Python就业班——Python基础知识

2021-05-04 01:28

阅读:428

标签:基础   员工信息   orm   阶乘   存在   put   items   版本   inter   

#!/usr/bin/env python3
# coding=utf-8
# Version:python3.6.1
__date__ = 2020/5/23 10:21
__author__ = Lgsp_Harold

# account = "888888"
# amt = 123456789
# # str = format(amt, "0,.2f")
# str = "请你向{}账户转账¥{:0,.2f}元".format(account, amt)
# print(str)
# str1 = "{p1},{p2}".format(p1=account, p2=amt)
# print(str1)
# ————————————————————————————
# num = int(input("输入阶乘数").strip())
# if num  100:
#     print("0-100之间的数字")
# else:
#     i = 1
#     result = 1
#     while i 
#         result = result * i
#         if i % 5 == 0:
#             print(i)
#         i = i + 1
#     print(result)
# ————————————————————————————

# num = 1000
# j = 2
#
# while j 
#     is_prime = True
#     i = 2
#     while i 
#         if j % i == 0:
#             is_prime = False
#             break
#         i += 1
#     if is_prime:
#         print("{}是质数".format(j))
#     j += 1

# ————————————————————————————

# list1 = [‘a‘, ‘b‘, ‘c‘, ‘d‘, ‘e‘, ‘f‘, ‘g‘]
# len1 = len(list1)
# print(len1)
# i = 0
# for l in list1:
#     if list1[i] == ‘c‘:
#         ri = len1 * -1 + i
#         print(l, i, ri)
#     i += 1
#
# i = 0
# len1 = len(list1)
# while i 
#     if list1[i] == ‘c‘:
#         ri = len1 * -1 + i
#         print(list1[i], i, ri)
#     i += 1

# ————————————————————————————

# emp_list = []
#
# while True:
#     info = input("请输入:")
#     if info == "":
#         break
#     info_list = info.split(",")
#     emp_list.append(info_list)
#     # print(emp_list)
#
#     for emp in emp_list:
#         print("姓名:{n},年龄:{a},薪资:{s}".format(n=emp[0],a=emp[1],s=emp[2]))

# ————————————————————————————

# # 创建字典
# dict1 = {"name": "aa", "age": 19, "sex": "1"}
# print(dict1)
# dict2 = dict(name="bb", age=45, sex="0")
# print(dict2)
# dict3 = dict.fromkeys([‘name‘, ‘age‘, ‘sex‘])
# print(dict3)
# dict4 = dict.fromkeys([‘name‘, ‘age‘], ‘N/A‘)
# print(dict4)
# # 查找
# print(dict1["name"])
# print(dict2.get("name"))
# print(dict2.get("dept"))
# print(dict2.get("dept", "客服部"))
# print("sex" in dict1)
# for key in dict1:
#     print(dict1[key])
#     print(dict1.get(key))
#
# for key, value in dict2.items():
#     print(key, value)
#
# # 新增、修改
# dict1["name"] = "cc"
# print(dict1)
# dict2.update(name="dd", age=99)
# print(dict2)
# dict2.update(dept="研发部")
# print(dict2)
# dict2.update(weight=80, dept="客服部")
# print(dict2)
# # 删除
# dict1.pop(‘age‘)
# print(dict1)
# dict1.popitem()
#
# print(dict1)
# dict2.clear()
# print(dict2)
#
# # 设置默认值
# dict4.setdefault(‘sex‘, 0)
# print(dict4)
#
# # 获取字典视图
# ks = dict4.keys()
# print(ks)
# vs = dict4.values()
# print(vs)
# its = dict4.items()
# print(its)
#
# dict2 = dict(name="bb", age=45, sex="0")
# # 老版本字符串格式化
# dict2_str = "姓名:%(name)s, 年龄:%(age)s, 性别:%(sex)s" %(dict2)
# print(dict2_str)
#
# # 新版本字符串格式化
# dict2_str = "姓名:{name}, 年龄:{age}, 性别:{sex}".format_map(dict2)
# print(dict2_str)

# ————————————————————————————

# # 散列值
# # (单次运行,多次生成一样)
# h1 = hash("abc")
# h2 = hash("abc")
# print(h1)
# print(h2)

# ————————————————————————————

# #处理员工数据
# source = "77,cl1,ma1,sal1,5000$76,cl2,ma2,sal2,4000$75,cl3,ma3,sal3,3000"
#
# emp_list = source.split("$")
# print(emp_list)
#
# # 保存所有解析后的员工信息,key是员工编号,value是包含员工完整信息的字典
# all_emp = {}
# for i in range(0, len(emp_list)):
#     print(i)
#     e = emp_list[i].split(",")
#     print(e)
#     # 创建员工字典
#     emp = {‘no‘: e[0], ‘name‘: e[1], ‘job‘: e[2], ‘depa‘: e[3], ‘salary‘: e[4]}
#     print(emp)
#     all_emp[emp[‘no‘]] = emp
# print(all_emp)
#
# empno = input("请输入员工编号:")
# if empno in all_emp:
#     emp = all_emp.get(empno)
#     print(emp)
#     print(type(emp))
#     print("工号:{no}, 姓名:{name}, 岗位:{job}, 部门:{depa}, 工资:{salary}".format_map(emp))
# else:
#     print("员工不存在")





