首页
学习
活动
专区
圈层
工具
发布
社区首页 >问答首页 >键错误:0尽管索引从0开始

问键错误:0尽管索引从0开始
EN

Stack Overflow用户
提问于 2020-06-10 16:07:19
回答 1查看 25关注 0票数 0

这是我的数据集。

​

我基本上想要单词列中每个列表中的单词数。这是我的密码

代码语言:javascript
复制
for ind,word in enumerate(brand6['words']):
  brand6['len']=len(brand6[ind][word])    

但是,我遇到了这样一个错误:

代码语言:javascript
复制
  Traceback (most recent call last):

  File "C:\Users\darshini.nagaraj\Anaconda3\lib\site-packages\pandas\core\indexes\base.py", line 2646, in get_loc
    return self._engine.get_loc(key)

  File "pandas\_libs\index.pyx", line 111, in pandas._libs.index.IndexEngine.get_loc

  File "pandas\_libs\index.pyx", line 138, in pandas._libs.index.IndexEngine.get_loc

  File "pandas\_libs\hashtable_class_helper.pxi", line 1618, in 
pandas._libs.hashtable.PyObjectHashTable.get_item


  File "pandas\_libs\hashtable_class_helper.pxi", line 1626, in 
pandas._libs.hashtable.PyObjectHashTable.get_item

KeyError: 0
    During handling of the above exception, another exception occurred:

Traceback (most recent call last):

  File "C:\Users\darshini.nagaraj\Desktop\example.py", line 28, in <module>
print((brand6[ind][word]))

  File "C:\Users\darshini.nagaraj\Anaconda3\lib\site-packages\pandas\core\frame.py", line 2800, in __getitem__
    indexer = self.columns.get_loc(key)

  File "C:\Users\darshini.nagaraj\Anaconda3\lib\site-packages\pandas\core\indexes\base.py", line 2648, in get_loc
    return self._engine.get_loc(self._maybe_cast_indexer(key))

  File "pandas\_libs\index.pyx", line 111, in pandas._libs.index.IndexEngine.get_loc

 File "pandas\_libs\index.pyx", line 138, in pandas._libs.index.IndexEngine.get_loc

  File "pandas\_libs\hashtable_class_helper.pxi", line 1618, in 
pandas._libs.hashtable.PyObjectHashTable.get_item

  File "pandas\_libs\hashtable_class_helper.pxi", line 1626, in 
pandas._libs.hashtable.PyObjectHashTable.get_item

KeyError: 0    

我已经为此工作了三天了。任何帮助都将不胜感激!提前谢谢!!

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2020-06-10 17:43:41

尝尝这个

代码语言:javascript
复制
word_count=[]
for i in range(brand6.shape[0]):
   word_count.append(len(brand6['words'][i]))
brand6['word_count']=word_count

最后一行将在数据集中创建一个新列为“word_count”。

票数 0
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/62308072

复制
相关文章

相似问题

领券
问题归档专栏文章快讯文章归档关键词归档开发者手册归档开发者手册 Section 归档