如何获取子进程的最后 N 行？ stderr 流输出？

Question

我是一名 Python 新手，正在编写一个 Python (2.7) 脚本，该脚本需要执行许多外部应用程序，其中一个应用程序将大量输出写入其 stderr 流。我试图找出一种简洁明了的方法（在 Python 中）从该子进程的 stderr 输出流中获取最后 N 行。

目前，我正在从我的 Python 脚本运行该外部应用程序，如下所示：

p = subprocess.Popen('/path/to/external-app.sh', stdout=subprocess.PIPE, stderr=subprocess.PIPE)
stdout, stderr = p.communicate()

if p.returncode != 0:
    print "ERROR: External app did not complete successfully (error code is " + str(p.returncode) + ")"
    print "Error/failure details: ", stderr
    status = False
else:
    status = True

我想从其 stderr 流中捕获最后 N 行输出，以便可以将它们写入日志文件或通过电子邮件发送等。

Answer 1

N = 3 # for 3 lines of output
p = subprocess.Popen(['/path/to/external-app.sh'], 
    stdout=subprocess.PIPE, stderr=subprocess.PIPE)
stdout, stderr = p.communicate()

if p.returncode != 0:
    print ("ERROR: External app did not complete successfully "
           "(error code is %s)" % p.returncode)
    print "Error/failure details: ", '\n'.join(stderr.splitlines()[-N:])
    status = False
else:
    status = True

Answer 2

如果整个输出无法存储在 RAM 中，则：

import sys

from collections import deque
from subprocess  import Popen, PIPE
from threading   import Thread

ON_POSIX = 'posix' in sys.builtin_module_names

def start_thread(func, *args):
    t = Thread(target=func, args=args)
    t.daemon = True
    t.start()
    return t

def consume(infile, output):
    for line in iter(infile.readline, ''):
        output(line)
    infile.close()

p = Popen(['cat', sys.argv[1]], stdout=PIPE, stderr=PIPE,
          bufsize=1, close_fds=ON_POSIX)

# preserve last N lines of stdout,  print stderr immediately
N = 100 
queue = deque(maxlen=N)
threads  = [start_thread(consume, *args)
            for args in (p.stdout, queue.append), (p.stderr, sys.stdout.write)]
for t in threads: t.join() # wait for IO completion

print ''.join(queue), # print last N lines
retcode = p.wait()