重新学习python21

 

## 0, 将fun_A函数转换为lambda表达式
def fun_A(x, y=3):
    return x * y

a = fun_A(5, )
print(a)

b  = fun_A(5, 5)
print(b)


##lambda表达式
lambda  x, y=3 : x * y

 

##1， lambda表达式 转化为 一般函数
lambda x : x if x % 2 else None

def fun_B(x) :
    if x % 2 == 0 :
        return x
    else :
        return None


'''

 

# 3 利用lambda表达式 和 filter查找100以内的3的倍数
def  add(a):
    return a%2==0
print(list(filter(add,[11,22,33,44])))


stu_score = {'xiaobai':50, 'xiaohei':30, 'xiaolan':80, 'xiaojun':100, 'xiaoming':60}
result=filter(lambda score:score>60,stu_score.values())
print(list(result))

print("第3题")
print(list(filter(lambda x : (x % 3 == 0), range(1,100))))

print(list(filter(lambda x : (x % 3 == 0), range(100))))

 

# 4, 使用列表推导式，代替filter 和 lambda组合, 将第3题转为列表推导式
print("第4题")
b = [ i for i in range(1, 100) if not(i % 3)]
print(b)

 

#5， 使用zip，将两数以元组形式绑定在一起
print("第5题")
c = list(zip([1, 3, 4, 7, 9], [2, 4, 6, 8,10]))
print(c)
d = list(zip([1, 3, 4, 7, 9], [2, 4, 6, 8]))
print(d)

##以map 和 lambda表达式转化为列表的形式
e = list(map(lambda x,y : [x, y], [1, 3, 5, 7, 9], [2, 4, 6, 8, 10]))
print(e)

 

 

##6 看一下输出打印
print("第6题")
def make_repeat(n):
    return lambda s : s * n

double = make_repeat(2)
print(double(8))
print(double('FishC'))

 

从上到下的输出：

[22, 44]
[80, 100]
第3题
[3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99]
[0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99]
第4题
[3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99]
第5题
[(1, 2), (3, 4), (4, 6), (7, 8), (9, 10)]
[(1, 2), (3, 4), (4, 6), (7, 8)]
[[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]
第6题
16
FishCFishC