import collections

#需求：lt=[1,2,3,4,5,6,7,8]-->效果：[2,4,6,8]

lt=[1,2,3,4,5,6,7,8]

#代码实现一：老技术

lt1=[]

for i in lt:

    if i%2==0:

        lt1.append(i)

print(lt1)

#代码实现二：新技术

def func(obj):

    if obj%2==0:

        return True

    return False

fo=filter(func,lt)

print(fo,type(fo))

# print(isinstance(fo,collections.Iterator))

# print(next(fo))

print(list(fo))

#终极版

print(list(filter(lambda x:x%2==0,lt)))

'''

需求：

lt = [123,'abcd',0,3.14,0.0,'haha','hehe','',True,False,(),[],{},[1,2,3],{11,22,33},{'name':'jack','age':23}]

得到效果如下：

[123,'abcd',3.14,'haha','hehe',True,[1,2,3],{11,22,33},{'name':'jack','age':23}]

'''

lt = [123,'abcd',0,3.14,0.0,'haha','hehe','',True,False,(),[],{},[1,2,3],{11,22,33},{'name':'jack','age':23}]

print(list(filter(lambda x:bool(x),lt)))

print(list(filter(bool,lt)))

'''

需求：

lt1 = ['aaaaaaaa','bbbb','cccccc','ddd']

得到效果如下：保留长度大于等于4的元素

['aaaaaaaa','bbbb','cccccc']

'''

lt1 = ['aaaaaaaa','bbbb','cccccc','ddd']

print(list(filter(lambda x:len(x) >= 4,lt1)))