Python字典中的键解析和排序

我创建了以下词典：

code dictionary =  {u'News; comment; negative': u'contradictory about news', u'News; comment': u'something about news'}

现在，我想编写一些Python代码，它遍历字典的键并分离代码及其对应的值。因此，对于字典中的第一个元素，我想以：

News: 'contradictory about news', 'something about news'
comment: 'contradictory about news', 'something about news'
negative: 'contradictory about news'

最终结果可以是字典、列表、制表符或逗号分隔的文本

您可以在此处看到我的尝试：

from bs4 import BeautifulSoup as Soup
f = open('transcript.xml','r')
soup = Soup(f)
#print soup.prettify()


#searches text for all w:commentrangestart tags and makes a dictionary that matches ids with text
textdict = {}
for i in soup.find_all('w:commentrangestart'):
        # variable 'key' is assigned to the tag id
        key = i.parent.contents[1].attrs['w:id']
        key = str(key)
        #variable 'value' is assigned to the tag's text
        value= ''.join(i.nextSibling.findAll(text=True))
        # key / value pairs are added to the dictionary 'textdict'
        textdict[key]=value
print "Transcript Text = " , textdict

# makes a dictionary that matches ids with codes        
codedict = {}
for i in soup.find_all('w:comment'):
        key = i.attrs['w:id']
        key = str(key)
        value= ''.join(i.findAll(text=True))
        codedict[key]=value
print "Codes = ", codedict

# makes a dictionary that matches all codes with text
output = {}
for key in set(textdict.keys()).union(codedict.keys()):
        print "key= ", key
        txt = textdict[key]
        print "txt = ", txt
        ct = codedict[key]
        print "ct= ", ct
        output[ct] = txt
        #print "output = ", output
print "All code dictionary = ", output

#codelist={}
#for key in output:
#   codelist =key.split(";")
#print "codelist= " , codelist


code_negative = {}
code_news = {}
print output.keys()
for i in output:
    if 'negative' in output.keys():
        print 'yay'
        code_negative[i]=textdict[i]
        print 'text coded negative: ' , code_negative
    if 'News' in i:
        code_news[i]=textdict[i]
        print 'text coded News: ' ,code_news

但出于某种原因，在运行最后一个函数时，我不断遇到一个键错误：

code_negative = {}
code_news = {}
for i in output:
    if 'negative' in output.keys():
        code_negative[i]=textdict[i]
    print 'text coded negative: ' , code_negative
if 'News' in i:
    code_news[i]=textdict[i]
    print 'text coded News: ' ,code_news

---

有什么想法吗？谢谢

如果我正确理解了问题，以下代码应该可以工作：

from collections import defaultdict

out = defaultdict(list)
for k, v in code_dictionary.viewitems():
    for item in k.split('; '):
        out[item].append(v)

---

使用split函数不断迭代所需内容…类似于for循环，然后返回i.split（“；”）。这应该允许您迭代您需要的内容

output = {u'News; comment; negative': u'contradictory about news', u'News; comment': u'something about news'}
negatives = []
comments = []
news = []
for k, v in output.items():
    key_parts = k.split('; ')
    key_parts = [part.lower() for part in key_parts]
    if 'negative' in key_parts:
        negatives.append(v)
    if 'news' in key_parts:
        news.append(v)
    if 'comment' in key_parts:
        comments.append(v)