在列表中过滤 pandas 并发送电子邮件

Question

我有一个如下所示的 pandas 数据框：-

    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...

我还有一个列表假设如下：-

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

现在，如果 my_list 中有匹配词，我想过滤行并获取整行并将其作为电子邮件发送或发送到 slack。

就像我应该得到第 no 行的输出是因为它里面有舞蹈词。

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

我尝试使用下面的代码进行过滤，但每次都给我一个空值

data[data['Tweets'].str.contains('my_list')]

此外，如果我有来自列表中的匹配词，我只想将电子邮件与正文发送到同一行，否则我不想要。

Answer 1

使用regex=True

例如：

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

Answer 2

这样就搞定了：

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]

在列表中过滤 pandas 并发送电子邮件

filtering pandas over a list and sending email

python

dataframe

python-3.x

pandas

slack-api