Pandas将用户代理列解析为多个列

时间:2017-01-24 23:31:11

标签: python pandas user-agent

我有一个http请求日志的数据框。唯一相关的列是我尝试解析的userAgent列。我正在使用ua_parser。这会将每个userAgent转换为嵌套字典,如下所示:

>>> from ua_parser import user_agent_parser
>>> user_agent_parser.Parse('Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.95 Safari/537.36')
{
     'device': {'brand': None, 
                'model': None, 
                'family': 'Other'}, 
     'os': {'major': '10', 
            'patch_minor': None, 
            'minor': '10', 
            'family': 'Mac OS X', 
            'patch': '5'}, 
     'user_agent': {'major': '55', 
                    'minor': '0', 
                    'family': 'Chrome', 
                    'patch': '2883'}, 
     'string': 'Mozilla/5.0 (Macintosh; Intel Mac OS X 10_10_5) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/55.0.2883.95 Safari/537.36'
}

我尝试使用user_agent_parser的结果在我的日志数据框架上创建4个额外的列。我喜欢device_brand,device_model,os_family和user_agent_family列。

不幸的是,当我将它存储为numpy数组时,我无法访问字典索引:

>>> parsed_ua = logs['userAgent'].apply(user_agent_parser.Parse)
>>> logs['device_brand'] = parsed_ua['device']['brand']
KeyError: 'device'

我尝试将其转换为数据框,因此我可以将parsed_ua与日志合并。不幸的是,这会将每个字典写入单个列

>>> pd.DataFrame(parsed_ua)
userAgent
0   {u'device': {u'brand': None, u'model': None, u...
1   {u'device': {u'brand': None, u'model': None, u...
2   {u'device': {u'brand': None, u'model': None, u...
3   {u'device': {u'brand': None, u'model': None, u...
4   {u'device': {u'brand': None, u'model': None, u...

如何解析userAgent列并将结果写入多列?

2 个答案:

答案 0 :(得分:2)

您可以使用json_normalize()方法:

In [146]: pd.io.json.json_normalize(parsed_ua)
Out[146]:
  device.brand device.family device.model os.family os.major os.minor  \
0         None         Other         None  Mac OS X       10       10

  os.patch os.patch_minor                                   string  \
0        5           None  Mozilla/5.0 (Macintosh; Intel Mac OS...

  user_agent.family user_agent.major user_agent.minor user_agent.patch
0            Chrome               55                0             2883

答案 1 :(得分:0)

除了您已完成的工作外,您还可以使用Series' apply的lambda:

ua = logs['userAgent'].apply(lambda ua: user_agent_parser.Parse(ua))

logs['device_brand'] = ua.apply(lambda x: x['device']['brand'])
logs['device_model'] = ua.apply(lambda x: x['device']['model'])
logs['os_family'] = ua.apply(lambda x: x['os']['family'])
logs['user_agent_family'] = ua.apply(lambda x: x['user_agent']['family'])