Python 在字典中迭代多个值？
我有一个单词列表和字典：Python 在字典中迭代多个值？,python,list,python-2.7,dictionary,Python,List,Python 2.7,Dictionary,我有一个单词列表和字典： word_list = ["it's","they're","there's","he's"] 以及一本包含单词列表中的单词在多个文档中出现频率信息的词典： 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
word_list = ["it's","they're","there's","he's"]
单词列表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})]
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
res = {}
for i in words_list:
count = 0
for j in dict.items():
if i == j:
count = count + 1
res[i,j] = count
单词。我有900多份文件
我在想下面的事情：
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
res = {}
for i in words_list:
count = 0
for j in dict.items():
if i == j:
count = count + 1
res[i,j] = count
从这里我可以去哪里？像这样的怎么样
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
像这样的怎么样
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
首先，你的dict不是一个dict，应该像这样构建
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}}
现在我们已经有了一个字典，我们可以使用pandas来构建一个数据帧，但是为了得到您想要的方式，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dataframe并标记列，然后进行排序
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
如果需要前n个匹配项，则可以在排序时使用groupby（）
，然后使用head（）或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
列表理解返回如下所示的列表列表
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]]
为了正确地构建词典，您只需使用以下内容
d['document1']['it\'s'] = 1
如果出于某种原因，你执意使用str和dict的元组列表，你可以使用这个列表
[[i[0],k1,v1] for i in d for k1,v1 in i[1].items()]
首先，你的dict不是一个dict，应该像这样构建
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}}
现在我们已经有了一个字典，我们可以使用pandas来构建一个数据帧，但是为了得到您想要的方式，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dataframe并标记列，然后进行排序
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
如果需要前n个匹配项，则可以在排序时使用groupby（）
，然后使用head（）或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
列表理解返回如下所示的列表列表
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]]
为了正确地构建词典，您只需使用以下内容
d['document1']['it\'s'] = 1
如果出于某种原因，你执意使用str和dict的元组列表，你可以使用这个列表
[[i[0],k1,v1] for i in d for k1,v1 in i[1].items()]
您应该使用Python熊猫库来创建您在文章中显示的数据帧类型。我从哪里开始呢？我应该考虑的任何方法？使一个名为dict
的变量将使内置的dict
函数无法访问。你应该将它重命名为其他名称。此外，它不是dict，而是字符串和dict的元组列表。您应该使用Python Pandas库来创建您在帖子中显示的数据帧类型。我从哪里开始？我应该考虑的任何方法？使一个名为dict
的变量将使内置的dict
函数无法访问。你应该将它重命名为其他名称。此外，它不是dict，而是字符串和dict的元组列表。我得到以下错误：TypeError:string索引必须是整数，而不是str
。我不能用这个词本身来表示你是否使用与我相同的数据运行代码？唯一可能失败的地方是文档[1][word]
，而文档[1]
中的所有键都是所提供数据中的字符串。不应该失败。编辑：再想一想，这个错误意味着您试图用另一个字符串访问字符串的一个元素。你的频率
是否包含任何原始字符串？我不这么认为。从字面上看是这样的，尽管从我使用的实际数据来看要简单得多。。它遵循完全相同的语法结构，但是frequencies
更容易谈论。是的，没有“count”的代码：document[1][word]
words很好。运行以下命令：[x代表x，如果类型（x[1]）是str的话
并查看是否有任何显示，我得到以下错误：TypeError:string索引必须是整数，而不是str
。我不能用这个词本身来表示你是否使用与我相同的数据运行代码？唯一可能失败的地方是文档[1][word]
，而文档[1]
中的所有键都是所提供数据中的字符串。不应该失败。编辑：再想一想，这个错误意味着您试图用另一个字符串访问字符串的一个元素。你的频率
是否包含任何原始字符串？我不这么认为。从字面上看是这样的，尽管从我使用的实际数据来看要简单得多。。它遵循完全相同的语法结构，但是frequencies
更容易谈论。是的，没有的代码“count”：document[1][word]
words很好。运行以下命令：[x for x in frequencies，如果type（x[1]）是str的话，看看是否有任何东西显示了很好的答案。一个问题：d.sort（['File'，'Count']，升序=[1,1]）
也会更改索引。有什么特别的原因让你这么做吗？@JoeR我只是把它改了，让文件从低到高，然后计数从低到高。没必要，但我觉得看起来好多了。回答得很好。一个问题：d.sort（['File'，'Count']，升序=[1,1]）
也会更改索引。有什么特别的原因让你这么做吗？@JoeR我只是把它改了，让文件从低到高，然后计数从低到高。没必要，但我觉得它看起来好多了。