获取文档中出现的词频的累积计数

Question

我一直在尝试检测 word/bigram 文本片段的趋势。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)..]

有什么想法吗？

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]

Answer 1

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)]

另一种处理重复项的方法是使用 pandas groupby() 函数，然后使用 sort() 函数按 count 和 word 排序，例如所以

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

获取文档中出现的词频的累积计数

Getting cumulative counts of word frequencies founds in a documents

python

nltk

trend

word-frequency