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我一直在尝试检测文本片段上的单词/bigram 趋势。到目前为止，我所做的是删除停用词、小写和获取词频，并将每个文本最常见的 30 个附加到列表中，

例如

[(u'seeing', 2), (u'said.', 2), (u'one', 2), (u'death', 2), (u'entertainment',   2), (u'it\u2019s', 2), (u'weiss', 2), (u'read', 2), (u'\u201cit', 1), (u'shot', 1), (u'show\u2019s', 1), (u'people', 1), (u'dead,\u201d', 1), (u'bloody', 1),...]

然后我将上面的列表转换为一个包含所有单词及其每个文档频率的巨大列表，我现在需要做的是取回一个排序列表，即：

[(u'snow', 32), (u'said.', 12), (u'GoT', 10), (u'death', 8), (u'entertainment', 4)..]

有任何想法吗？

代码：

fdists = []
for i in texts:
    words = FreqDist(w.lower() for w in i.split() if w.lower() not in    stopwords)
    fdists.append(words.most_common(30))

all_in_one = [item for sublist in fdists for item in sublist]
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1 回答 1

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如果您只想对列表进行排序，您可以使用

import operator

fdists = [(u'seeing', 2), (u'said.', 2), (u'one', 2), (u'death', 2), (u'entertainment',   2), (u'it\u2019s', 2), (u'weiss', 2), (u'read', 2), (u'\u201cit', 1), (u'shot', 1), (u'show\u2019s', 1), (u'people', 1), (u'dead,\u201d', 1), (u'bloody', 1)]
fdists2 = [(u'seeing', 3), (u'said.', 4), (u'one', 2), (u'death', 2), (u'entertainment',   2), (u'it\u2019s', 2), (u'weiss', 2), (u'read', 2)]
fdists += fdists2

fdict = {}
for i in fdists:
    if i[0] in fdict:
        fdict[i[0]] += i[1]
    else:
        fdict[i[0]] = i[1]

sorted_f = sorted(fdict.items(), key=operator.itemgetter(1), reverse=True)
print sorted_f[:30]

[(u'said.', 6), (u'seeing', 5), (u'death', 4), (u'entertainment', 4), (u'read', 4), (u'it\u2019s', 4), (u'weiss', 4), (u'one', 4), (u'\u201cit', 1), (u'shot', 1), (u'show\u2019s', 1), (u'people', 1), (u'dead,\u201d', 1), (u'bloody', 1)]

处理重复项的另一种方法是使用 pandasgroupby()函数，然后使用该sort()函数进行排序count，word就像这样

from pandas import *
import pandas as pd

fdists = [(u'seeing', 2), (u'said.', 2), (u'one', 2), (u'death', 2), (u'entertainment',   2), (u'it\u2019s', 2), (u'weiss', 2), (u'read', 2), (u'\u201cit', 1), (u'shot', 1), (u'show\u2019s', 1), (u'people', 1), (u'dead,\u201d', 1), (u'bloody', 1)]
fdists2 = [(u'seeing', 3), (u'said.', 4), (u'one', 2), (u'death', 2), (u'entertainment',   2), (u'it\u2019s', 2), (u'weiss', 2), (u'read', 2)]
fdists += fdists2

df = DataFrame(data = fdists, columns = ['word','count'])
df= DataFrame([{'word': k, 'count': (v['count'].sum())} for k,v in df.groupby(['word'])], columns = ['word','count'])

Sorted = df.sort(['count','word'], ascending = [0,1])
print Sorted[:30]

             word  count
8           said.      6
9          seeing      5
2           death      4
3   entertainment      4
4            it’s      4
5             one      4
7            read      4
12          weiss      4
0          bloody      1
1          dead,”      1
6          people      1
10           shot      1
11         show’s      1
13            “it      1
于 2015-10-26T15:06:04.570 回答