内置函数

# print('123'.__len__())
# print(len('123'))


# 排序:sorted
# dic = {
# 'owen': (1, 88888),
# 'egon': (2, 300000),
# 'liuXX': (3, 99999)
# }

# 总结:排序的可迭代对象,排序的规则,是否反转
# res = sorted(dic, key=lambda k: dic[k][1], reverse=True) # 按薪资排序的人名list
# for k in res:
# print(k, dic[k][1])

# dic={
# 'owen':(1,888),
# 'gegon':(2,333),
# 'linux':(3,999)
# }
#
# res=sorted(dic,key=lambda k:dic[k][1],reverse=True)
#
# for k in res:
# print(k,dic[k][1])


# map:映射 - 格式化每一次的遍历结果
# names = ['Owen', 'Egon', 'Liuxx']
# def fn(x):
# # print(x)
# # 将所有名字全小写
# return x.lower()

# res = map(fn, names)
# print(list(res))

names = ['owen','Egon','Linux']

def fn(x):
print(x)
return x.lower()
res= map(fn,names)
print(list(res))





# ls = [88888, 300000, 99999]
# # 薪资加一元
# res = map(lambda x: x + 1, ls)
# print(list(res))
#
# dic1 = {
# 'owen': 88888,
# 'egon': 300000,
# 'liuXX': 99999
# }
# def fn1(x):
# dic1[x] += 1
# return 10000
# # 总结:遍历第二个参数(可迭代对象),将遍历的结果丢给第一个函数,
# # 函数有一个参数,就是一一遍历的值
# # map的作用(返回值):在当前数据基础上改变值(可以任意修改)
# res = map(fn1, dic1)
# print(list(res))
# print(dic1)
#
#
# # 合并:reduce
# from functools import reduce
# # 求[1, 3, 4, 2, 10]所有元素的总和
# res = reduce(lambda x, y: x + y, [1, 3, 4, 2, 10])
# print(res)
#
#
#
# # 已见过的
# # 1.类型转换:int() tuple()
# # 2.常规使用:print() input() len() next() iter() open() range() enumerate() id()
# # 3.进制转换:bin() oct() hex() 将10进制转换为2 | 8 | 16进制
# print(bin(10)) # 0b1010
# print(oct(10)) # 0o12
# print(hex(10)) # 0xa
#
# # 3.运算:abs()
# print(abs(-1)) # 绝对值
# print(chr(9326)) # 将ASCII转换为字符
# print(ord('①')) # 逆运算
# print(pow(2, 3)) # 2的3次方
# print(pow(2, 3, 3)) # 2的3次方对3求余
# print(sum([1, 2, 3])) # 求和
#
# # 4.反射:getattr() delattr() hasattr() setattr()
#
# # 5.面向对象的相关方法:super() staticmethod() classmethod()
# def fn():pass
# print(callable(fn)) # 对象能不能被调用
#
# # 6.原义字符串
# print('a\nb')
# s = ascii('a\nb')
# print(s)
# s = repr('a\nb')
# print(s)
# print(r'a\nb')
#
# print(all([1, 0, 0]))
# print(any([0, 0, 1]))
#
# # compile() exec() eval()
#
#
posted @ 2019-04-28 22:03  zhouyuqiang  阅读(118)  评论(0编辑  收藏  举报