Python - 在打印前按级别对日志消息进行排序

Question

假设我的 Python 脚本中有这样的日志记录设置：

import logging

logging.basicConfig(level=logging.DEBUG, stream=sys.stdout,
                    format='%(asctime)s %(levelname)s %(message)s',
                    datefmt='%a, %d %b %Y %H:%M:%S')
logging.info('info')
logging.error('error...')
logging.debug('debug...')

有没有办法让它等待打印到标准输出，直到脚本完成运行并在打印前按级别对日志消息进行排序？

Answer 1

看起来，传递给stream的对象只需要一个write方法。这意味着您可以创建一个类似列表的对象，该对象在调用 write 时附加数据——然后您可以在打印之前对类似列表的对象进行简单的排序。

import logging

class LogList(list):
   def write(self,data):
       self.append(data)

LL = LogList()
logging.basicConfig(stream = LL,level=logging.DEBUG)
logging.debug('This message should go to the log file')
logging.info('So should this')
logging.error('Wow, this is bad')
logging.warning('And this, too')
logging.debug('foobar')
logging.warning('baz')

for line in sorted(LL):
   print line[:-1]

当然，您可能需要使您的排序键多一点才能选择不同的级别。比如：

levels = {'DEBUG':0,'INFO':1,'WARNING':2,'ERROR':3}
LL.sort(key = lambda x: levels[x.split(':')[0]])

Answer 2

这很 hack-y，但您可以登录到 StringIO 对象，拆分行，对它们进行排序，然后将结果写入文件。

import logging
import cStringIO as StringIO

logStrObj = StringIO.StringIO()

logging.basicConfig(level=logging.DEBUG, stream=logStrObj,
                    format='%(asctime)s %(levelname)s %(message)s',
                    datefmt='%a, %d %b %Y %H:%M:%S')

logging.info('info')
logging.error('error...')
logging.info('info 2')
logging.debug('debug...')

# sort the contents of logStrObj
logList = logStrObj.getvalue().split('\n')
infoLogList = []
debugLogList = []
warningLogList = []
errorLogList = []
criticalLogList = []
for log in logList:
    if 'INFO' in log:
        infoLogList.append(log)
    elif 'DEBUG' in log:
        debugLogList.append(log)
    elif 'WARNING' in log:
        warningLogList.append(log)
    elif 'ERROR' in log:
        errorLogList.append(log)
    elif 'CRITICAL' in log:
        criticalLogList.append(log)
logList = infoLogList + debugLogList + warningLogList + errorLogList + criticalLogList

# write to a file (or print or whatever you want)
for line in logList:
    print line

Answer 3

这是一种不涉及事后排序的方法（因此效率更高）。它使用来自this question 的单级过滤器，并为每种类型的错误使用单独的处理程序。这样一来，日志肯定是按类型组织的，不会出现只检查字符串来确定日志类型的问题。

import logging
import cStringIO as StringIO

class SingleLevelFilter(logging.Filter):
    '''This single level logging filter is from https://stackoverflow.com/a/1383365/1460235'''
    def __init__(self, passlevel, reject):
        self.passlevel = passlevel
        self.reject = reject

    def filter(self, record):
        if self.reject:
            return (record.levelno != self.passlevel)
        else:
            return (record.levelno == self.passlevel)

# Use this formatter and logLevel in place of setting the global ones
formatter = logging.Formatter(fmt='%(asctime)s %(levelname)s %(message)s',
                    datefmt='%a, %d %b %Y %H:%M:%S')
globalLogLevel = logging.INFO

# build handlers/ StringIO logs for each type
rootLogger = logging.getLogger()
rootLogger.setLevel(globalLogLevel)
logStrObjList = []
handlers = []
i = 0
for logLevel in [logging.INFO, logging.DEBUG, logging.WARNING, logging.ERROR, logging.CRITICAL]:
    logStrObjList.append(StringIO.StringIO())
    handlers.append(logging.StreamHandler(logStrObjList[i]))
    handlers[i].addFilter(SingleLevelFilter(logLevel, False))
    handlers[i].setFormatter(formatter)
    handlers[i].setLevel(globalLogLevel)
    rootLogger.addHandler(handlers[i])
    i += 1

logging.critical('bad news bears')
logging.info('info')
logging.error('error...')
logging.info('info 2')
logging.debug('debug...')
logging.error('another errooo')