Python的排序問題 - 給出的列表[ '網址'， '標籤1'， '標籤2'，..] S和搜索規範[ '標籤3'， '標籤1'，...]，返回相關的URL列表

Question

我m對於編程來說相當新穎，所以我確信有一種更好的方式來構成這個問題，但我正在嘗試創建一個個人書籤程序。給定多個網址，每個網址都有一個按相關性排序的標籤列表，我希望能夠創建一個由一系列標籤組成的搜索，這些標籤返回最相關的url列表。下面我的第一個解決方案是讓第一個標籤的值爲1，第二個爲&，讓python list sort函數完成剩下的工作。 2個問題：

1）是否有更加優雅/有效的方式來做到這一點（令我難堪！） 2）通過給定上述輸入問題的相關性排序的任何其他一般方法？

非常感激。

# Given a list of saved urls each with a corresponding user-generated taglist 
# (ordered by relevance), the user enters a "search" list-of-tags, and is 
# returned a sorted list of urls. 

# Generate sample "content" linked-list-dictionary. The rationale is to 
# be able to add things like 'title' etc at later stages and to 
# treat each url/note as in independent entity. But a single dictionary 
# approach like "note['url1']=['b','a','c','d']" might work better? 

content = [] 
note = {'url':'url1', 'taglist':['b','a','c','d']} 
content.append(note) 
note = {'url':'url2', 'taglist':['c','a','b','d']} 
content.append(note) 
note = {'url':'url3', 'taglist':['a','b','c','d']} 
content.append(note) 
note = {'url':'url4', 'taglist':['a','b','d','c']} 
content.append(note) 
note = {'url':'url5', 'taglist':['d','a','c','b']} 
content.append(note) 

# An example search term of tags, ordered by importance 
# I'm using a dictionary with an ordinal number system 
# This seems clumsy 
search = {'d':1,'a':2,'b':3} 

# Create a tagCloud with one entry for each tag that occurs 
tagCloud = [] 
for note in content: 
    for tag in note['taglist']: 
     if tagCloud.count(tag) == 0: 
      tagCloud.append(tag) 

# Create a dictionary that associates an integer value denoting 
# relevance (1 is most relevant etc) for each existing tag 

d={}    
for tag in tagCloud: 
    try: 
     d[tag]=search[tag] 
    except KeyError: 
     d[tag]=100 

# Create a [[relevance, tag],[],[],...] result list & sort 
result=[]  
for note in content: 
    resultNote=[] 
    for tag in note['taglist']: 
     resultNote.append([d[tag],tag]) 
    resultNote.append(note['url']) 
    result.append(resultNote) 
result.sort() 

# Remove the relevance values & recreate a list containing 
# the url string followed by corresponding tags. 
# Its so hacky i've forgotten how it works! 
# It's mostly for display, but suggestions on "best-practice" 
# intermediate-form data storage? 

finalResult=[] 
for note in result: 
    temp=[] 
    temp.append(note.pop()) 
    for tag in note: 
     temp.append(tag[1]) 
    finalResult.append(temp) 

print "Content: ", content 
print "Search: ", search 
print "Final Result: ", finalResult 

Answer 1

1）是否有這樣做的更優雅/有效的方式（讓我難堪！）

當然可以。基本思路：不要試圖告訴Python該做什麼，只要問它想要什麼。

content = [ 
    {'url':'url1', 'taglist':['b','a','c','d']}, 
    {'url':'url2', 'taglist':['c','a','b','d']}, 
    {'url':'url3', 'taglist':['a','b','c','d']}, 
    {'url':'url4', 'taglist':['a','b','d','c']}, 
    {'url':'url5', 'taglist':['d','a','c','b']} 
] 

search = {'d' : 1, 'a' : 2, 'b' : 3} 

# We can create the tag cloud like this: 
# tagCloud = set(sum((note['taglist'] for note in content), [])) 
# But we don't actually need it: instead, we'll just use a default value 
# when looking things up in the 'search' dict. 

# Create a [[relevance, tag],[],[],...] result list & sort 
result = sorted(
    [ 
     [search.get(tag, 100), tag] 
     for tag in note['taglist'] 
    ] + [[note['url']]] 
    # The result will look like [ [relevance, tag],... , [url] ] 
    # Note that the url is wrapped in a list too. This makes the 
    # last processing step easier: we just take the last element of 
    # each nested list. 
    for note in content 
) 

# Remove the relevance values & recreate a list containing 
# the url string followed by corresponding tags. 
finalResult = [ 
    [x[-1] for x in note] 
    for note in result 
] 

print "Content: ", content 
print "Search: ", search 
print "Final Result: ", finalResult 

Answer 2

我建議你也給一個權重給每個標籤，這取決於它是多麼難得的（例如，「狼蛛」標籤將重量超過「自然」tag¹更多）。對於給定的URL，罕見的標記，是常見的其他網址應當標註較強的相關性，而給定的URL 不存在於另一網址經常使用的標籤應當標註下的相關性。

可以很容易地轉換我上面描述的作爲每隔URL數值相關的計算規則。

¹除非您的所有網址都與「狼蛛」相關，當然:)