Roadmap¶
- 10/27 練習題
- 流程敘述回顧
- 更多變數型態
- 古典加密法
10/27 練習題¶
- 產生無重複數字的三位數
- 列出100~999的數字
- 利用除法、取餘數，取得各位數的值 or 轉成字串，取各位數
- 比較各位數，要三個位數都不相等
In [28]:
for number in range(100,999+1):
str_number = str(number)
if str_number[0]!=str_number[1] and str_number[0]!=str_number[2] and str_number[1]!=str_number[2]:
print (number)
if 學生:
price = 500
elif 博愛票:
price = 300
elif 身高120以下:
price = 200
else:
price = 700
if 早鳥票:
price = int(price*0.8)
while not 我很滿意你的照顧:
想辦法讓對方的心情變好
while 血量>0:
繼續戰鬥
今日客人人數 = 0
while 營業時間:
if 有人進來:
今日客人人數 += 1
結算今日客人人數
In [29]:
my_list = [5,8,1,4,2]
index = 0 # 從第1個
while index<len(my_list): # 直到拿到最後一個
print (my_list[index])
index+=1 # 每次只拿1個，這次的拿完了，換下一個
In [25]:
my_list = [5,8,1,4,2]
for index in range(0,len(my_list)):
print (my_list[index])
In [26]:
my_list = [5,8,1,4,2]
for each_value in my_list:
print (each_value)
In [30]:
my_string = "PyLadies"
for each_alpha in my_string:
print (each_alpha)
更多變數型態¶
bool¶
- 布林值：條件判斷
- True / False
In [76]:
print (2>1)
my_status = False
print (my_status)
應用情境¶
In [104]:
# 找出是否有出現三次以上的數字
found = False
my_list = [2,8,6,7,6,8,9,5566,8]
for i in my_list:
if my_list.count(i)>=3:
found = True
if found:
print ("有出現三次以上的數字")
else:
print ("沒有出現三次以上的數字")
ord¶
ord("字母")：回傳該字母的編碼值- 只能處理長度為1的字串
In [41]:
print (ord("A"))
print (ord("Z"))
print (ord("a"))
print (ord("z"))
print (ord("Ａ"))
chr¶
chr(數字)：回傳該數字編碼後的字母
In [43]:
print (chr(65))
print (chr(122))
print (chr(65313))
應用情境¶
In [51]:
# 資料正規化 - 舊版
brand = "ＡＰＰＬＥ"
brand = brand.replace("Ａ","A").replace("Ｐ","P").replace("Ｌ","L").replace("Ｅ","E")
print (brand)
In [1]:
# 資料正規化 - 新版
brand = "ＡＰＰＬＥ"
for i in range(ord("Ａ"),ord("Ｚ")+1):
brand = brand.replace(chr(i),chr(i-(ord("Ａ")-ord("A"))))
print (brand)
In [7]:
# 資料正規化 - 月份
month_list = ["Jan","Feb","Mar","Apr","May","Jun","Jul","Aug","Sep","Oct","Nov","Dec"]
ptt_datatime = "Tue Oct 18 23:22:05 2016"
ptt_split_list = ptt_datatime.split(" ")
ptt_year = ptt_split_list[-1]
ptt_month = ptt_split_list[1]
for index in range(0,len(month_list)):
ptt_month = ptt_month.replace(month_list[index],str(index+1))
print (ptt_month)
古典加密法-凱撒密碼¶
think¶
- 可以直接對全文用replace嗎？
- Y -> ？
In [5]:
print (ord("M")-ord("K"))
print (chr(ord("K")+2))
In [8]:
article = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
new_article = ""
code = 2
for i in article:
if ord(i) in range(ord("a"),ord("z")+1):
if (ord(i)+code) >= ord("z"):
new_article+= chr(ord(i)+code-26)
else:
new_article+= chr(ord(i)+code)
else:
new_article+= i
print (new_article)
think¶
- 如果沒有解密表...
印出每個字母出現幾次¶
- 同一個字母只會印一次
- 只印出英文字母
In [72]:
s = "g fmnc wms bgblr rpylqjyrc gr zw fylb. rfyrq ufyr amknsrcpq ypc dmp. bmgle gr gl zw fylb gq glcddgagclr ylb rfyr'q ufw rfgq rcvr gq qm jmle. sqgle qrpgle.kyicrpylq() gq pcamkkclbcb. lmu ynnjw ml rfc spj."
unique_word_list=[]
for i in s:
if (ord(i) in range(ord("a"),ord("z")+1)) and (i not in unique_word_list):
unique_word_list.append(i)
print (i,s.count(i))
到目前為止教的
- 變數、型別轉換、串列
- 判斷式(if/elif/else)
- 迴圈(while/for)
- 輸入與輸出(input/print)
都是程式設計中最基礎需要熟悉的，
要解決生活上的問題，都需要這些基礎
CheckIO¶
- Even the last
- Digits Multiplication
- Right to Left
- Non-unique Elements
- Verify anagrams
- Secret Message
以上都是今天教得語法可以解的題目
沒有頭緒嗎？下一頁有提示用哪些語法去解題¶
CheckIO-Hint¶
- Even the last：for、%、變數累加、if
- Digits Multiplication：for、型別轉換、if
- Right to Left：for、replace
- Non-unique Elements：for、if、count、append
- Verify anagrams：for、if、in、len、count
- Secret Message：for、if、split、字串位置、字串相加、ord、range