使用Python正则表达式将注释分割成数据框。

Question

使用Python正则表达式将注释分割成数据框。

3

我有一堆由用户输入的字符串，这些字符串是不同评论串联在一起的。如果有多天的评论，他们有时会输入日期。我正在尝试找到一种方法来分离每个日期和相应的评论。文本评论可能像这样：

raw_text = ['3/30: The dog is red. 4/01: The dog is blue', 'there is a green door', '3-25:Foobar baz']

I would like to transform that text to:

df = pd.DataFrame([[0,'3/30','The dog is red.'],[0,'4/01','The dog is blue'],[1,np.nan,'there is a green door'],[2,'3-25','Foobar baz']],columns = 'row_id','date','text')

print(df)

   row_id  date                   text
0       0  3/30        The dog is red.
1       0  4/01        The dog is blue
2       1   NaN  there is a green door
3       2  3-25             Foobar baz

我认为我需要做的是找到分号，然后回溯到分号之前的第一个数字来确定日期（有时使用 / 分隔，有时使用 -）。

如果能提供使用正则表达式的方法将会非常感激——这已经超出了我的简单切割/查找知识范围。

谢谢！

- flyingmeatball

2个回答

0

数据

df=pd.DataFrame({'raw_text':['3/30: The dog is red.', '4/01: The dog is blue', 'there is a green door', '3-25:Foobar baz']})
df

创建日期列

df['date']=df.raw_text.str.extract(r"([\d+\/\-+\d+]+(?=\:))")
df

创建文本列

df['text']=df.raw_text.str.extract(r"((?:-)?[^\s:][A-Za-z\s]+[^s])", expand=True)
df

创建行ID列 匹配文本'The dog'并创建临时列索引 k= 'The dog'

dicto ={'The dog':0}
df['index']=df['raw_text'].str.extract('('+ k + ')', expand=False).map(dicto)
df

使用 index 列来输入 row_id

df['row_id']=df['index'].isna().astype('int64')

将包含文本'The dog'的行进行掩码处理，并对其余行自动递增添加数字

    m=df['row_id']!=0
    df.loc[m,'row_id']=np.arange(start=1, stop=3,step=1)# please note the stop may need to be increased if df is longer
df.drop(columns=['index'], inplace=True)

输出

- wwnde

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- It_is_Chris · Accepted Answer

我不是很擅长正则表达式（所以可能有更好的解决方案），但这似乎是有效的...

# sample list
raw_text = ['10-30: The dog is red. 4/01: The dog is blue', 'there is a green door',
            '3-25:Foobar baz', '11-25:Foobar baz. 12/20: something else']

# create regex (e.g., the variable 'n' in the comment below represents a number)
# if 'nn/nn' OR 'nn-nn' OR ' n-nn' OR ' n/nn' OR ' nn-nn' OR ' nn/nn' OR string starts with a number
regex = r'(?=\d\d/\d\d:)|(?=\d\d-\d\d:)|(?= \d-\d\d:)|(?= \d/\d\d:)|(?= \d\d-\d\d:)|(?= \d\d/\d\d:)|(?=^\d)'
# if string starts with alpha characters or there is a ':'
regex2 = r'(?=^\D)|:'

# create a Series by splitting on regex and explode
s = pd.DataFrame(raw_text)[0].str.split(regex).explode()
# boolean indexing to remove blanks
s2 = s[(s != '') & (s != ' ')]

# strip leading or trailing white space then split on regex2
df = s2.str.strip().str.split(regex2, expand=True).reset_index()
# rename columns
df.columns = ['row_id', 'date', 'text']


   row_id   date                         text
0       0  10-30   The dog is red until 5/15.
1       0   4/01              The dog is blue
2       1               there is a green door
3       2   3-25                   Foobar baz
4       3  11-25                  Foobar baz.
5       3  12/20               something else