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)))

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