我需要處理電子郵件的列表，切斷他們，使輸出與迭代計數器指數
MailList = [[email protected], [email protected], [email protected]] # I have a list with emails
# I need to process them and get next output: "User%number% is email1" I'm using something like:
for c in MailList:
print('user'+ ' is '+c[0:7]) # what I need to insert here to get desirable counter for my output?
如果我理解正確的話，你想這樣：
MailList = ["[email protected]", "[email protected]", "[email protected]"]
for index, value in enumerate(MailList):
print('user {} is {}'.format(index, value.split("@")[0]))
見the docs爲01細節。
您需要@拆分郵件：
>>> MailList = ['[email protected]', '[email protected]', '[email protected]']
>>> for i in MailList :
... print ('user'+ ' is {}'.format(i.split('@')[0]))
...
user is email1
user is email2
user is email3
使用itertools.takewhile：
>>> import itertools
>>> MailList = ['[email protected]', '[email protected]', '[email protected]']
>>> for x in MailList:
... print("user is {}".format("".join(itertools.takewhile(lambda x:x!='@',x))))
...
user is email1
user is email2
user is email3
使用index：
>>> for x in MailList:
... print("user is {}".format(x[:x.index('@')]))
...
user is email1
user is email2
user is email3
使用find：
>>> for x in MailList:
... print("user is {}".format(x[:x.find('@')]))
...
user is email1
user is email2
user is email3
謝謝！它有幫助。 – 2014-12-05 08:40:56
@AntonJakimenko我很高興我能幫助你！如果這個答案（或另一個）解決了你的問題，那麼如果你能接受它，那將是非常好的。 – rlms 2014-12-05 18:32:32