活動地址：CSDN21天學習挑戰賽
學習的最大理由是想擺脫平庸，早一天就多一份人生的精彩；遲一天就多一天平庸的困擾。各位小伙伴，如果您：
想系統/深入學習某技術知識點…
一個人摸索學習很難堅持，想組團高效學習…
想寫博客但無從下手，急需寫作干貨注入能量…
熱愛寫作，願意讓自己成為更好的人…
…
歡迎參與CSDN學習挑戰賽，成為更好的自己，請參考活動中各位優質專欄博主的免費高質量專欄資源（這部分優質資源是活動限時免費開放喔~），按照自身的學習領域和學習進度學習並記錄自己的學習過程。您可以從以下3個方面任選其一著手（不強制），或者按照自己的理解發布專欄學習作品，參考如下：
**
1，學習目標
提示：可以添加學習目標
例如： 一周掌握 Python 入門知識
2，學習內容
提示：可以添加要學的內容
例如：
A，掌握 Python 基本語法
B，掌握 Python 列表、字典、元組等集合類型的用法
C，掌握 Python 函數
D，掌握 Python 常用模塊
E, 掌握 Python 模塊與包
F，掌握 Python 類與對象
3，學習時間
每天拿出來兩個小時
4，學習產出
CSDN技術博客 每天一篇
name1 = "張新曉"
name2 = "潘磊"
# 列表的構建 語法格式 [數據對象,...,]
# 特點1： 列表中的元素可以是任意數據類型
# 特點2： 列表中的元素個數沒有限制
l1 = [100, "hello", True]
print(type(l1)) # list
l2 = [100, 2, 34, 5]
print(len(l2)) # 4
names = ["", "", "", ]
l = [10, 20, 30, 40, 50]
# 列表是一個序列類型
# 一 支持正負索引 數據對象[索引]
# print(l[1])
# print(l[-3])
print(id(l))
l[0] = 100
print(l)
print(id(l))
# 二 支持切片: 數據對象[開始索引:結束索引:步長] 顧頭不顧尾
# print(l[1:4]) # [20, 30, 40]
# print(l[-4:-1]) # [20, 30, 40]
# print(l[1:]) # [20, 30, 40, 50]
# print(l[:3]) # [20, 30, 40, 50]
# print(l[:]) # [10, 20, 30, 40, 50]
# print(l[-2:-5:-1]) # [40, 30, 20]
# 三 in操作: 某個值是否是列表中的某個元素值
# names = ["rain", "yuan", "alvin"]
# print("rai" in names)
# print("rai" in "rain")
# print(1 in ["1", 2, 3])
# # 案例
# n = 0
# while n < 100:
#
# # if n != 88 and n != 66:
# # print(n)
#
# if n not in [66, 88, 33, 44, 55]:
# print(n)
#
# n += 1
# 拼接 +
l1 = [1, 2, 3]
l2 = [4, 5, 6]
print(l1 + l2) # [1, 2, 3, 4, 5, 6]
print(l1)
print(l2)
# 列表內置方法： 列表對象.方法()
# 列表的內置方法是用來管理列表中的數據：增刪改查
# 一 增：append:追加 insert:向任意位置插入值 extend
# print("hello".upper())
# 案例1： append
# l = [10, 20, 30, 40]
# ret = l.append(50)
# print(l) # [10, 20, 30, 40,50]
# print(ret) # None
# 案例2: insert(索引,值)
# l = [10, 20, 30, 40]
# l.insert(0, 100)
# print(l) # [100, 10, 20, 30, 40]
# 案例3: extend
# l1 = [10, 20, 30, [4, 5, 6]]
# 如何獲取5的值
# print(l1[3][1])
# print(len(l1))
# l2 = [1, 2, 3]
# l3 = [4, 5, 6]
# l2.append(l3)
# print(l2) # [1,2,3,[4,5,6]]
# print(len(l2))
# print(id(l2))
# ret = l2.extend(l3)
# print("ret:", ret)
# print(l2) # [1, 2, 3, 4, 5, 6]
# print(id(l2))
# 二 刪 pop():按索引刪除 remove()：按值刪除 clear()：清空
# names = ["張三", "李四", "王五"]
# 按索引刪除
# del_name = names.pop(0) # pop返回刪除元素值
# print(names)
# print("將%s刪除掉了" % del_name)
# 按元素值刪除
# ret = names.remove("李四")
# print(ret) # None
# print(names)
# 清空列表
# names.clear()
# print(names) # []
# del names
# print(names)
# 三 改: 沒有內置方法修改元素，只能通過索引完成
# l = [1, 2, 3]
# l[0] = 100
# print(l) # [100, 2, 3]
# 四 查 index count sort reverse
# names = ["張三", "李四", "王五", "李四"]
# print(names.index("李四"))
