機試題

lis = [['哇',['how',{'good':['cd',1000,'99']},'大帥哥'],'I']] （總分2分）

• 列表lis中的'cd'全部變成大寫。(1分)

• 列表中的1000通過數字相加在轉換成字符串的方式變成'10086'。(1分)

  lis = [['哇',['how',{'good':['cd',1000,'99']},'大帥哥'],'I']] lis[0][1][1]["good"][0] =lis[0][1][1]["good"][0].upper() print(lis) lis[0][1][1]["good"][1] = str(lis[0][1][1]["good"][1]+9086) print(lis)

dic = {'k1':'v1','k2':['alex','sb'],(1,2,3,):{'k3':['2',100,'wer']}} （總3分）

• 將'k3'對應的值的最后面添加一個元素(1,2,3)。(1分)

• 將'k2'對應的值的第2個位置前插入元素{'a'}。(1分)

• 將(1,2,3,)對應的值添加一個鍵值對key:(1,)。(1分)

  dic = {'k1':'v1','k2':['alex','sb'],(1,2,3,):{'k3':['2',100,'wer']}} dic[(1,2,3)]["k3"].append((1,2,3)) dic["k2"].insert(1,{"a"}) dic[(1,2,3)].update({"key":(1,)}) print(dic)

敲七游戲. 從1開始數數. 遇到7或者7的倍數包(含17,27,這種數）要在桌上敲?下. 編程來完成敲七. 給出?個任意的數字n. 從1開始數. 數到n結束. 把每個數字都放在列表中, 在數的過程中出現7或 者7的倍數（包含17,27,這種數）.則向列表中添加?個'咣' 例如, 輸?20.

lst = [1, 2, 3, 4, 5, 6, '咣', 8, 9, 10,11,12,13,'咣',15,16,'咣',18,19,20]

while True:s = input("請輸入一個正整數:")if s.isdecimal() and int(s)>0:lst = []for i in range(1,int(s)+1):if i % 7 == 0 or "7" in str(i):lst.append("咣")else:lst.append(i)print(lst)breakelse:print("你輸入的格式有誤,請重新輸入!")

完成36選7的操作, 1- 50之間隨機產生36個數,從這36個數中選擇7個不重復的數添加到列表中(5分)

from random import randint print(randint(1,20)) # 從1到20之間隨機產生1個整數 from random import randint lst = [] for i in range(36):lst.append(randint(1, 50)) lst = list(set(lst))[:7] print(lst)

有文件t1.txt里面的內容為:（5分）

1,alex,22,13651054608,IT 2,wusir,23,13304320533,Tearcher 3,taibai,18,1333235322,IT利用文件操作，將其構造成如下數據類型。 [{'id':'1','name':'alex','age':'22','phone':'13651054608','job':'IT',{'id':'2','name':'wusir','age':'23','phone':'13304320533','job':'Tearcher'},{'id':'3','name':'taibai','age':'18','phone':'1333235322','job':'IT'}] with open("t1.txt","r",encoding="utf-8") as f:lst = f.readlines() lst_job = [] for i in lst:lst1 = i.strip().split(",")dic = {"id":lst1[0],"name":lst1[1],"age":lst1[2],"phone":lst1[3],"job":lst1[4]}lst_job.append(dic) print(lst_job)

有如下車牌和車輛歸屬地,形成一個新的字典,顯示每個歸屬地的車輛共有多少：(8分)

cars = ['魯A32444','魯A12333','湘B8989M','?A49678','?B46555','滬B25044','冀A11111',"京A01101"] locals = {'冀':{"A":"石家莊","B":"唐山"},'?':{"A":"哈爾濱","B":"齊齊哈爾"},'魯':{"A":"濟南","B":"青島"},'鄂':{"A":"武漢","B":"黃石"},'湘':{"A":"長沙","B":"株洲"},}結果: {'濟南':2, '株洲': 1, '哈爾濱': 1,'齊齊哈爾':1,'石家莊':1} cars = ['魯A32444','魯A12333','湘B8989M','?A49678','?B46555','滬B25044','冀A11111',"京A01101"] locals = {'冀':{"A":"石家莊","B":"唐山"},'?':{"A":"哈爾濱","B":"齊齊哈爾"},'魯':{"A":"濟南","B":"青島"},'鄂':{"A":"武漢","B":"黃石"},'湘':{"A":"長沙","B":"株洲"},}dic ={} for i in cars:if i[0] in locals:dic[locals[i[0]][i[1]]] = dic.get(locals[i[0]][i[1]],0)+1 print(dic)

有如下值li 將所有的數字保存至字典的第一個key中，將所有的字符串保存至第二個key的值中(4分)

li= ["a","b",11,22,33,44,55,"66",["alex","baoyuan","taibai"],"77","88","99"] li = ["a","b",11,22,33,44,55,"66",["alex","baoyuan","taibai"],"77","88","99"] dic = {"key1":[],"key2":[]} for i in li:if type(i) ==list:for j in i:if type(j) == int:dic["key1"].append(j)else:dic["key2"].append(j)elif type(i) == int:dic["key1"].append(i)else:dic["key2"].append(i) print(dic)

userinfo.txt 文件中存放以下結構:(總分8分)

alex:alex3714 wusir:123456 meet:meet123

1.讓用戶選擇:

1.注冊 2.登錄

2.用戶選擇注冊就將賬號和密碼添加到userinfo.txt中,如果用戶名存在就提示用戶名存在,不存在就進行添加(2分)

3.用戶選擇登錄,就驗證用戶的賬號和密碼是否與userinfo.txt一致,如果一致終止循環提示登錄成功(3分)

4.讓用戶登錄三次,三次錯誤將用戶進行鎖定提示用戶名已鎖定,并打印錯誤次數(使用字符串格式化)(3分)

with open("userinfo.txt ","r",encoding="utf-8")as f1:lst = f1.readlines()dic = {} for i in lst:k,v = i.strip().split(":")dic[k]=v while True:chose = input("請輸入你的選擇(1.注冊 2.登錄)\n>>>")if chose == "1":name = input("請輸入用戶名:")psd = input("請輸入密碼:")psd1 = input("請再次輸入密碼:")if name in dic:print("用戶名已存在,注冊失敗!")continueelif psd != psd1:print("兩次密碼輸入不一致,請重新注冊!")else:dic[name] = psdprint("注冊成功!")with open("userinfo.txt","a",encoding="utf-8")as f2:f2.write(f"\n{name}:{psd}")breakelif chose == "2":count = 3while count :count -=1name2 = input("請輸入用戶名:")psd2 = input("請輸入密碼:")if name2 in dic :if psd2 == dic[name2]:print("登陸成功!")breakelse:print("密碼錯誤!")if count:print(f"你還有{count}次機會")else:print("此用戶名已鎖定!")else:print("賬號不存在!")if count:print(f"你還有{count}次機會")else:print("此用戶名已鎖定!")breakelse :print("請輸入正確的選擇")

轉載于:https://www.cnblogs.com/ciquankun/p/11197576.html

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