day11 filter函数

场景模拟：我想判断某个列表里面的某个元素怎么怎么样

基础方法，如果需要判断多次则重复代码

ret = []
move_peole = ["alex","sb_wupeiqi","yangtuo","sb_yuanhao"]
for p in move_peole:
    if    not p.startswith("sb"):
        ret.append(p)

print(ret)

函数方法
利用参数更换来多次代入执行，判断方式单一

move_peole = ["alex","sb_wupeiqi","yangtuo","sb_yuanhao"]
move_peole1 = ["alex","wupeiqi_sb","yangtuo","yuanhao_sb"]
def    filter_test(array):
    ret = []
    for p in array:
        if not p.startswith("sb"):
            ret.append(p)
    return ret

print(filter_test(move_peole))

双函数参数调用
解决了方法限定的问题但是代码复杂，代码量大

move_peole1 = ["alex","wupeiqi_sb","yangtuo","yuanhao_sb"]
def sb_show(n):
    return n.endswith("sb")
def    filter_test(funk,array):
    ret = []
    for p in array:
        if not funk(p):
            ret.append(p)
    return ret

print(filter_test(sb_show,move_peole1))

终极版本目前可以做到的最好的版本

def sb_show(n):
    return n.endswith("sb")
move_peole1 = ["alex","wupeiqi_sb","yangtuo","yuanhao_sb"]
lambda x:x.endswith("sb")
def    filter_test(funk,array):
    ret = []
    for p in array:
        if not funk(p):
            ret.append(p)
    return ret
print(filter_test(lambda x:x.endswith("sb"),move_peole1))

filter 函数，计算布尔值，得出为true则保留
利用filter可以实现的效果 
格式：filter(方法,参数)

move_peole1 = ["alex","wupeiqi_sb","yangtuo","yuanhao_sb"]
print(filter (lambda x:not x.endswith("sb"),move_peole1))
print(list(filter (lambda x:not x.endswith("sb"),move_peole1)))

---