python - 如何使用反向分割线？

Question

我想解析变量output并打印最新的标签(按时间)，目前我正在打印最旧的标签，有没有办法使用reverse分割线？

import re
output="""Fri Apr 27 18:43:20 2018 username3 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="572224322539363a32641734383a2736392e7934383a" rel="noreferrer noopener nofollow">[email protected]</a>> submitted CoreCapture-274.1 to release1
Fri Apr 27 18:43:20 2018 username3 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="493c3a2c3b2728242c7a092a262439282730672a2624" rel="noreferrer noopener nofollow">[email protected]</a>> forwarded CoreCapture-274.1 to release2, release2E, release1E, release3, and release3E
Fri May 18 20:09:22 2018 username2 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="fe8b8d9b8c909f939bccbe9d91938e9f9087d09d9193" rel="noreferrer noopener nofollow">[email protected]</a>> submitted CoreCapture-274.2 to release1
Fri May 18 20:09:22 2018 username2 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="c6b3b5a3b4a8a7aba3f486a5a9abb6a7a8bfe8a5a9ab" rel="noreferrer noopener nofollow">[email protected]</a>> forwarded CoreCapture-274.2 to release2, release2E, release1E, release3, and release3E
Fri Jun  8 13:35:44 2018 username1 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="01747264736f606c643041626e6c71606f782f626e6c" rel="noreferrer noopener nofollow">[email protected]</a>> submitted CoreCapture-274.3 to release1
Fri Jun  8 13:35:44 2018 username1 <<a href="https://stackoverflow.com/cdn-cgi/l/email-protection" class="__cf_email__" data-cfemail="9feaecfaedf1fef2faaedffcf0f2effef1e6b1fcf0f2" rel="noreferrer noopener nofollow">[email protected]</a>> forwarded CoreCapture-274.3 to release2, release2E, release1E, release3, and release3E"""

train = "release2"
for line in output.splitlines():
    #print "line"
    #print line
    if re.match(r'.*(submitted|forwarded)(.*) to .*%s.*'%train,line):
        print line
        break

输出:-

CoreCapture-274.1

预期输出:-

CoreCapture-274.3

Answer 1

将 splitlines 调用包装在 reversed 中，它将以相反的顺序迭代结果行(无需额外的临时 列表):

for line in reversed(output.splitlines()):

或者，如果您需要相反顺序的列表本身，您可以将其分成两行:

lines = output.splitlines()
lines.reverse()  # Reverses in place

或者用一个反向切片将其单行(以额外的临时列表为代价):

lines = output.splitlines()[::-1]  # Returns reversed list