如何防止字典中的元素不斷更新？

這是完整的代碼：

import re

f = open('movies.item','r') 
# First three item of movies.item below:
#1|Toy Story (1995)|01-Jan-1995||http://us.imdb.com/M/title-exact?Toy Story (1995)|0|0|0|1|1|1|0|0|0|0|0|0|0|0|0|0|0|0|0 
#2|GoldenEye (1995)|01-Jan-1995||http://us.imdb.com/M/title-exact?GoldenEye (1995)|0|1|1|0|0|0|0|0|0|0|0|0|0|0|0|0|1|0|0
#3|Four Rooms (1995)|01-Jan-1995||http://us.imdb.com/M/title-exact?Four Rooms (1995)|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|0|1|0|0
empty_list = []
for items in f:
    new_item = re.sub(r'\n', '', items)
    empty_list.append(new_item)
movie_names = []
splitted_list = None
for i in range(len(empty_list)):

    splitted_list = empty_list[i].split("|")
    movie_names.append(splitted_list[1])

genres = ["Unknown", "Action", "Adventure", "Animation", "Children's","Comedy", "Crime", "Documentary", "Drama",
"Fantasy", "Film-Noir", "Horror", "Musical", "Mystery","Romance", "Sci-Fi", "Thriller", "War", "Western"]
genres.reverse()
genredict = {}
last_dict = {}
reversegenresum = []
for i in range(len(empty_list)):
    x = list(empty_list[i])
    claer_list = []
    for k in range(len(x)):
        if x[k] != "|":
            claer_list.append(x[k])

    claer_list.reverse()
    reverse_genre_data = claer_list[0:19]
    reversegenresum.append(reverse_genre_data)
    

for i in range(3): #trying for 3 movie
    for j in range(len(genres)): 
        if reversegenresum[i][j] == '1':      
            genredict[genres[j]] = '1'
        last_dict[movie_names[i]] = genredict    


print(last_dict) 

我想做什么？我嘗試匹配名為“movies.item”的檔案中的資料。有電影及其資料資訊，如'0|0|1|0'。如果資料的值等于 1 我需要將其與相應的類別進行匹配。但我只能為 1 部電影做到這一點。盡管我嘗試以其他方式執行此操作時不會出現錯誤，但我的所有資料都是根據最后一個資料進行整形的。如果您不明白我的意思，請復制代碼并自己嘗試。

輸入 ：

{'Toy Story (1995)': {'Comedy': '1', "Children's": '1', 'Animation': '1', 'Thriller': '1', 'Adventure': '1', 'Action': '1'}, 
 'GoldenEye (1995)': {'Comedy': '1', "Children's": '1', 'Animation': '1', 'Thriller': '1', 'Adventure': '1', 'Action': '1'}, 
 'Four Rooms (1995)': {'Comedy': '1', "Children's": '1', 'Animation': '1', 'Thriller': '1', 'Adventure': '1', 'Action': '1'}}  

我想要的是：

{'Toy Story (1995)': {'Animation': 1, "Children's": 1, 'Comedy': 1}, 
 'GoldenEye (1995)': {'Action': 1, 'Adventure': 1, 'Thriller': 1}, 
 'Four Rooms (1995)': {'Thriller': 1},

uj5u.com熱心網友回復：

問題

genredict您不斷為每部電影保存完全相同的字典

last_dict[movie_names[i]] = genredict  

簡單修復

j為每部電影使用一個新的字典，并在回圈之后分配它就足夠了

for i in range(3):  
    genredict = {}
    for j in range(len(genres)):
        if reversegenresum[i][j] == '1':
            genredict[genres[j]] = '1'
    last_dict[movie_names[i]] = genredict

提升

您基本上有 4 個回圈遍歷同一件事：電影，而不是在每部電影上一個接一個地執行動作，而是在電影上一個接一個地執行它們

genres = ["Unknown", "Action", "Adventure", "Animation", "Children's", "Comedy", "Crime", "Documentary", "Drama",
          "Fantasy", "Film-Noir", "Horror", "Musical", "Mystery", "Romance", "Sci-Fi", "Thriller", "War", "Western"]
result = {}
with open('movies.item', 'r') as f:
    for items in f:
        index, name, date, url, _, *values = items.rstrip("\n").split("|")
        item_genre = dict(zip(genres, values))
        result[name] = {genre: value for genre, value in item_genre.items() if value == '1'}

• 將行拆分一次并檢索您需要的所有元素：名稱和值
• dict(zip( , ))將流派和價值觀配對
• {genre: value for genre, value in item_genre.items() if value == '1'}只保留流派1

請注意，最后一行應該更好，您不需要所有值都相同的字典（1），只需保留一個串列

result[name] = [genre for genre, value in item_genre.items() if value == '1']

# {'Toy Story (1995)': ['Animation', "Children's", 'Comedy'], 'GoldenEye (1995)': ['Action', 'Adventure', 'Thriller'], 'Four Rooms (1995)': ['Thriller']}

標籤：Python 列表 字典 集体智慧

上一篇：Python如何將帶有特殊字符的字串轉換為python中的dict

下一篇：如何除錯倒排索引？