AttributeError：“系列”对象在发送电子邮件时没有属性“拆分”错误答案

【问题标题】：AttributeError: 'Series' object has no attribute 'split' error in sending emailsAttributeError：“系列”对象在发送电子邮件时没有属性“拆分”错误
【发布时间】：2021-07-27 08:11:52
【问题描述】：

如何解决以下错误。用分号分隔测试电子邮件时的消息如下？理想情况下，我应该从 Sendfrom 中的相应电子邮件发送电子邮件。

测试

SENDFROM        Test 
xx@gmail.com  xxxx@vvv.com;yyy@gggfg.com;tiitioo@ggg.com
yy@xxx.com     ggg@vvv.com;yyy@gggfg.com;vvv@ggg.com

AttributeError: 'Series' object has no attribute 'split'

我的代码如下：

import smtplib, ssl
from email.message import EmailMessage
import getpass
email_pass = getpass.getpass() #Office 365 password 
# email_pass = input() #Office 365 password 
context=ssl.create_default_context()
for idx, row in test.iterrows():
    
    emails = test['Test']
    sender_list  = test["SENDFROM"]
    
    smtp_ssl_host = 'smtp.office365.com'
    smtp_ssl_port = 587
    email_login = "xx@xx.com"
    email_from = sender_list
    email_to = emails
    msg2 = MIMEMultipart()
    msg2['Subject'] = "xxx"
    msg2['From'] = sender_list

    msg2['To'] = ", ".join(email_to.split(";"))
    msg2['X-Priority'] = '2'

    text = ("xxxx")
        
    msg2.attach(MIMEText(text))
    s2 = smtplib.SMTP(smtp_ssl_host, smtp_ssl_port)
    s2.starttls(context=context)
    s2.login(email_login, email_pass) 

      
    s2.send_message(msg2)
        
    s2.quit()

【问题讨论】：

标签： python pandas smtp

【解决方案1】：

让我们试试str.split 和str.join：

import pandas as pd

df = pd.DataFrame({'SENDFROM': {0: 'xx@gmail.com', 1: 'yy@xxx.com'},
                   'Test': {0: 'xxxx@vvv.com;yyy@gggfg.com;tiitioo@ggg.com',
                            1: 'ggg@vvv.com;yyy@gggfg.com;vvv@ggg.com'}})

# Use str.split and str.join and astype
df['Test'] = df['Test'].str.split(';').str.join(',')

print(df.to_string())

输出：

       SENDFROM                                        Test
0  xx@gmail.com  xxxx@vvv.com,yyy@gggfg.com,tiitioo@ggg.com
1    yy@xxx.com       ggg@vvv.com,yyy@gggfg.com,vvv@ggg.com

【讨论】：

【解决方案2】：

email_to 对象显然是 Series，而不是字符串，因此它没有 split() 方法。 Series 已经是一个类似序列的对象，所以无论如何都不需要拆分它。发送type(email_to) 以确认这一点。

【讨论】：

是的，我该如何解决？

【解决方案3】：

您不能将split 用于Series Object。据我了解，您想做这样的事情：

import pandas as pd

test  = pd.Series(['xxxx@vvv.com;yyy@gggfg.com;tiitioo@ggg.com'])

print(test)

>> 0    xxxx@vvv.com;yyy@gggfg.com;tiitioo@ggg.com
>> dtype: object

# You can see that the only and first row of Series s is a string of all 
# emails you want to split by ';'. Here you can do:

# Apply split to string in first row of Series: returns a list
print(test[0].split(';'))

>> ['xxxx@vvv.com', 'yyy@gggfg.com', 'tiitioo@ggg.com']

# I believe you can solve your problem with this list of emails. 
# However you should code a loop to iterate for the remaing rows of initial Series. 
# ----------------------------------------------------------------------------------
# Furthermore, you can explode your pandas Series. 
# This will return you a DataFrame (ser), from which you can extract the info you want.

t = pd.concat([pd.Series(test[0], test[0].split(';'))              
                    for _, row in test.iteritems()]).reset_index()

# Remove weird column
t.drop(labels=[0], axis=1, inplace=True)

# Convert DataFrame back to Series
t = t.squeeze()

# The info you probably want:
print(t)

>> 0       xxxx@vvv.com
>> 1      yyy@gggfg.com
>> 2    tiitioo@ggg.com
>> Name: index, dtype: object

大声疾呼：Split (explode) pandas dataframe string entry to separate rows

【讨论】：

刚刚我意识到您必须在您的系列中迭代并按每一行拆分电子邮件

【解决方案4】：

您可以使用pandas.Series.replace() 将; 替换为,

df['Test'] = df['Test'].replace(';', ',')

【讨论】：