字典与Python中不同类型的数组作为值 [英] Dictionary with array of different types as value in Python

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问题描述

我正在学习Python,我有一本这样的字典:

  category = {World news:[0] ,政治:[0],...} 

文件。每篇文章都有一个类别。所以我想要0我必须每个键增加1,当一个类别满足,并增加数组每个键与它发现的文章。所以输出将是:

  category = {世界新闻:[2,foo bla content of,content第二篇],
政治:[1,只有1篇文章],
...
}
pre>

这是我现在所在的:

  iter_reader:
category [row [4]] [0] = category [row [4]] [0] + 1
category [row [4]]。append(category [row [3]] )

但我认为它将文章解释为关键字,并引发相关错误:


文件./wordCloud.py,第41行,
category [row [4]]。append(category [row [3 ]])
KeyError:当苹果公司在2010年1月推出时。







行[4] 是文章的类别。
row [3] 是本文的内容。



正如Bryan Oakley所说,不需要一个柜台,所以我现在正在这个方向思考...

解决方案

在第三行,你是附加 category [row [3]] 您应该可以简单地附加 row [3] 。您目前正在使用 row [3] 作为类别列表的索引,但我怀疑行[3] 是实际上你想追加的内容。


I am learning Python and I have a dictionary like this:

category = { "World news": [0], "Politics": [0], ... }

where now I am reading articles from a file. Every article has a category. So I would like the 0 I have to every key to be incremented by 1 when a category is meet and to augment the array every key has with the article it found. So the output would be:

category = {  "World news": [2, "foo bla content of", "content of 2nd article"],
              "Politics": [1, "only 1 article here"],
              ...
}

Here is what I have now:

for row in iter_reader:
    category[row[4]][0] = category[row[4]][0] + 1
    category[row[4]].append(category[row[3]])

but I think it interprets an article as a key and throws a relevant error:

File "./wordCloud.py", line 41, in category[row[4]].append(category[row[3]]) KeyError: 'When Apple unveiled its in January 2010..

Any ideas please?


row[4] is the category of the article. row[3] is the content of the article.

As Bryan Oakley said, I do not need to keep a counter, so I am now thinking in this direction...

解决方案

In the third line, you are appending category[row[3]] where you should probably be appending simply row[3]. You are currently using row[3] as an index for the category list, but I suspect row[3] is actually the content you want to append.

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