# print(names.count("李四"))
#
# age_list = [20, 22, 18, 45, 32]
# # age_list.reverse() # age_list :[32,45,18,22,20]
#
# age_list.sort(reverse=True)
# print(age_list) # [45, 32, 22, 20, 18]
#
# age_list = ["100", "10", "12"]
# age_list.sort()
# print(age_list) # ['10', '100', '32']
# 練習題
# 將字符串 "20,15,34,188,26"中的所有數字進行從大到小排序按"-"作為分隔符組成新的字符串
# 方案1：
# s = "20,15,34,188,26"
# l1 = s.split(",")
# print(l1) # ['20', '15', '34', '188', '26']
# l2 = []
# for i in l1:
# l2.append(int(i))
#
# print(l2)
# l2.sort(reverse=True)
# print(l2)
# l3 = []
# for i in l2: # l2: [15, 20, 26, 34, 188]
# l3.append(str(i))
# print(l3) # ['15', '20', '26', '34', '188']
#
# print("-".join(l3)) # 15-20-26-34-188
# 方案2
s = "20,15,34,188,26"
l1 = s.split(",") # ['20', '15', '34', '188', '26']
l2 = [int(i) for i in l1]
l2.sort(reverse=True)
l3 = [str(i) for i in l2]
print("-".join(l3)) # 188-34-26-20-15
# 列表內置方法： 列表對象.方法()
# 列表的內置方法是用來管理列表中的數據：增刪改查
# 一 增：append:追加 insert:向任意位置插入值 extend
# print("hello".upper())
# 案例1： append
# l = [10, 20, 30, 40]
# ret = l.append(50)
# print(l) # [10, 20, 30, 40,50]
# print(ret) # None
# 案例2: insert(索引,值)
# l = [10, 20, 30, 40]
# l.insert(0, 100)
# print(l) # [100, 10, 20, 30, 40]
# 案例3: extend
# l1 = [10, 20, 30, [4, 5, 6]]
# 如何獲取5的值
# print(l1[3][1])
# print(len(l1))
# l2 = [1, 2, 3]
# l3 = [4, 5, 6]
# l2.append(l3)
# print(l2) # [1,2,3,[4,5,6]]
# print(len(l2))
# print(id(l2))
# ret = l2.extend(l3)
# print("ret:", ret)
# print(l2) # [1, 2, 3, 4, 5, 6]
# print(id(l2))
# 二 刪 pop():按索引刪除 remove()：按值刪除 clear()：清空
# names = ["張三", "李四", "王五"]
# 按索引刪除
# del_name = names.pop(0) # pop返回刪除元素值
# print(names)
# print("將%s刪除掉了" % del_name)
# 按元素值刪除
# ret = names.remove("李四")
# print(ret) # None
# print(names)
# 清空列表
# names.clear()
# print(names) # []
# del names
# print(names)
# 三 改: 沒有內置方法修改元素，只能通過索引完成
# l = [1, 2, 3]
# l[0] = 100
# print(l) # [100, 2, 3]
# 四 查 index count sort reverse
# names = ["張三", "李四", "王五", "李四"]
# print(names.index("李四"))
# print(names.count("李四"))
#
# age_list = [20, 22, 18, 45, 32]
# # age_list.reverse() # age_list :[32,45,18,22,20]
#
# age_list.sort(reverse=True)
# print(age_list) # [45, 32, 22, 20, 18]
#
# age_list = ["100", "10", "12"]
# age_list.sort()
# print(age_list) # ['10', '100', '32']
# 練習題
# 將字符串 "20,15,34,188,26"中的所有數字進行從大到小排序按"-"作為分隔符組成新的字符串
# 方案1：
# s = "20,15,34,188,26"
# l1 = s.split(",")
# print(l1) # ['20', '15', '34', '188', '26']
# l2 = []
# for i in l1:
# l2.append(int(i))
#
# print(l2)
# l2.sort(reverse=True)
# print(l2)
# l3 = []
# for i in l2: # l2: [15, 20, 26, 34, 188]
# l3.append(str(i))
# print(l3) # ['15', '20', '26', '34', '188']
#
# print("-".join(l3)) # 15-20-26-34-188
# 方案2
s = "20,15,34,188,26"
