在数据框内搜索并拆分文本

Question

I have a dataframe that consists of two columns, id and text.

I want to retrieve rows that have a text length larger than 2, as an example.

The text length is the number of words in the text rather than the number of chars.

I did the following:

df = pd.DataFrame([{'id': 1, 'text': 'Connected to hgfxg debugger'},
                   {'id': 2, 'text': 'fdss debugger - process 6384 is connecting'},
                   {'id': 3, 'text': 'we are'},
                   ])
df = df[df['text'].str.len() > 2]
print(df) #<-- it will print all the sentences above

But this retrieve the sentences that have more than 2 chars (in our case, all the sentences above).

How can I achieve what I want in one code line? possible?

I can do it with more than one, like:

df['text_len'] = df['text'].map(lambda x: len(str(x).split()))
df = df[df['text_len'] > 2]
print(df) #<-- will print the first two sentences

Answer 1

Just think about another way , you want more than 2 sentence , so that you need two ' ' in the string , and here we just count the ' ' is more than 2

df[df['text'].str.count(' ')>2]
Out[230]: 
   id                                        text
0   1                 Connected to hgfxg debugger
1   2  fdss debugger - process 6384 is connecting

Answer 2

您还可以使用：

df[df.text.str.split('\s+').str.len().gt(2)]