迭代字典中的多个值？

Question

我有一个单词列表和字典：

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

并作为在words_list的话如何频繁地出现在几个文件包含信息的字典：

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})] 

我想开发一个数据结构（数据帧，也许？），看起来像如下：

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使用的是。我有900多个文件。

我在考虑类似如下：

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

我在哪里可以从这里走？

Answer 1

好第一件事情，你的字典是不是一个字典，并且现在应建设成为一个像这样

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}} 

有，我们实际上我们可以用大熊猫建立一个数据帧一本字典，而是在为了以你想要的方式获得它，我们将不得不从字典中建立一个列表清单。然后，我们将创建一个数据框和标记列，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 

如果你想与前n次出现，那么你可以使用groupby()，然后要么排序

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 

时head() or tail()列表理解返回名单列表，看起来像这样

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]] 

为了正确地建立字典，你只需要使用一些东西克

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

如果由于某种原因，你使用STR的元组和类型的字典的列表，你可以使用这个列表理解，而不是

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

Answer 2

这样的事情呢？第一

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