# 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() # #