Question

我有一张包含以下内容的表：

table = [[1, 'FANTASTIC FOUR', 'EXOTIC SPACE'],[4, 'CRIMSON PEAK', 'MINIONS','SUPERMAN'],[20, 'FANTASTIC FOUR', 'EXOTIC SPACE']]

我正在编写一个python函数来遍历表格，查找字符串元素中的相似之处并以格式打印出来：

Movie: FANTASTIC FOUR, EXOTIC SPACE
UserID: 1,20   #since user 1 and user 20 both watch exactly the same movie

我试过写作：

i = 0
while i<len(table)-1:
    g = table[i][1:]
    if g == table[i+1][1:]:
        print(table[i][0],table[i+1][0])
    i+=1

但它的效果不佳。我并不擅长使用while循环进行打印，所以我会对此有所帮助。

Answer 1

Python中的循环通常不会非常使用i。试试这个：

table = [[1, 'FANTASTIC FOUR', 'EXOTIC SPACE'],[4, 'CRIMSON PEAK', 'MINIONS','SUPERMAN'],[20, 'FANTASTIC FOUR', 'EXOTIC SPACE']]

watcher = {}

for x in table:
    for movie in x[1:]:
        watcher_for_movie = watcher.get(movie, [])
        watcher_for_movie.append(x[0])
        watcher[movie] = watcher_for_movie

print(watcher)

输出：

{'EXOTIC SPACE': [1, 20], 'CRIMSON PEAK': [4], 'MINIONS': [4], 'SUPERMAN': [4], 'FANTASTIC FOUR': [1, 20]}

Answer 2

以下是使用itertool.combinations和词典的一种解决方案。

对于字典键使用集合或frozensets是最佳选择，因为无论您是查找(1, 20)还是(20, 1)，都希望获得相同的结果。

from itertools import combinations

table = [[1, 'FANTASTIC FOUR', 'EXOTIC SPACE'],
         [4, 'CRIMSON PEAK', 'MINIONS','SUPERMAN'],
         [20, 'FANTASTIC FOUR', 'EXOTIC SPACE']]

d = {k: set(v) for k, *v in table}

common = {frozenset((i, j)): d[i] & d[j] for i, j in \
          combinations(d, 2) if d[i] & d[j]}

# {frozenset({1, 20}): {'EXOTIC SPACE', 'FANTASTIC FOUR'}}

扭转这种映射也是微不足道的：

common_movies = {frozenset(v): set(k) for k, v in common.items()}

# {frozenset({'EXOTIC SPACE', 'FANTASTIC FOUR'}): {1, 20}}

Answer 3

您可以使用词典来吸引观看同一部电影的用户table

table = [[1, 'FANTASTIC FOUR', 'EXOTIC SPACE'],[4, 'CRIMSON PEAK', 'MINIONS','SUPERMAN'],[20, 'FANTASTIC FOUR', 'EXOTIC SPACE']]
movie_user_mapping = dict() # Create an empty dictionary

# Iterate over every item in table
for item in table:
     # Loop over the movies i.e excluding the first element
     for movie in item[1:]:
         # Check if movie is present as key in the dictionary, if not create a new key with the movie name and assign it an empty list
         if movie not in movie_user_mapping:
             movie_user_mapping[movie] = []
         # Check if user is already added to the list for the movie, if not add the user
         if item[0] not in movie_user_mapping[movie]:
             movie_user_mapping[movie].append(item[0])

# Print the result
print(movie_user_mapping)

输出：

{'FANTASTIC FOUR': [1, 20], 'EXOTIC SPACE': [1, 20], 'CRIMSON PEAK': [4], 'MINIONS': [4], 'SUPERMAN': [4]}

在列表列表中搜索类似元素

3 个答案: