我有 5 個單詞串列，它們基本上充當字典中的值，其中鍵是檔案的 ID。
對于每個檔案，我想應用一些計算并在嵌套字典中顯示計算的值和結果。
到目前為止一切順利，我設法做所有事情，但我在最簡單的部分失敗了。
當顯示生成的嵌套字典時，似乎它只是迭代5 個串列中每個串列的最后一個元素，因此沒有顯示所有元素......
誰能解釋我在哪里失敗？
這是原始字典 data_docs：
{'doc01': ['simpl', 'hello', 'world', 'test', 'python', 'code'],
'doc02': ['today', 'wonder', 'day'],
'doc03': ['studi', 'pac', 'today'],
'doc04': ['write', 'need', 'cup', 'coffe'],
'doc05': ['finish', 'pac', 'use', 'python']}
這是我得到的結果（例如 doc01 中缺少 'simpl'、'hello'、'world'、'test'、'python'）：
{'doc01': {'code': 0.6989700043360189},
'doc02': {'day': 0.6989700043360189},
'doc03': {'today': 0.3979400086720376},
'doc04': {'coffe': 0.6989700043360189},
'doc05': {'python': 0.3979400086720376}}
這是代碼：
def tfidf (data, idf_score): #function, 2 dictionaries as parameters
tfidf = {} #dict for output
for word, val in data.items(): #for each word and value in data_docs(first dict)
for v in val: #for each value in each list
a = val.count(v) #count the number of times that appears in that list
scores = {v :a * idf_score[v]} # dictionary that will act as value in the nested
tfidf[word] = scores #final dictionary, the key is doc01,doc02... and the value the above dict
return tfidf
tfidf(data_docs, idf_score)
謝謝，
uj5u.com熱心網友回復：
你的意思是這樣做嗎？
def tfidf(data, idf_score): # function, 2 dictionaries as parameters
tfidf = {} # dict for output
for word, val in data.items(): # for each word and value in data_docs(first dict)
scores = {} # <---- a new dict for each outer iteration
for v in val: # for each value in each list
a = val.count(v) # count the number of times that appears in that list
scores[v] = a * idf_score[v] # <---- keep adding items to the dictionary
tfidf[word] = scores # final dictionary, the key is doc01,doc02... and the value the above dict
return tfidf
...用 <----- 箭頭查看我的更改 :) 回傳：
{'doc01': {'simpl': 1,
'hello': 1,
'world': 1,
'test': 1,
'python': 1,
'code': 1},
'doc02': {'today': 1, 'wonder': 1, 'day': 1},
'doc03': {'studi': 1, 'pac': 1, 'today': 1},
'doc04': {'write': 1, 'need': 1, 'cup': 1, 'coffe': 1},
'doc05': {'finish': 1, 'pac': 1, 'use': 1, 'python': 1}}
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