l1 = s.split(",") # ['20', '15', '34', '188', '26']
l2 = [int(i) for i in l1]
l2.sort(reverse=True)
l3 = [str(i) for i in l2]
print("-".join(l3)) # 188-34-26-20-15
# l = [1, 2, 3, 4, 5]
# l2 = []
# for i in l:
# l2.append(i * i)
# print(l2)
# 列表推導式：[表達式 for i in 可迭代對象]
l = [1, 2, 3, 4, 5]
l2 = [i * i for i in l]
print(l2) # [1,4,9,16,25]
# 字典是python的唯一一個hash映射的數據類型
l = ["rain", 22, "male"]
# 字典中一個鍵值對是一個元素
# 字典的鍵必須是不可變數據類型（除了列表和字典數據類型），常用的就是字符串和整型
# 字典的值可以是任意數據類型
# 字典的鍵值對原則上沒有限制
# 字典特點：無序，鍵不能重復
d = {
"name": "rain", "age": 22, "gender": "male"}
print(d, type(d)) # dict
stu = {
"name": "yuan", "age": 22}
# print(len(stu))
# 一 通過鍵查詢值 列表對象[索引] 字典對象[鍵]
print(stu["name"]) # yuan
print(stu["age"]) # yuan
print(stu["xxx"]) # 鍵不存在報錯
# 二 添加或修改
stu["gender"] = "male" # 鍵不存在，則是添加鍵值對
stu["age"] = 24 # 鍵存在，則修改鍵值對
print(stu)
# 刪除鍵值對 del命令
del stu["gender"]
print(stu)
# 字典的內置方法：對字典對象的數據管理
stu = {
"name": "yuan", "age": 22, "gender": "female"}
# 一 查看鍵值對
# print(stu.get("name"))
# print(stu.get("age"))
# print(stu.get("gender", "male"))
# print(stu.keys()) # ['name', 'age']
# print(stu.values()) # ['yuan', 22]
# print(stu.items()) # [('name', 'yuan'), ('age', 22)]
# 二 添加或修改鍵值對：update
# stu.update({"age":32,"gender":"male"})
# ret = stu.setdefault("gender", "male") # 獲取name鍵的值
# print(ret)
# print(stu)
# 三 刪除鍵值對：pop
# ret = stu.pop("age")
# stu.popitem()
# print("age",ret)
# print("stu",stu)
stu = {
"name": "yuan", "age": 22, "gender": "female"}
# print(stu.get("name"))
# k = "age"
# print(stu.get(k))
# 方式1
# for i in stu:
# print(i, stu.get(i)) # 遍歷字典的鍵
# 方式2
# x = [1, 2, 3]
# print(x, type(x)) # tuple
# a, *b = [1, 2, 3]
# print(a, b)
for k,v in stu.items(): # [("name","yuan")],("age",22),("gender","male)]
print(k,v)
# 元組稱之為只讀列表
l = [1, 2, 3]
l[0] = 100
t = (1, 2, 3)
print(t, type(t)) # tuple
print(t[1])
print(t[1:])
print(1 in t)
t.count(1)
t.index(1)
for i in t:
print(i)
# 練習1
stu = {
"name": "yuan", "score": [100, 90, 80]}
# 獲取外語的成績
l = stu.get("score") # [100, 90, 80]
print(l[-1])
print(stu.get("score")[-1])
# 練習2
stu = {
"name": "yuan", "score": {
"chinese": 100, "math": 90, "english": 80}}
print(len(stu))
print(stu.get("score").get("english"))
# 練習3
data = [
{
"sid":1001,"name": "rain", "age": 22},
{
"sid":1003,"name": "eric", "age": 32},
{
"sid":1002,"name": "alvin", "age": 24},
]
for stu_dict in data:
name = stu_dict.get("name")
age = stu_dict.get("age")
