Roadmap¶
- 11/17 練習題
- 排序
- 函式
- 字典
- 檔案IO
In [ ]:
article = """
Marvel braves a stumble with each dip into its grab bag of second-tier, less-recognizable characters.
But it won't come with "Doctor Strange," an extremely entertaining and sure-footed adaptation that manages to conjure more than enough magic to easily pass its spell check.
Much of that has to do with a splendid cast, led by Benedict Cumberbatch.
Having top-flight actors -- including Chiwetel Ejiofor, Tilda Swinton and Mads Mikkelsen -- is especially helpful with this enterprise.
That's because some of the mystical-mumbo-jumbo dialogue -- as in, "Dormammu dwells in the dark dimension" -- could sound stilted or silly tripping off less talented tongues.
Largely faithful to the comics, the film tells a straightforward origin story -- in hindsight, an area where Marvel has generally excelled.
The first "Iron Man," for example, is far more satisfying than either of its sequels.
A brilliant neurosurgeon, Dr. Stephen Strange has his million-dollar hands mangled in a terrible car accident.
The closing credits might elicit the biggest laugh with a disclaimer warning about the dangers of distracted driving.
Desperate to be restored, Strange begins exploring alternative means of healing, a search that ultimately leads him to Nepal, a hidden retreat and a bald sage known as the Ancient One (Tilda Swinton, in glorious-weirdness mode).
Aided by her chief lieutenant Mordo (Ejiofor), they train him in the mystic arts.
"""
In [30]:
_symbol = [",",":",";","?","!","(",")","\"","$","[","]","{","}","\n","\t","-","."]
clean_article = article
# 去除所有的標點符號
for i in _symbol:
clean_article = clean_article.replace(i," ")
# 把大寫改成小寫
for i in range(ord('A'),ord('Z')+1):
clean_article = clean_article.replace(chr(i),chr(i+32)) # ord("A")-ord("a") = -32
# 把文章切成單字
split_article = clean_article.split(" ")
# 計算文章出現最多的單字
max_count = 0
max_word = ""
for word in split_article:
if word and split_article.count(word)>max_count:
max_count = split_article.count(word)
max_word = word
print (max_word,max_count)
think¶
- 如果要拿第二多的、第五多的...
In [36]:
simple_list = [1,5,3,4,2]
print (sorted(simple_list))
print (sorted(simple_list,reverse=True))
In [47]:
nested_list_number = [[5,4],[1,6],[3,7],[4,2]]
nested_list_string = [["Python",4],["Hello",6],["World",7],["PyLadies",2]]
nested_list_ch_string = [["中文",4],["台北",6],["高雄",7],["台中",2]]
print (sorted(nested_list_number))
print (sorted(nested_list_string))
print (sorted(nested_list_ch_string))
print (ord("中"),ord("台"),ord("中"),ord("北"),ord("高"))
[Coding Time]¶
把文章的單字、頻率以串列型別儲存
[ ['marvel', 2], ['braves', 1], ...... ]不要存到空白字
- 不要存到重複的單字！
- 可以紀錄已經出現哪些單字
- 可以試著使用
sorted()排序看看
In [4]:
_symbol = [",",":",";","?","!","(",")","\"","$","[","]","{","}","\n","\t","-","."]
clean_article = article
for i in _symbol:
clean_article = clean_article.replace(i," ")
for i in range(ord('A'),ord('Z')+1):
clean_article = clean_article.replace(chr(i),chr(i+32))
split_article = clean_article.split(" ")
# [Coding Time]
my_list = []
word_list = []
for word in split_article:
if (word) and (word not in word_list):
word_list.append(word)
count = split_article.count(word)
my_list.append([word,count])
print (my_list)
In [5]:
print (sorted(my_list))
去咖啡機弄一杯咖啡¶
- 打開開關
- 放下杯子
- 選擇「咖啡類別」，按下製作
- 咖啡機「製作咖啡」
- 得到一杯咖啡
def 製作咖啡(咖啡類別):
加入濃縮咖啡
if 咖啡類別=="美式咖啡":
加入很多的水
elif 咖啡類別=="拿鐵" or 咖啡類別=="Latte":
加入很多的牛奶，還有一些奶泡
elif 咖啡類別=="卡布奇諾":
加入一些牛奶，還有一些奶泡
.
.
