Question

我有一个单词列表和词典：

word_list = ["it's","they're","there's","he's"]

一个字典，其中包含words_list中的单词出现在多个文档中的频率的信息：

dict = [('document1',{"it's": 0,"they're": 2,"there's": 5,"he's": 1}),
('document2',{"it's": 4,"they're": 2,"there's": 3,"he's": 0}),
('document3',{"it's": 7,"they're": 0,"there's": 4,"he's": 1})]

我想开发一个如下所示的数据结构（可能是数据框？）：

file       word       count
document1  it's        0
document1  they're     2
document1  there's     5
document1  he's        1
document2  it's        4
document2  they're     2
document2  there's     3
document2  he's        0
document3  it's        7
document3  they're     0
document3  there's     4
document3  he's        1

我试图找到这些文档中最常用的words。我有900多份文件。

我正在考虑以下内容：

res = {}
for i in words_list:
    count = 0
    for j in dict.items():
         if i == j:
              count = count + 1
              res[i,j] = count

我可以从哪里离开？

Answer 1

首先，首先，你的dict不是dict，应该像这样建立

d = {'document1':{"it's": 0,"they're": 2,"there's": 5,"he's": 1},
    'document2':{"it's": 4,"they're": 2,"there's": 3,"he's": 0},
    'document3':{"it's": 7,"they're": 0,"there's": 4,"he's": 1}}

现在我们实际上有一个字典，我们可以使用pandas来构建数据帧，但为了按照你想要的方式获取它，我们必须从字典中构建一个列表列表。然后我们将创建一个数据框并标记列，然后排序

import collections
import pandas as pd

d = {'document1':{"it's": 0,"they're": 2,"there's": 5,"he's": 1},
    'document2':{"it's": 4,"they're": 2,"there's": 3,"he's": 0},
    'document3':{"it's": 7,"they're": 0,"there's": 4,"he's": 1}}

d = pd.DataFrame([[k,k1,v1] for k,v in d.items() for k1,v1 in v.items()], columns = ['File','Words','Count'])
print d.sort(['File','Count'], ascending=[1,1])

         File    Words  Count
1   document1     it's      0
0   document1     he's      1
3   document1  they're      2
2   document1  there's      5
4   document2     he's      0
7   document2  they're      2
6   document2  there's      3
5   document2     it's      4
11  document3  they're      0
8   document3     he's      1
10  document3  there's      4
9   document3     it's      7

如果您想要前n次出现，那么您可以使用groupby()，然后在排序时使用head() or tail()

d = d.sort(['File','Count'], ascending=[1,1]).groupby('File').head(2)

         File    Words  Count
1   document1     it's      0
0   document1     he's      1
4   document2     he's      0
7   document2  they're      2
11  document3  they're      0
8   document3     he's      1

list comprehension返回一个看起来像这样的列表列表

d = [['document1', "he's", 1], ['document1', "it's", 0], ['document1', "there's", 5], ['document1', "they're", 2], ['document2', "he's", 0], ['document2', "it's", 4], ['document2', "there's", 3], ['document2', "they're", 2], ['document3', "he's", 1], ['document3', "it's", 7], ['document3', "there's", 4], ['document3', "they're", 0]]

为了正确构建字典，您只需使用

行

d['document1']['it\'s'] = 1

如果由于某种原因你没有使用str和dicts的元组列表，你可以使用这个列表理解

[[i[0],k1,v1] for i in d for k1,v1 in i[1].items()]

Answer 2

这样的事情怎么样？

word_list = ["it's","they're","there's","he's"]

frequencies = [('document1',{"it's": 0,"they're": 2,"there's": 5,"he's": 1}),
('document2',{"it's": 4,"they're": 2,"there's": 3,"he's": 0}),
('document3',{"it's": 7,"they're": 0,"there's": 4,"he's": 1})]

result = []
for document in frequencies:
    for word in word_list:
        result.append({"file":document[0], "word":word,"count":document[1][word]})

print result

迭代字典中的多个值？

2 个答案: