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  • python中的基础坑

    v = [lambda :x  for x in range(10)]
    print(v)    #[lambda :x,lambda :x....]10个匿名函数
    print(v[0]) #lambda :x
    print(v[0]())   #9
    

      

    v = (lambda :x for x in range(10))
    print(v)		#生成器
    # print(v[0])	    #报错:生成器不能索引取值
    # print(v[0]())   #报错:生成器不能索引取值
    print(next(v))	#lambda :x
    print(next(v)()) #1
    

      

    def extendList(val, list=[]):
        list.append(val)
        return list
    
    
    list1 = extendList(10)
    list2 = extendList(123, [])
    list3 = extendList('a')
    
    print('list1=%s' % list1)  # list1=[10,'a']
    print('list2=%s' % list2)  # list2=[123]
    print('list3=%s' % list3)  # list3=[10,'a']
    

      

    def add(a, b):
        return a + b
    
    
    def test():
        for i in range(4):
            yield i
    
    
    g = test()  # 生成器
    
    for n in [2, 10]:
        g = (add(n, i) for i in g)# 生成器
        
    print(list(g))#[20,21,22,13]
    #n=2:
            # 循环体执行时i对应的值 (0,1,2,3)
            # 循环体执行之后:g类似于[add(n, 0),add(n, 1),add(n, 2),add(n, 3)]
        #n=10时:
            # 循环体执行时i 的值 (10,11,12,13)-----对n=2循环之后的g进行取值
            # 循环体执行之后:list(g)=(add(n, 10),add(n, 11),add(n, 12),add(n, 13))
    

      

    li = [7,-8,5,4,0,-2,-5]
    print(sorted(li,key=lambda x:(x<0,abs(x))))
    

      

    print(5/2)
    print(5.0/2)
    print(5/2.0)
    print(5.0/2.0)
    print(5//2)
    print(5.0//2)
    print(5//2.0)
    print(5.0//2.0)
    
    #python3中   2.5     2.5     2.5     2.5     2   2.0     2.0     2.0
    #python2中   2       2.5     2.5     2.5     2   2.0     2.0     2.0
    

      

    dic={x:y for x in ['Male','Female'] for y in ['Red','Black']}
    print(dic)
    # {'Male':'Black','Female':'Black'}
    
    li=[(x,y) for x in ['Male','Female'] for y in ['Red','Black']]
    print(li)
    #[('Male','Red'),('Male','Black'),('Female','Red'),('Female','Black')]
    

      

    li=[[]]*5           #[[], [], [], [], []]
    li[0].append(1)     #[[1], [1], [1], [1], [1]]
    li[3].append(0)     #[[1, 0], [1, 0], [1, 0], [1, 0], [1, 0]]
    li.append(8)        #[[1, 0], [1, 0], [1, 0], [1, 0], [1, 0],8]
    print(li)
    

      

    t=((1,2))*5         #(1, 2, 1, 2, 1, 2, 1, 2, 1, 2)
    t1=((1,2),)*5       #((1, 2), (1, 2), (1, 2), (1, 2), (1, 2))
    print(t)
    print(t1)
    

      

    for x in range(5):
        pass
    print(x)      #4
    
    def f():
        x = 0
        for i in range(5):
            x += i
    f()
    print(x)
    

      

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  • 原文地址:https://www.cnblogs.com/open-yang/p/11425593.html
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