如何通过过滤和排序“重复”值来比较两个列表

Question

我有以下用于电子邮件活动的 act2.txt 文件：

2021-04-02//email@example.com//Enhance your presentation skills in 15 minutes//Open
2021-04-11//email@example.com//Enroll in the presentations skills - FREE WEBINAR//Open
2021-04-11//email@example.com//Enroll in the presentations skills - FREE WEBINAR//Delivered
2021-04-11//email@example.com//Enroll in the presentations skills - FREE WEBINAR//Delivered
2021-04-11//email@example.com//Enroll in the presentations skills - FREE WEBINAR//Delivered
2021-04-16//email@example.com//YOU ARE INVITED TO THIS PROGRAMMING EVENT//Delivered
2021-04-01//email@example.com//Enhance your presentation skills in 15 minutes//Delivered
2021-04-09//email@example.com//we are here to help you improve your skills//Delivered
2021-04-12//email@example.com//(1st meeting) here is our recorded presentation skills webinar//Delivered
2021-04-13//email@example.com//YOU ARE INVITED TO THIS PROGRAMMING EVENT//Delivered

我想按客户跟踪电子邮件活动 - 我计算了已发送的电子邮件、已发送的电子邮件然后打开率。

我生成了两个列表，一个用于发送的电子邮件，另一个用于打开的电子邮件：

import re
from pprint import pprint

#read the file with activities separated by //
afile = "act2.txt"
afile_read = open(afile,"r")
lines = afile_read.readlines()

activityList = []
for activities in lines:
            activity = activities.split("//")
            date = activity[0]
            customer_email = activity[1]
            email_title = activity[2]
            action = activity[3]
            stripped_line = [s.rstrip() for s in activity]
            activityList.append(stripped_line)

#print (activityList)


stripped_email = 'email@example.com'
email_actions = [x for x in activityList if stripped_email in x[1]]
delivered = [x for x in email_actions if 'Delivered' in x]
Opened = [x for x in email_actions if 'Open' in x]
delcount = (len(delivered))
opencount = (len(Opened))
try:
  Open_rate =  opencount / delcount * 100
except ZeroDivisionError:
  Open_rate = 0
print (stripped_email,",", delcount,",", opencount,",", Open_rate,"%")

pprint(delivered)
pprint (Opened)

交付清单：

[['2021-04-11',
  'email@example.com',
  'Enroll in the presentations skills - FREE WEBINAR',
  'Delivered'],
 ['2021-04-11',
  'email@example.com',
  'Enroll in the presentations skills - FREE WEBINAR',
  'Delivered'],
 ['2021-04-11',
  'email@example.com',
  'Enroll in the presentations skills - FREE WEBINAR',
  'Delivered'],
 ['2021-04-16',
  'email@example.com',
  'YOU ARE INVITED TO THIS PROGRAMMING EVENT',
  'Delivered'],
 ['2021-04-01',
  'email@example.com',
  'Enhance your presentation skills in 15 minutes',
  'Delivered'],
 ['2021-04-09',
  'email@example.com',
  'we are here to help you improve your skills',
  'Delivered'],
 ['2021-04-12',
  'email@example.com',
  '(1st meeting) here is our recorded presentation skills webinar',
  'Delivered'],
 ['2021-04-13',
  'email@example.com',
  'YOU ARE INVITED TO THIS PROGRAMMING EVENT',
  'Delivered']]

打开列表：

[['2021-04-02',
  'email@example.com',
  'Enhance your presentation skills in 15 minutes',
  'Open'],
 ['2021-04-11',
  'email@example.com',
  'Enroll in the presentations skills - FREE WEBINAR',
  'Open']]

我想比较两个列表并生成第三个列表（组合活动），按电子邮件主题过滤 - 如果主题在已发送列表和打开列表中，那么它将被计为一个活动。但是，邮件主题可以重复，就像邮件发送了 3 次但只打开了一次。因为我还在学习 python，所以我找不到合适的逻辑。

为更清楚而编辑：

如果在按标题过滤的打开列表中找到一封电子邮件，则应在最后日期之前从已发送列表中删除相同的标题，并生成包含组合活动的新列表。

Answer 1

您需要以不同的方式考虑这一点，您不是在组合列表。

如果打开了一封电子邮件，则表示它也已收到。这意味着您打开的列表也是您的组合列表。

意识到这一点后，您所要做的就是将未打开的电子邮件复制到未打开电子邮件的结果列表中。

查看打开的邮件列表并将主题复制到一个集合中，然后查看收到的电子邮件并检查主题是否在集合中，如果是则什么都不做。如果主题不在集合中，则将其复制到未打开的电子邮件列表中。

这是一段非常简单的代码：

opened_subjects = set()
unopened = []
for email in opened:
    opened_subjects.add(email[2])

unopened_subjects = set()
for email in received:
    if all(email[2] not in subj_set 
           for subj_set in (opened_subjects, unopened_subjects)):
        unopened.append(email)
        unopened_subjects.add(email[2])

print('Both received and opened:', opened)
print('Unopened emails:', unopened)

一个小笔记 -
每一组的原因都不一样。第一个集合opened_subjects 之所以存在，是因为set 能够仅包含唯一项目，而这正是本例所需要的。第二组unopened_subjects 在那里，因为检查一个项目是否在一个集合中比在一个列表中更快，因为我在以任何方式添加到集合之前检查，所以不需要集合存储唯一的能力仅限。