Pandas NLTK 标记“不可散列的类型：'列表'”

Question

下面这个例子： 使用 Python 和 Gephi 进行 Twitter 数据挖掘：案例合成生物学

CSV to: df['Country', 'Responses']

'Country'
Italy
Italy
France
Germany

'Responses' 
"Loren ipsum..."
"Loren ipsum..."
"Loren ipsum..."
"Loren ipsum..."

1. 标记“响应”中的文本
2. 删除 100 个最常用的单词（基于 brown.corpus）
3. 找出剩下的 100 个最常用的词

我可以完成第 1 步和第 2 步，但在第 3 步中出现错误：

TypeError: unhashable type: 'list'

我相信这是因为我在数据帧中工作并且进行了这个（可能是错误的）修改：

原始示例：

#divide to words
tokenizer = RegexpTokenizer(r'\w+')
words = tokenizer.tokenize(tweets)

我的代码：

#divide to words
tokenizer = RegexpTokenizer(r'\w+')
df['tokenized_sents'] = df['Responses'].apply(nltk.word_tokenize)

我的完整代码：

df = pd.read_csv('CountryResponses.csv', encoding='utf-8', skiprows=0, error_bad_lines=False)

tokenizer = RegexpTokenizer(r'\w+')
df['tokenized_sents'] = df['Responses'].apply(nltk.word_tokenize)

words =  df['tokenized_sents']

#remove 100 most common words based on Brown corpus
fdist = FreqDist(brown.words())
mostcommon = fdist.most_common(100)
mclist = []
for i in range(len(mostcommon)):
    mclist.append(mostcommon[i][0])
words = [w for w in words if w not in mclist]

Out: ['the',
 ',',
 '.',
 'of',
 'and',
...]

#keep only most common words
fdist = FreqDist(words)
mostcommon = fdist.most_common(100)
mclist = []
for i in range(len(mostcommon)):
    mclist.append(mostcommon[i][0])
words = [w for w in words if w not in mclist]

TypeError: unhashable type: 'list'

关于不可散列的列表有很多问题，但我认为没有一个是完全相同的。 有什么建议吗？ 谢谢。

---

追溯

TypeError                                 Traceback (most recent call last)
<ipython-input-164-a0d17b850b10> in <module>()
  1 #keep only most common words
----> 2 fdist = FreqDist(words)
  3 mostcommon = fdist.most_common(100)
  4 mclist = []
  5 for i in range(len(mostcommon)):

/home/*******/anaconda3/envs/*******/lib/python3.5/site-packages/nltk/probability.py in __init__(self, samples)
    104         :type samples: Sequence
    105         """
--> 106         Counter.__init__(self, samples)
    107 
    108     def N(self):

/home/******/anaconda3/envs/******/lib/python3.5/collections/__init__.py in __init__(*args, **kwds)
    521             raise TypeError('expected at most 1 arguments, got %d' % len(args))
    522         super(Counter, self).__init__()
--> 523         self.update(*args, **kwds)
    524 
    525     def __missing__(self, key):

/home/******/anaconda3/envs/******/lib/python3.5/collections/__init__.py in update(*args, **kwds)
    608                     super(Counter, self).update(iterable) # fast path when counter is empty
    609             else:
--> 610                 _count_elements(self, iterable)
    611         if kwds:
    612             self.update(kwds)

TypeError: unhashable type: 'list'

Answer 1

FreqDist函数接受一个可迭代的可散列对象（制成字符串，但它可能适用于任何东西）。 你得到的错误是因为你传入了一个可迭代的列表。 正如您所建议的，这是因为您所做的更改：

df['tokenized_sents'] = df['Responses'].apply(nltk.word_tokenize)

如果我正确理解Pandas 应用函数文档，那一行就是将nltk.word_tokenize函数应用于某些系列。 word-tokenize返回一个单词列表。

作为解决方案，只需在尝试应用FreqDist之前将列表添加在一起，如下所示：

allWords = []
for wordList in words:
    allWords += wordList
FreqDist(allWords)

一个更完整的修订来做你想做的事。 如果您只需要识别第二组 100，请注意mclist将第二次识别。

df = pd.read_csv('CountryResponses.csv', encoding='utf-8', skiprows=0, error_bad_lines=False)

tokenizer = RegexpTokenizer(r'\w+')
df['tokenized_sents'] = df['Responses'].apply(nltk.word_tokenize)

lists =  df['tokenized_sents']
words = []
for wordList in lists:
    words += wordList

#remove 100 most common words based on Brown corpus
fdist = FreqDist(brown.words())
mostcommon = fdist.most_common(100)
mclist = []
for i in range(len(mostcommon)):
    mclist.append(mostcommon[i][0])
words = [w for w in words if w not in mclist]

Out: ['the',
 ',',
 '.',
 'of',
 'and',
...]

#keep only most common words
fdist = FreqDist(words)
mostcommon = fdist.most_common(100)
mclist = []
for i in range(len(mostcommon)):
    mclist.append(mostcommon[i][0])
# mclist contains second-most common set of 100 words
words = [w for w in words if w in mclist]
# this will keep ALL occurrences of the words in mclist