Question

我有一个如下所示的熊猫数据框：-

    [Route("api/[controller]/[action]")]
public class XmlController : Controller
{
    private readonly ITimerService _timerService;

    public XmlController(ITimerService timerService)
    {
        //Injected in
        _timerService = timerService;
    }

    [HttpGet]
    public IActionResult ProccessXML(object someXMLObject)
    {
        _timerService.StopTimer();
        SomeMethodWithXml(someXMLObject)
        //Reset Timer
        _timerService.StartTimer();
        return Ok();
    }


}

我也有一个类似下面的列表：-

    Tweets
0   RT @cizzorz: THE CHILLER TRAP *TEMPLE RUN* OBS...
1   Disco Domination receives a change in order to...
2   It's time for the Week 3 #FallSkirmish Trials!...
3   Dance your way to victory in the new Disco Dom...
4   Patch v6.02 is available now with a return fro...
5   Downtime for patch v6.02 has begun. Find out a...
6   ⛏️... soon
7   Launch into patch v6.02 Wednesday, October 10!...
8   Righteous Fury.\n\nThe Wukong and Dark Vanguar...
9   RT @wbgames: WB Games is happy to bring @Fortn...

现在，如果要从my_list中找到匹配的单词，我想过滤行，并获得整行并将其作为电子邮件发送或发送给空闲人员。

就像我应该在第no行那样得到输出是因为它里面有Dance字。

my_list = ['Launch', 'Dance', 'Issue']

我尝试下面的代码进行过滤，但是每次给我一个空值

3   Dance your way to victory in the new Disco Dom..

如果我有列表中的匹配词，我只想发送与正文相同的电子邮件。

Answer 1

使用regex=True

例如：

data[data['Tweets'].str.contains("|".join(my_list), regex=True)]

Answer 2

这将完成它：

import pandas as pd
import numpy as np
from io import StringIO

s = '''
"RT @cizzorz: THE CHILLER TRAP *TEMPLE RUN* OBS..."
"Disco Domination receives a change in order to..."
"It's time for the Week 3 #FallSkirmish Trials!..."
"Dance your way to victory in the new Disco Dom..."
"Patch v6.02 is available now with a return fro..."
"Downtime for patch v6.02 has begun. Find out a..."
"⛏️... soon"
"Launch into patch v6.02 Wednesday, October 10!..."
"Righteous Fury.\n\nThe Wukong and Dark Vanguar..."
"RT @wbgames: WB Games is happy to bring @Fortn...       plane     5         [20 , 12, 30]"
      '''

ss = StringIO(s)
df = pd.read_csv(ss, sep=r'\s+', names=['Data'])

my_list = ['Launch', 'Dance', 'Issue']

cond = df.Data.str.contains(my_list[0])

for x in my_list[1:]:
    cond = cond | df.Data.str.contains(x)

df[cond]

通过列表过滤熊猫并发送电子邮件

2 个答案: