python - 删除列表中的所有电子邮件，不包括 gmails

Question

我正在尝试过滤 emailList 以仅打印 gmail。该列表包含每个人的多封电子邮件，以逗号分隔。我想在保持相应名称顺序的同时检索每个人的 gmail。

class engineerInfo:
firstName = ""
lastName = ""
email = ""
title = ""

engineers = []

for col in rows:
    e = engineerInfo()
    e.firstName = col[0]
    e.lastName = col[1]
    e.email = col[2]
    e.title = col[3]
    engineers.append(e)

while True:
    print("1- Print gmails of software engineers")
    choice = int(input("Choose from the menu:"))

if choice == 1:
    emailList = []
    for i in engineers:
        if i.email not in emailList:
            emailList.append(i.email)

    gmailList = []
    for i in emailList:
        if i != 'gmail.com':
            continue
        else:
            gmailList.append(i)    
    print(gmailList)

Answer 1

检查应该是 .endswith('@gmail.com')。但是，电子邮件地址不区分大小写，例如(@kwhicks 的荣誉):

gmailList = []
for i in emailList:
    if <b>i.lower().endswith('@gmail.com')</b>:
        gmailList.append(i)
print(gmailList)

但是您最好为此使用列表理解(替换整个 for 循环):

gmailList = [i for i in emailList if i.lower().endswith('@gmail.com')]