语言中静态本地函数变量的类似的一些功能。但是，当你第一次碰到它时会觉得这很奇怪，
要摆脱这样的行为，在函数开始的地方用切片或者方法来创建默认参数的副本，或者将默认值的表达式移到函数里面；只要每次函数调用时这些值在函数里，就会每次都得到一个新的对象：
>>> def saver(x=None):
... if x is None: x = [] # 没有传入参数？
... x.append(1) # 改变新的列表
... print x
...
>>> saver([2]) # 没有使用默认值
[2, 1]
>>> saver() # 这次不会变了
[1]
>>> saver()
[1]
假设你有如下一段代码：
>>> try:... l = ["a", "b"]... int(l[2])... except ValueError, IndexError: # To catch both exceptions, right?... passTraceback (most recent call last): File "<stdin>", line 3, in <module>IndexError: list index out of range这里的问题在于 except 语句并不接受以这种方式指定的异常列表。相反，在Python 2.x中，使用语法 except Exception, e 是将一个异常对象绑定到第二个可选参数（在这个例子中是 e）上，以便在后面使用。
所以，在上面这个例子中，IndexError 这个异常并不是被except语句捕捉到的，而是被绑定到一个名叫 IndexError的参数上时引发的。
在一个except语句中捕获多个异常的正确做法是将第一个参数指定为一个含有所有要捕获异常的元组。并且，为了代码的可移植性，要使用as关键词，因为Python 2 和Python 3都支持这种语法：
>>> try:... l = ["a", "b"]... int(l[2])... except (ValueError, IndexError) as e: ... pass>>>>>> def foo1():
... lst.append(5) # 没有问题...
...
>>> foo1()
>>> lst
[1, 2, 3, 5]
>>> lst = [1, 2, 3]
>>> def foo2():
... lst += [5] # ... 但是这里有问题!
...
>>> foo2()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in foo
UnboundLocalError: local variable 'lst' referenced before assignment
嗯？为什么 foo2 报错，而foo1没有问题呢？
原因和之前那个例子的一样，不过更加令人难以捉摸。foo1 没有对 lst 进行赋值操作，而 foo2 做了。要知道， lst += [5] 是 lst = lst + [5] 的缩写，我们试图对 lst 进行赋值操作（Python把他当成了局部变量）。
>>> # INITIALIZING AN ARRAY -- METHOD 1
>>> x = [[1,2,3,4]] * 3
>>> x
[[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]
>>> # INITIALIZING AN ARRAY -- METHOD 2
>>> y = [[1,2,3,4] for _ in range(3)]
>>> y
[[1, 2, 3, 4], [1, 2, 3, 4], [1, 2, 3, 4]]
>>> x[0][3] = 99
>>> x
[[1, 2, 3, 99], [1, 2, 3, 99], [1, 2, 3, 99]]
>>> y[0][3] = 99
>>> y
[[1, 2, 3, 99], [1, 2, 3, 4], [1, 2, 3, 4]]
- 简单地交换2个元素
>>> x,y = y,x
Q: What will be printed out by the last statement below?
>>> flist = []
>>> for i in range(3):
... flist.append(lambda: i)
...
>>> [f() for f in flist] # what will this print out?
In any closure in Python, variables are bound by name. Thus, the above line of code will print out the following:
[2, 2, 2]
Presumably not what the author of the above code intended!
A workaround is to either create a separate function or to pass the args by name; e.g.:
>>> flist = []
>>> for i in range(3):
... flist.append(lambda i = i : i)
...
>>> [f() for f in flist]
[0, 1, 2]