.
return 咖啡
In [58]:
# 以第二個值排序
def my_value_sort(item):
return item[1]
nested_list_number = [[5,4],[1,6],[3,7],[4,2]]
print (sorted(nested_list_number,key=my_value_sort))
In [61]:
# 不分大小寫排序
def my_string_sort_old(item):
for i in range(ord('A'),ord('Z')+1):
item = item.replace(chr(i),chr(i+32))
return item
def my_string_sort(item):
return item.lower()
list_string = ["hi","Hello","HAPPY"]
print (sorted(list_string,key=my_string_sort_old))
print (sorted(list_string,key=my_string_sort))
In [65]:
def my_sort(item):
return item[1]
_symbol = [",",":",";","?","!","(",")","\"","$","[","]","{","}","\n","\t","-","."]
clean_article = article
for i in _symbol:
clean_article = clean_article.replace(i," ")
clean_article = clean_article.lower()
split_article = clean_article.split(" ")
my_list = []
word_list = []
for word in split_article:
if (word) and (word not in word_list):
word_list.append(word)
count = split_article.count(word)
my_list.append([word,count])
# [Coding Time]
for i in sorted(my_list,key=my_sort,reverse=True):
print (i[0],i[1])
如果頻率相同，以字母排序¶
In [66]:
def my_sort(item):
return item[1],item[0]
# 中略
# 中略
for i in sorted(my_list,key=my_sort,reverse=True):
print (i[0],i[1])
出現頻率多在前，同樣頻率按字母順序排序¶
- reverse 是直接把排序好的結果變成降冪
In [67]:
def my_sort(item):
return -item[1],item[0]
# 中略
# 中略
for i in sorted(my_list,key=my_sort):
print (i[0],i[1])
think¶
- 如果我要從串列中找「strange」這個單字出現幾次
In [69]:
for i in my_list:
if i[0]=="strange":
print (i[0],i[1])
In [81]:
my_dict = {}
my_dict["the"] = 10
my_dict["a"] = 9
my_dict["of"] = 6
my_dict["in"] = 6
my_dict2 = {"the":10,"a":9,"of":6,"in":6}
print (my_dict)
print (my_dict2)
print (my_dict["the"])
!注意¶
- 直接使用字典不存在的key去找資料會有錯誤！
In [86]:
my_dict = {"the":10,"a":9,"of":6,"in":6}
print ("with" in my_dict)
print (my_dict["with"])
[Coding Time]¶
- 把文章的單字、頻率以字典型別儲存
In [3]:
_symbol = [",",":",";","?","!","(",")","\"","$","[","]","{","}","\n","\t","-","."]
clean_article = article.lower()
for i in _symbol:
clean_article = clean_article.replace(i," ")
word_dict = {}
split_article = clean_article.split(" ")
for word in split_article:
if (word) and (word not in word_dict):
word_dict[word] = split_article.count(word)
print (word_dict)
字典的排序¶
In [84]:
my_dict = {"the":10,"a":9,"of":6,"in":6}
print (my_dict.items())
for i in my_dict.items():
print (i)
for i in my_dict.items():
print (i[0],i[1])
In [90]:
def my_sort(item):
return -item[1],item[0]
my_dict = {"the":10,"a":9,"of":6,"in":6}
for i in sorted(my_dict.items(),key=my_sort):
print (i[0],i[1])
In [91]:
def my_sort(item):
return -item[1],item[0]
my_dict2 = {
"the":[10,1],
"a": {"b":1},
"of":6,
"in":6
}
for i in sorted(my_dict2.items(),key=my_sort):
print (i[0],i[1])
In [101]:
def my_sort(item):
if type(item[1])==type({}) or type(item[1])==type([]) :
return 0,item[0]
else:
return -item[1],item[0]
my_dict2 = {
"the":[10,1],
"a": {"b":1},
"of":6,
"in":6
}
for i in sorted(my_dict2.items(),key=my_sort):
print (i[0],i[1])
字典與串列的比較¶
- 串列：索引index => 值value
- 字典：鍵值key => 值value
In [109]:
my_dict = {"the":10,"a":9,"of":6,"in":6}
print (list(my_dict.items()))
print (type(my_dict.items()))
my_list = [10,9,6,6]
print (list(enumerate(my_list)))
print (type(enumerate(my_list)))
In [111]:
fileout = open("test.txt",'w')
fileout.write("Hello"+"\n")
fileout.write("Python")
fileout.close()
filein = open("test.txt")
content = filein.read()
filein.close()
print (content)
分析電影¶
IMDB電影資料集(需要註冊才能下載)
- 建立一個資料集：以不同的資料類別當成鍵值/索引
- 電影名稱(movie_title)、票房(gross)、類別(genres)、成本(budget)、IMDB評分(imdb_score)
- 演員名稱(actor_x_name)、FB按讚數(actor_x_facebook_likes)
- 試著以票房排序看看
- 看看熱賣的電影
- 都屬於哪些類別(ex.動作片)
- 是不是有明星加持：明星可能是以按讚數、或是演的電影總數
- 評分就會高嗎？
- 是否賺錢(票房-成本)
- 演員演的電影，會依電影類別有票房差嗎？跟熱賣電影的類別是一樣的嗎？
- 看看熱賣的電影
!注意¶
- csv是以,分開每一欄的資料，所以可以利用split(",")去分割，並用串列的索引值去拿取想要的欄位
In [6]:
my_data = """
movie1,2
movie2,5
movie1,1
movie3,7
"""
my_dict = {}
for i in my_data.split("\n"):
print (i)
if i:
temp = i.split(",")
movie = temp[0]
count = int(temp[1])
if movie not in my_dict:
my_dict[movie]=count
else:
my_dict[movie]+=count
print (my_dict)