Question

我有一个像这样的数据列：

df['zone'].unique()

out[4]: 

array(['BROOKLYN', 'BRONX', '07 BRONX', 'Unspecified', '05 BRONX',
       'QUEENS', 'MANHATTAN', '07 MANHATTAN', 'STATEN ISLAND',
       '17 BROOKLYN', '0 Unspecified', 'Unspecified MANHATTAN',
       '12 BROOKLYN', '07 BROOKLYN', '09 MANHATTAN', '01 STATEN ISLAND',
       '12 MANHATTAN', '04 QUEENS', '06 BROOKLYN',
       '01/04/2016 01:45:00 PM', '01/02/2016 05:43:34 AM', '07 QUEENS',
       '11 BRONX', '01/04/2016 03:45:00 PM', '10 MANHATTAN', '03 BRONX',
       '04 BRONX', ' or 311 Online."', '01/13/2016 12:00:00 AM',
       '04 BROOKLYN', '03 BROOKLYN', '01 QUEENS',
       '01/04/2016 03:34:55 PM', '08 MANHATTAN', '14 BROOKLYN',
       '10 QUEENS', 'Unspecified STATEN ISLAND', '02 BRONX', '09 BRONX',
       '08 QUEENS', '10 BRONX', '03 MANHATTAN', '12 QUEENS',
       ' please call (212) NEW-YORK (212-639-9675)."',
       'Unspecified BROOKLYN', '01/11/2016 04:45:00 PM', '04 MANHATTAN',
       '01 BRONX', '09 BROOKLYN', '01/05/2016 07:00:00 AM', '18 BROOKLYN',
       '01/08/2016 09:00:00 AM', '01 BROOKLYN', '06 BRONX',
       '01 MANHATTAN', '01/06/2016 12:15:00 PM', '02/04/2016 08:45:00 PM',
       '01/05/2016 12:45:00 PM', ' no action was taken."', '05 BROOKLYN',
       '08 BROOKLYN', 'Unspecified QUEENS', '01/08/2016 03:00:00 PM',
       '08/22/2016 12:00:00 AM', '13 BROOKLYN', '02 QUEENS', '14 QUEENS',
       '01/05/2016 08:45:00 AM', '11 QUEENS', '02 MANHATTAN',
       '01/08/2016 10:05:00 AM', '01/05/2016 01:05:00 PM',
       'Unspecified BRONX', '06 QUEENS', '09 QUEENS', '15 BROOKLYN',
       '01/07/2016 09:25:00 AM', '02 STATEN ISLAND',
       '01/02/2016 12:00:00 PM', '01/06/2016 08:45:00 PM',
       '04/04/2016 12:00:00 AM', '01/06/2016 08:30:00 AM'])

如您所见，我在那里有很多混合类型，所有东西都被熊猫归类为字符串对象。我已经在pd.read_csv命令中尝试过一些参数，例如low_memory = False，chunksize等，都没有成功。

尽管如此，我真正需要做的是将该列映射为以下格式：

(Manhattan -> 1, Brooklyn -> 2, Queens -> 3, Staten Island -> 4, Bronx -> 5, Other -> 0)

我还需要包含字符串'07 BRONX'作为bronx，而不是其他或未知字符串。

我已经考虑过使用.map()方法，但是由于该列确实是混合类型的混乱，因此我不确定我的选择是什么。

在这里，我将不胜感激。

非常感谢

Answer 1

通过字典的extract键创建值的字典，其中|到map的OR，最后fillna的所有不匹配值都映射到{{1} }：

a = np.array(['BROOKLYN', 'BRONX', '07 BRONX', 'Unspecified', '05 BRONX',
       'QUEENS', 'MANHATTAN', '07 MANHATTAN', 'STATEN ISLAND',
       '17 BROOKLYN', '0 Unspecified', 'Unspecified MANHATTAN',
       '12 BROOKLYN', '07 BROOKLYN', '09 MANHATTAN', '01 STATEN ISLAND',
       '12 MANHATTAN', '04 QUEENS', '06 BROOKLYN',
       '01/04/2016 01:45:00 PM', '01/02/2016 05:43:34 AM', '07 QUEENS',
       '11 BRONX', '01/04/2016 03:45:00 PM', '10 MANHATTAN', '03 BRONX',
       '04 BRONX', ' or 311 Online."', '01/13/2016 12:00:00 AM',
       '04 BROOKLYN', '03 BROOKLYN', '01 QUEENS',
       '01/04/2016 03:34:55 PM', '08 MANHATTAN', '14 BROOKLYN',
       '10 QUEENS', 'Unspecified STATEN ISLAND', '02 BRONX', '09 BRONX',
       '08 QUEENS', '10 BRONX', '03 MANHATTAN', '12 QUEENS',
       ' please call (212) NEW-YORK (212-639-9675)."',
       'Unspecified BROOKLYN', '01/11/2016 04:45:00 PM', '04 MANHATTAN',
       '01 BRONX', '09 BROOKLYN', '01/05/2016 07:00:00 AM', '18 BROOKLYN',
       '01/08/2016 09:00:00 AM', '01 BROOKLYN', '06 BRONX',
       '01 MANHATTAN', '01/06/2016 12:15:00 PM', '02/04/2016 08:45:00 PM',
       '01/05/2016 12:45:00 PM', ' no action was taken."', '05 BROOKLYN',
       '08 BROOKLYN', 'Unspecified QUEENS', '01/08/2016 03:00:00 PM',
       '08/22/2016 12:00:00 AM', '13 BROOKLYN', '02 QUEENS', '14 QUEENS',
       '01/05/2016 08:45:00 AM', '11 QUEENS', '02 MANHATTAN',
       '01/08/2016 10:05:00 AM', '01/05/2016 01:05:00 PM',
       'Unspecified BRONX', '06 QUEENS', '09 QUEENS', '15 BROOKLYN',
       '01/07/2016 09:25:00 AM', '02 STATEN ISLAND',
       '01/02/2016 12:00:00 PM', '01/06/2016 08:45:00 PM',
       '04/04/2016 12:00:00 AM', '01/06/2016 08:30:00 AM'])
df=pd.DataFrame({ 'zone':a })

多字符串列映射和清理

1 个答案: