一些练习题

找了一些题目，想用Python解决一下。

反向迭代一个序列

# 如果是一个list，最快的方法使用reverse
>>> tempList = [1, 2, 3, 4]
>>> tempList.reverse()
>>> t
[4, 3, 2, 1]

# 如果不是list，需要手动重排
>>> temp = (1, 2, 3, 4)
>>> for i in range(len(temp) - 1, -1, -1):
>>>	print(temp[i])
4
3
2
1

查询替换文本中的字符串

# 最简单的方法使用replace()
>>> tempstr = "hello you hello python are you ok"
>>> tempstr = tempstr.replace("you", "python") # 把所有的you替换成python
'hello python hello python are python ok'

# 还可以使用正则
>>> tempstr = "hello you hello python are you ok"
>>> import re
>>> rex = r'(hello|Use)'
>>> print(re.sub(rex, "Bye", tempstr))	# 把所有hello替换成Bye
'Bye you Bye python are you ok'

python实现单例模式

# 方法一：可以使用__new__方法。
# 在__new__方法中把类实例绑定到类变量_instance上，如果cls._instance为None表示该类没有实例化过，实例化该类并返回。反之直接返回。

>>> class SingleTon(object):
        def __new__(cls, *args, **kwargs):
   	    if not hasattr(cls, '_instance'):
	        cls._instance = object.__new__(cls, *args, *kwargs)
	    return cls._instance

# 测试
>>> class TestClass(SingleTon):
        a = 1

>>> test1 = TestClass()
>>> test2 = TestClass()
>>> print(test1.a, test2.a)
1 1
>>> test1.a = 2
>>> print(test1.a, test2.a)
2 2
>>> print(id(test1), id(test2))
140666831839864 140666831839864

# 方法二：使用装饰器，建立过实例就放到instances里面，下次建立的时候先检查此里面是否有实例

>>> def SingleTon(cls, *args, **kwargs):
        instances = {}
        def _singleton():
	    if cls not in instances:
	        instances[cls] = cls(*args, **kwargs)
            return instances[cls]
        return _singleton

>>> @SingleTon
    class LastClass(object):
        a = 1

>>> test1 = LastClass()
>>> test2 = LastClass()
>>> print(test1.a, test2.a)
1 1
>>> print(id(test1), id(test2))
140666831841376 140666831841376

# 方法三：共享属性

>>> class SignalTon(object):
        _state = {}
        def __new__(cls, *args, **kwargs):
	    obj = object.__new__(cls, *args, **kwargs)
	    obj.__dict__ = cls._state
	    return obj

>>> class TestClass(SignalTon):
        a = 1
>>> test1 = TestClass()
>>> test2 = TestClass()
>>> test1.a = 2
>>> print(test1.a, test2.a)
2 2

重新实现str.strip()

>>> def rightStrip(tempStr, splitStr):
        endindex = tempStr.rfind(splitStr)
        while endindex != -1 and endindex == len(tempStr)-1:
	    tempStr = tempStr[:endindex]
	    endindex = tempStr.rfind(splitStr)
        return tempStr

>>> def leftStrip(tempStr, splitStr):
        startindex = tempStr.find(splitStr)
        while startindex == 0:
	    tempStr = tempStr[startindex+1:]
	    startindex = tempStr.find(splitStr)
        return tempStr

>>> str = " H "
>>> print(str)
 H 
>>> print(leftStrip(str, " "))
H 
>>> print(rightStrip(str, " "))
 H

给列表中字典排序

>>> alist = [{"name": "a", "age": 20},
	    {"name": "b", "age": 30},
	    {"name": "c", "age": 25}]

>>> alist.sort(key=lambda x:x.get("age"))
>>> print(alist)
[{'name': 'a', 'age': 20}, {'name': 'c', 'age': 25}, {'name': 'b', 'age': 30}]

打乱一个排序好的列表

>>> from random import shuffle
>>> alist = range(10)
>>> v = list(alist)
>>> shuffle(v)
>>> print(v)
[6, 3, 4, 5, 1, 9, 7, 8, 2, 0]

获取最大公约数

>>> a = 25
>>> b = 15
>>> def max_common(a, b):
	while b:
	    a, b = b, a%b
	return a

获取两个数的最小公倍数

a = 25
b = 15
def min_common(a, b):
    c = a * b
    while b:
	a, b = b, a%b
    return c//a

实现一个简单的栈结构

class Stack(object):
    def __init__(self):
        self.value = []
    def push(self, x):
	self.value.append(x)
    def pop(self):
	s = self.value.pop()
	return s

st = Stack()
stack.push(1)
stack.push(2)
stack.push(3)
print(stack.value)
stack.pop()
print(stack.value)

合并两个列表，排除重复元素

>>> a = ['x', 'y', 'z', 'e', 'f']
>>> b = ['a', 'b', 'c', 'd', 'e', 'f']

>>> def merge_list(*args):
        s = set()
        for i in args:
	    s = s.union(i)
        return s

>>> merge_list(a, b)
{'d', 'y', 'b', 'c', 'e', 'a', 'z', 'x', 'f'}