我有一个包含如下输入的文件：
host1 192.168.100.24
user1@abc.com host2 192.168.100.45 host7 192.168.100.40 host3 192.168.100.34 host4 192.168.100.20
user2@xyz.com host8 192.168.100.48 host6 192.168.100.43 host10 192.168.100.37
host5 192.168.100.24 host9 192.168.100.33
预期产出：
no_email:
host1 192.168.100.24
host5 192.168.100.24
host9 192.168.100.33
user1@abc.com:
host2 192.168.100.45
host7 192.168.100.40
host3 192.168.100.34
host4 192.168.100.20
user2@xyz.com:
host8 192.168.100.48
host6 192.168.100.43
host10 192.168.100.37
代码：
def get_contacts(filename):
emails = []
hostname = []
ip = []
with open(filename,'r') as contacts_file:
for a_contact in contacts_file:
match = re.match('^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4})$', a_contact.split()[0])
if match == None:
emails.append('no_email')
hostname.append(a_contact.split()[0])
ip.append(a_contact.split()[1])
line_length = a_contact.count(' ')
elif line_length > 1:
emails.append(a_contact.split()[0])
hostname.append(a_contact.split()[1])
ip.append(a_contact.split()[2])
else:
emails.append(a_contact.split()[0])
hostname.append(a_contact.split()[1])
ip.append(a_contact.split()[2])
return emails, hostname, ip
我只想返回主机名和IP列表，用于发送到列表返回的指定电子邮件地址。任何人都可以帮助我轻松完成它吗？感谢。
答案 0 :(得分：2)
首先安装validate_email模块：
$pip3 install validate_email
然后：
from validate_email import validate_email
result = {}
with open('file.txt') as f:
for line in f:
words = line.split()
if validate_email(words[0]): # If first word of the line is a valid email, lets store data on the result dict using the email as key.
email = words[0]
words = words[1:]
else:
email = 'no_email'
hosts_emails = [(words[i], words[i+1]) for i in range(0, len(words) - 1, 2)]
(result.setdefault(email, [])).append(hosts_emails)
print(result)
<强>输出：
{'no_email': [[('host1', '192.168.100.24')], [('host5', '192.168.100.24'), ('host9', '192.168.100.33')]], 'user1@abc.com': [[('host2', '192.168.100.45'), ('host7', '192.168.100.40'), ('host3', '192.168.100.34'), ('host4', '192.168.100.20')]], 'user2@xyz.com': [[('host8', '192.168.100.48'), ('host6', '192.168.100.43'), ('host10', '192.168.100.37')]]}
答案 1 :(得分：0)
希望这会有所帮助。使用字典是明智的，其中密钥可以是no_email或email_id（如果与电子邮件正则表达式匹配）。对于每次迭代，我们首先将to_update变量设置为no-email，并且只有在找到匹配的电子邮件时才更改它。因此，host_and_ip变量被设置为仅使用主机和ips获取每行的部分，即，当检测到匹配的电子邮件时剥离电子邮件地址。如果检测到电子邮件，我们会看到我们的字典dicto中是否已存在相同的电子邮件，如果是，我们只需更新主机和ips，否则我们将正确初始化电子邮件列表（作为新密钥）。
import re
def get_contacts(filename):
dicto={}
dicto['no_email']=[]
with open(filename,'r') as contacts_file:
for a_contact in contacts_file:
match = re.match('^[_a-z0-9-]+(\.[_a-z0-9-]+)*@[a-z0-9-]+(\.[a-z0-9-]+)*(\.[a-z]{2,4})$', a_contact.split()[0])
to_update = 'no_email' #by default to_update is set to no_email
if match == None:
host_and_ip = a_contact.split() #grab all as host and ip
else:
curr_email = a_contact.split()[0]
if curr_email not in dicto.keys():
dicto[curr_email]=[] #initialize for new email
host_and_ip = a_contact.split()[1:] #grab leaving one behind i.e. the email
to_update = curr_email #to be updated to the email
for i in range(len(host_and_ip)//2):
dicto[to_update]+=[[host_and_ip[2*i],host_and_ip[2*i+1]]]
return dicto
print(get_contacts('test.txt'))
该函数将提供如下字典：
{'no_email': [['host1', '192.168.100.24'], ['host4', '192.168.100.20'], ['host5', '192.168.100.24'], ['host9', '192.168.100.33']], 'user1@abc.com': [['host2', '192.168.100.45'], ['host7', '192.168.100.40'], ['host3', '192.168.100.34']], 'user2@xyz.com': [['host8', '192.168.100.48'], ['host6', '192.168.100.43'], ['host10', '192.168.100.37']]}
您可以轻松访问特定电子邮件ID的主机和IP列表，如下所示：
get_contacts('test.txt')['user1@abc.com']将返回主机和ips列表。
答案 2 :(得分：0)
我使用第三方库more_itertools来帮助实现grouper itertools配方。这可以通过pip install more_itertools安装。
import more_itertools as mit
dd = ct.defaultdict(list)
with open(filename, "r") as f:
for line in f.readlines():
parts = line.split()
if "@" not in parts[0]:
dd["no email"].extend(list(mit.grouper(2, parts)))
else:
name = parts[0]
dd[name].extend(list(mit.grouper(2, parts[1:])))
dd
输出
defaultdict(list,
{'no email': [
('host1', '192.168.100.24'),
('host5', '192.168.100.24'),
('host9', '192.168.100.33')],
'user1@abc.com': [
('host2', '192.168.100.45'),
('host7', '192.168.100.40'),
('host3', '192.168.100.34'),
('host4', '192.168.100.20')],
'user2@xyz.com': [
('host8', '192.168.100.48'),
('host6', '192.168.100.43'),
('host10', '192.168.100.37')]})
grouper配方帮助重新组合（主机，IP），每行后面都用空格分隔。
您可以选择不安装more_itertools来实施此配方。
来自itertools recipes（在Python 3中）：
from itertools import zip_longest
def grouper(iterable, n, fillvalue=None):
"Collect data into fixed-length chunks or blocks"
# grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return zip_longest(*args, fillvalue=fillvalue)
答案 3 :(得分：0)
一种方法是拆分每一行，并确定第一个条目中是否有Request.GetOwinContext().Request.Headers.Remove("X-XSRF-TOKEN");
个字符。然后使用切片来提取剩余的条目：
@这将显示：
def get_contacts(filename):
no_email = []
users = []
with open(filename) as f_contacts:
for row in f_contacts:
entries = row.split()
if '@' in entries[0]:
pairs = [entries[i:i+2] for i in range(1, len(entries), 2)]
users.append([entries[0], pairs])
else:
for i in range(0, len(entries), 2):
no_email.append(entries[i:i+2])
return no_email, users
no_email, users = get_contacts('contacts.txt')
print "no_email:"
for host, ip in no_email:
print " {} {}".format(host, ip)
for user_entry in users:
print "{}:".format(user_entry[0])
for host, ip in user_entry[1]:
print " {} {}".format(host, ip)
no_email:
host1 192.168.100.24
host5 192.168.100.24
host9 192.168.100.33
user1@abc.com:
host2 192.168.100.45
host7 192.168.100.40
host3 192.168.100.34
host4 192.168.100.20
user2@xyz.com:
host8 192.168.100.48
host6 192.168.100.43
host10 192.168.100.37
以users
如果您的文件对同一个用户有多行，则需要使用["username", [["host1", "ip1"], ["host2, "ip2"]]]在同一位置存储同一用户的所有条目。