print("姓名：%s，年齡：%d" % (name, age))
data2 = {
1001: {
"name": "rain", "age": 22},
1002: {
"name": "eric", "age": 32},
1003: {
"name": "alvin", "age": 24},
}
print(data2.get(1001).get("name"))
for stu_id,stu_info in data2.items():
name = stu_info.get("name")
age = stu_info.get("age")
print("學號：%s，姓名：%s，年齡：%d" % (stu_id,name, age))
# 案例1
# l1 = [1, 2, 3]
# l2 = [4, 5, l1]
# print(l2) # [4, 5, [1, 2, 3]]
# l1[0] = 100
# print(l2)
# 案例2
stu01 = {
"name": "rain"}
stus_dict = {
1001: stu01}
print(stus_dict)
stu01["name"] = "alvin"
print(stus_dict)
print(id(stus_dict[1001]))
stus_dict[1001] = {
"name": "eric"}
print(id(stus_dict[1001]))
print(id(stu01))
# 確定數據以什麼數據類型和格式進行存儲
students_dict = {
1001: {
"name": "yuan",
"scores": {
"chinese": 100,
"math": 89,
"english": 100,
}
},
1002: {
"name": "rain",
"scores": {
"chinese": 100,
"math": 100,
"english": 100,
}
},
}
while 1:
print(''' 1. 查看所有學生成績 2. 添加一個學生成績 3. 修改一個學生成績 4. 刪除一個學生成績 5. 退出程序 ''')
choice = input("請輸入您的選擇：")
if choice == "1":
# 查看所有學生信息
print("*" * 60)
for sid, stu_dic in students_dict.items():
# print(sid,stu_dic)
name = stu_dic.get("name")
chinese = stu_dic.get("scores").get("chinese")
math = stu_dic.get("scores").get("math")
english = stu_dic.get("scores").get("english")
print("學號:%4s 姓名:%4s 語文成績：%4s 數學成績%4s 英文成績:%4s" % (sid, name, chinese, math, english))
print("*" * 60)
elif choice == "2":
while 1:
sid = input("請輸入學生學號>>>")
# 判斷該學號是否存在
if int(sid) in students_dict: # 該學號已經存在！
print("該學號已經存在！")
else: # # 該學號不存在！
break
name = input("請輸入學生姓名>>>")
chinese_score = input("請輸入學生語文成績>>>")
math_score = input("請輸入學生數學成績>>>")
english_score = input("請輸入學生英語成績>>>")
# 構建學生字典
scores_dict = {
"chinese": chinese_score,
"math": math_score,
"english": english_score,
}
stu_dic = {
"name": name,
"scores": scores_dict
}
print("stu_dic", stu_dic)
students_dict[int(sid)] = stu_dic
elif choice == "3":
while 1:
sid = input("請輸入學生學號>>>")
# 判斷該學號是否存在
if int(sid) in students_dict: # 該學號已經存在！
break
else: # # 該學號不存在！
print("該修改學號不存在！")
chinese_score = input("請輸入學生語文成績>>>")
math_score = input("請輸入學生數學成績>>>")
english_score = input("請輸入學生英語成績>>>")
# 修改學生成績
scores_dict = {
"chinese": chinese_score,
"math": math_score,
"english": english_score,
}
students_dict.get(int(sid)).update({
"scores": scores_dict})
print("修改成功")
elif choice == "4":
while 1:
sid = input("請輸入學生學號>>>")
# 判斷該學號是否存在
if int(sid) in students_dict: # 該學號已經存在！
break
else: # # 該學號不存在！
print("該修改學號不存在！")
students_dict.pop(int(sid))
print("刪除成功")
elif choice == "5":
# 退出程序
break
else:
print("輸入有誤！")