# source = "77,cl1,ma1,sal1,5000$76,cl2,ma2,sal2,4000$75,cl3,ma3,sal3,3000"
# emp_list = source.split("$")
# print(emp_list)
# print(type(emp_list))
#
# emp_all = {}
# for i in range(0, len(emp_list)):
#     print(i)
#     print(emp_list[i])
#     e = emp_list[i].split(",")
#     print(e)
#     print(type(e))
#     emp = {"no": e[0], "name": e[1], ‘a‘: e[2], ‘b‘: e[3], ‘c‘: e[4]}
#     print(emp)
#     emp_all[emp[‘no‘]] = emp
#
# print(emp_all)



# source = ‘77,cl1,ma1,sal1,5000$76,cl2,ma2,sal2,4000$75,cl3,ma3,sal3,3000‘
# emp_list = source.split(‘$‘)
# all_emp = {}
# for i in range(0, len(emp_list)):
#     e = emp_list[i].split(‘,‘)
#     emp = {‘id‘: e[0], ‘A‘: e[1], ‘B‘: e[2]}
#     all_emp[emp[‘id‘]] = emp
#
# print(all_emp)

# ————————————————————————————

# c = ‘abcdef‘
#
# for i in range(0, len(c)):
#     letter = c[i]
#     print(letter)

# ————————————————————————————

# # 斐波那契数列
# result = []
# for i in range(0, 50):
#     if i == 0 or i == 1:
#         result.append(1)
#     else:
#         result.append(result[i-2] + result[i-1])
#
# print(result)

# # 判断质数
#
# l = 776351721
#
# is_prime = True
# for i in range(2, l):
#     if l % i == 0:
#         print(i)
#         is_prime = False
#         break
#
# if is_prime == True:
#     print("{0}是质数".format(l))
# else:
#     print("{0}不是质数".format(l))

# ————————————————————————————

# 序列相互转换

# l1 = [‘a‘, ‘b‘, ‘c‘]
# t1 = (‘a‘, ‘b‘, ‘c‘)
# s1 = ‘abc123‘
# s2 = ‘abc,123‘
# r1 = range(1,4)
#
# print(list(t1))
# print(list(s1))
# print(s2.split(‘,‘))
# print(list(r1))
#
# print(tuple(l1))
# print(tuple(s1))
# print(tuple(s2.split(‘,‘)))
# print(tuple(r1))
#
# print(str(l1))
# print(type(str(l1)))
# print(‘‘.join(t1))
# print(‘‘.join(r1)) # join必须要求所有元素都是字符串
‘‘‘
# 将包含数字的序列输出
s3 = ‘‘
for i in r1:
    s3 += str(i)
print(s3)
‘‘‘

# ————————————————————————————

# college1 = {‘A‘, ‘B‘, ‘C‘, ‘D‘}
# print(college1)
# college2 = set(["E", "F", "G", "H"])
# print(college2)
# college3 = set(‘中国广西南宁‘)
# print(college3)
#
# college4 = set()

# 集合的数学运算
# college1 = {‘A‘, ‘B‘, ‘C‘, ‘D‘}
# college2 = set(["E", "F", "G", "H", "A", "C"])
# 交集
# c3 = college1.intersection(college2)
# print(c3)
# college1.intersection_update(college2)
# print(college1)
# ————————————————————————————
# 并集
# c4 = college1.union(college2)
# print(c4)
# ————————————————————————————
# 差集
# 单项差集
# c5 = college1.difference(college2)
# print(c5)
# c6 = college2.difference(college1)
# print(c6)
# college1.difference_update(college2)
# ————————————————————————————
# # 双项差集
# c7 = college1.symmetric_difference(college2)
# print(c7)
# college1.symmetric_difference_update(college2)
# ————————————————————————————
# 判断是否为子集issubset:college1.issubset(college2)
# 判断是否为父集issuperset:college1.issuperset(college2)
# 判断两个集合是否存在重复元素isdisjoint(True为不存在,False为存在):college1.isdisjoint(college2)
# ————————————————————————————
# 新增
# college1.add(‘AAAA‘)
# college1.update([‘CCC‘, ‘DDD‘, ‘BBB‘])
# college1.update((‘CC‘, ‘DD‘, ‘BB‘))
# 删除,(remove)如不存在,报错;(discard)如不存在,忽略删除操作
# college1.remove(‘DD‘)
# college1.discard(‘DDDDDDDD‘)
# 不支持更新,须删除后再添加

# ————————————————————————————

# 三种内置生成式

# 生成式语法:[被追加的数据 循环语句 循环或者判断语句]  、 {}
# 列表生成式
# ‘‘‘
# lst = []
# for i in range(10, 20):
#     lst.append(i * 10)
# ‘‘‘
# lst = [i * 10 for i in range(10, 20)]
# lst1 = [i * 10 for i in range(10, 20) if i % 2 == 0]
# print(lst1)
# lst2 = [i * j for i in range(1, 5) for j in range(1, 5)]
# print(lst2)
# ————————————————————————————
# 字典生成式
# lst5 = [‘AA‘, ‘BB‘, ‘CC‘, ‘DD‘]
# ‘‘‘
# for i in range(0, len(lst5)):
#     dict[i+1] = lst5[i]
# ‘‘‘
# dict1 = {i+1: lst5[i] for i in range(0, len(lst5))}
# print(dict1)
# ————————————————————————————
# 集合生成式
# set1 = {i * j for i in range(10, 20) for j in range(10, 20) if i == j}
# print(set1)

if __name__ == __main__:
    pass

 

Python就业班——Python基础知识

标签:基础   员工信息   orm   阶乘   存在   put   items   版本   inter   

原文地址:https://www.cnblogs.com/Harold-Hua/p/13196824.html


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