函数式编程->map

# 函数式编程之map
"""
num_l = [1, 2, 10, 5, 3, 7]
# 需求1:对每个之执行平方
# 方法1:for 循环
ret = []
for i in num_l:
    ret.append(i**2)
# 方法2:对上面的过程进行封装
def map_test(array):
    ret = []
    for i in array:
        ret.append(i**2)
    return ret

# 需求2:队列表的每个值自增1
def map_test(array):
    ret = []
    for i in array:
        ret.append(i+1)
    return ret
# 需求3:队列表的每个值自减1
def map_test(array):
    ret = []
    for i in array:
        ret.append(i-1)
    return ret
# 需求N:....

# 怎么应对上面的情况?
def add_one(i):
    return i + 1
def reduce_one(i):
    return i - 1
def map_test(func, array):
    ret = []
    for i in array:
        ret.append( func(i))
    return ret
print(map_test(add_one, [1, 2, 3, 4])) # 这也是一种高阶函数, [2, 3, 4, 5]
# add_one改写
print(map_test(lambda x:x+1, [1, 2, 3, 4])) # [2, 3, 4, 5]
print(map_test(lambda x:x-1, [1, 2, 3, 4])) # [0, 1, 2, 3]

# map就是map_test的功能
print(map(lambda x:x+1, [1, 2, 3, 4])) # <map object at 0x102979f60>
res = map(lambda x:x+1, [1, 2, 3, 4])
for i in res:
    print(i)
# print(list(res)) 两种输出都可以,notes:只能迭代一次.
print(list(map(lambda x:x+1, [1, 2, 3, 4])))
print(list(map(add_one, [1, 2, 3, 4]))) # 实现简单,难读
# notes, map的第二个参数是可迭代对象即可.
# map的第一个参数可以传入函数,或者是lamda.
print(list(map(lambda x:x.upper(), "supter"))) # ['S', 'U', 'P', 'T', 'E', 'R']
"""




"""
# map filter
movie_people = ["sb_alex", "sb_wupeiqi", "sb_yuanhao", "haha1", "hsha2"]
ret = []
for p in movie_people:
    if not p.startswith("sb"):
        ret.append(p)
print(ret)

# 函数封装
def filter_test(array):
    ret = []
    for p in array:
        if not p.startswith("sb"):
            ret.append(p)
    return ret
print(filter_test(movie_people))
"""
# 改造方式和上面一样,最终的程序结果是
movie_people = ["sb_alex", "sb_wupeiqi", "sb_yuanhao", "haha1", "hsha2"]
def filter_test(func,array):
    ret = []
    for p in array:
        if not func(p):
            ret.append(p)
    return ret
print(filter_test(lambda n:n.startswith("sb"),movie_people)) # ['haha1', 'hsha2']

# filter
movie_people = ["sb_alex", "sb_wupeiqi", "sb_yuanhao", "haha1", "hsha2"]
print("filter", list(filter(lambda n:n.startswith("sb"), movie_people)))