Question

编辑： @rong @shaik moeed 这是生成数据框架和我面临的问题的一部分的代码：

temp = [[1, 'blblblblblb. The quaity of research was good. blblblblb'],
        [2, 'blblblblblb. The quaity of research was average. blblblblb'],
        [3, 'blblblblblb. The quaity of research was poor. blblblblb'],
        [4, 'blblblblblb. The quaity of research was good. blblblblb']
        ]
Data = pd.DataFrame(temp,columns=['ID','Report'])
Data['Sentence']=Data['Report'].str.extract(r"([^.]*?The quaity of research was [^.]*\.)")

Quality_dic=dict([(1, 'excellent'), (2, 'good'),  (3, 'average') , (4, 'poor'), (5, 'unassessable')])



Data['Quality']=[k for k,v in Quality_dic.items() if v in  Data['Sentence'].str.split()]

不幸的是，建议的解决方案仍然无法正常工作。

关于如何解决此问题的任何想法？谢谢大家的时间和投入

Answer 1

我已经创建了一个df作为您的数据，并根据需要完全实现了。

在Quality_dic中，您为Good和Unassessable使用相同的密钥。因此Good将被Unassessable覆盖。

立即尝试

>>> temp = [[1, 'blblblblblb. The quaity of research was good. blblblblb'],
        [2, 'blblblblblb. The quaity of research was average. blblblblb'],
        [3, 'blblblblblb. The quaity of research was poor. blblblblb'],
        [4, 'blblblblblb. The quaity of research was good. blblblblb']
        ]

>>> Data = pd.DataFrame(temp,columns=['ID','Report'])

>>> Data['Sentence']=Data['Report'].str.extract(r"([^.]*?The quaity of research was [^.]*\.)")

>>> index_col = []

>>> for index, row in Data.iterrows():
        index_col.append([k for k,v in Quality_dic.items() if v.lower() in  row['Sentence'].replace('.','').split()][0])
>>> Data["index_col"]=index_col

输出：

>>> Data

   ID    ...    index_col
0   1    ...            2
1   2    ...            3
2   3    ...            4
3   4    ...            2

[4 rows x 4 columns]

注意：

... - means columns are hiding as there is no space to display.

Answer 2

quality_dic = dict([(1, 'Excellent'), (2, 'Good'), (3, 'Average') , (4, 'Poor'), (2, 'Unassessable')])

sentence = 'The quality of the research was Poor' # note that 'Poor' here is capitalized

for rating in quality_dic:
    if quality_dic[rating] in sentence:
        print(quality_dic[rating]) # df['Quality'] = quality_dic[rating]

# or if you want a one-liner:
df['Quality'] = [quality_dic[rating] in sentence for rating in quality_dic]

搜索字符串中的单词列表并返回匹配的单词

2 个答案: