Python，pandas如何通过查找特定单词而不是“，”或“\”和

Question

我很难从字符串中提取id号。你知道吗

我可以使用索引获得它，但是对于数据帧的其他行，它将失败。你知道吗

如何提取campaignid=351154190，这种方法适用于所有行。你知道吗

唯一的模式是单词campaignid，需要提取并存储在数据帧的新列中。在这项任务中，表现并不重要。你知道吗

原始字符串

https:_utm_source=googlebrand&utm_medium=ppc&utm_campaign=brand&utm_campaignid=3
51154190&keyword=aihdisadjiajdutm_matchtype=e&device=m&utm_network=g&utm_adposit
ion=1t1&geo=9027258&gclid=CjwKCsadjjsaopdl[psdklksfdosjfidj9FOk033DKW1xoCXlwQAvD
_BwE&affiliate_id=asdaskdosjadiasjdisaj-asdhasuigdyusagdyusagyk033DKW1xoCXlwQAvD_BwE&utm_content=search&utm_contentid=1251489456158180&placement&extension

拆分字符串

x= cw.captureurl.str.split('&').str[:-1]

打印一行

print(x[25])

['https:_utm_source=googlebrand', 'utm_medium=ppc', 'utm_campaign=brand', 
'utm_campaignid=35119190', 'keyword=co',
 'utm_matchtype=e', 'device=m', 'utm_network=g', 'utm_adposition=1t1',
 'geo=9027258', 'gclid=CjwKCAjwnMTqBRAzEiwAEF3ndo3-
CNOsp1VT5OIxm0BuUcSWQEwtJSR5KLiJzrvjjc9FOk033DKW1xoCXlwQAvD_BwE',
 'affiliate_id=CjwKCAjwnMTqBRAzEiwAEF3ndo3-
CNOsp1VT5OIxm0BuUcSWQEwtJSR5KLiJzrvjjc9FOk033DKW1xoCXlwQAvD_BwE', 
'utm_content=search', 'utm_contentid=1211732930', 'placement']

如果我能使用一些搜索单词“campaignid”（我的目标是什么）的工具，那就太好了

然后将其存储在某个数据帧的另一列中。你知道吗

我试过一次又一次的分开，但没有成功 我尝试使用for循环with if语句，但也不起作用。你知道吗

Answer 1

使用正则表达式：

campaign_id = cw['captureurl'].str.extract('campaignid=(\\d+)')[0]

Answer 2

我建议使用urllib。特别是，parse_qs函数将获得字符串参数的字典。https://docs.python.org/3/library/urllib.parse.html

使用您的示例URL，我们得到：

from urllib.parse import parse_qs

test = 'https:_utm_source=googlebrand&utm_medium=ppc&utm_campaign=brand&utm_campaignid=351154190&keyword=aihdisadjiajdutm_matchtype=e&device=m&utm_network=g&utm_adposition=1t1&geo=9027258&gclid=CjwKCsadjjsaopdl[psdklksfdosjfidj9FOk033DKW1xoCXlwQAvD_BwE&affiliate_id=asdaskdosjadiasjdisaj-asdhasuigdyusagdyusagyk033DKW1xoCXlwQAvD_BwE&utm_content=search&utm_contentid=1251489456158180&placement&extension'

print(parse_qs(test))

{'https:_utm_source': ['googlebrand'],
 'utm_medium': ['ppc'],
 'utm_campaign': ['brand'],
 'utm_campaignid': ['351154190'],
 'keyword': ['aihdisadjiajdutm_matchtype=e'],
 'device': ['m'],
 'utm_network': ['g'],
 'utm_adposition': ['1t1'],
 'geo': ['9027258'],
 'gclid': ['CjwKCsadjjsaopdl[psdklksfdosjfidj9FOk033DKW1xoCXlwQAvD_BwE'],
 'affiliate_id': ['asdaskdosjadiasjdisaj-asdhasuigdyusagdyusagyk033DKW1xoCXlwQAvD_BwE'],
 'utm_content': ['search'],
 'utm_contentid': ['1251489456158180']}

要获取整个数据帧的活动ID，我们可以使用.apply来完成：

# After parsing each url's arguments, we extract the first campaignid from the dictionary's list.
df['utm_campaignid'] = df['url'].apply(lambda x: parse_qs(x)['utm_campaignid'][0])
df.head()

    url utm_campaignid
0   https:_utm_source=googlebrand&utm_medium=ppc&u...   351154190