Question

我有一个嵌套的字符串列表，我想提取它们的日期。日期格式为：

两个数字（从01到12）连字符树字母（有效月份）连字符两个数字，例如：08-Jan—07或03-Oct—01

我尝试使用以下正则表达式：

r'\d{2}(—|-)(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)-\d{2,4}'

然后我按如下方式测试：

import pandas as pd
df = pd.DataFrame({'blobs':['6-Feb- 1 4 Facebook’s virtual-reality division created a 3-EBÚ7 11 network of 500 free demo stations in Best Buy stores to give people a taste of VR using the Oculus Rift 90 GT 48 headset. But according to a Wednesday report from Business Insider, about 200 of the demo stations will close after low interest from consumers. 17-Feb-2014',
                         'I think in a store environment getting people to sit down and go through that experience of getting a headset on and getting set up is quite a difficult thing to achieve,” said Geoff Blaber, a CCS Insight analyst. 29—Oct-2012 Blaber 32 FAX 2978 expects that it will get easier when companies can convince  18-Oct-12 credit cards. '
                            ]})
df

然后：

df['blobs'].str.extractall(r'\d{2}(—|-)(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)-\d{2,4}')

然而，他们没有工作。以前的正则表达式没有给我任何东西（即只是夸大-）：

    Col
0   NaN
1    -
2    -
3   NaN
4   NaN
5    -
...
n    -

如何修复它们才能获得？：

           Col
0 6-Feb-14, 17-Feb-2014
1 29—Oct-2012, 18-Oct-12

更新

我也试过：

import re
df['col'] = df.blobs.apply(lambda x: re.findall('\d{2}(—|-)(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)-\d{2,4}',x))
s = df.apply(lambda x: pd.Series(x['col']),axis=1).stack().reset_index(level=1, drop=True)
s.name = "col"
df = df.drop('col')
df

然而我也得到了：

ValueError                                Traceback (most recent call last)
<ipython-input-4-5e9a34bd159f> in <module>()
      3 s = df.apply(lambda x: pd.Series(x['col']),axis=1).stack().reset_index(level=1, drop=True)
      4 s.name = "col"
----> 5 df = df.drop('col')
      6 df

/usr/local/lib/python3.5/site-packages/pandas/core/generic.py in drop(self, labels, axis, level, inplace, errors)
   1905                 new_axis = axis.drop(labels, level=level, errors=errors)
   1906             else:
-> 1907                 new_axis = axis.drop(labels, errors=errors)
   1908             dropped = self.reindex(**{axis_name: new_axis})
   1909             try:

/usr/local/lib/python3.5/site-packages/pandas/indexes/base.py in drop(self, labels, errors)
   3260             if errors != 'ignore':
   3261                 raise ValueError('labels %s not contained in axis' %
-> 3262                                  labels[mask])
   3263             indexer = indexer[~mask]
   3264         return self.delete(indexer)

ValueError: labels ['col'] not contained in axis

Answer 1

当您使用Series.str.extract或Series.str.extractall时，会返回捕获的子字符串，而不是整个匹配项。因此，您需要确保捕获（即添加(和)）您需要抓取的模式部分。

现在，您行中的几个预期匹配项使extractall更难处理，如果没有捕获，您可能会使用可能会返回整个匹配的Series.str.findall group在模式中定义。

使用

rx = r'\b\d{1,2}[-–—](?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec)[-–—](?:\d{4}|\d{2})\b'
df['Col'] = df['blobs'].str.findall(rx).apply(','.join)

.apply(','.join)会将列表转换为Col列中以逗号分隔的字符串。

模式意味着：

\b - 字边界
\d{1,2} - 1或2位数字
[-–—] - 连字符，em-或en-dash
(?:Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) - 12个月缩短名称中的任何一个
[-–—] - 连字符，em-或en-dash
(?:\d{4}|\d{2}) - 4位或2位
\b - 字边界

pandas extractall（）是不是在提供正则表达式的情况下提取所有案例？

1 个答案: