我需要使用Python代码获取一些服务信息。我希望以非常低的CPU消耗运行这段代码
以下是简化代码:
import subprocess, time
start_time = time.time()
command1 = "systemctl show alsa-restore.service --property=LoadState,UnitFileState,ActiveState,SubState"
command2 = "systemctl show apt-daily.service --property=LoadState,UnitFileState,ActiveState,SubState"
command3 = "systemctl show cryptdisks-early.service --property=LoadState,UnitFileState,ActiveState,SubState"
commands = command1 + "\n" + command2 + "\n" + command3
output = subprocess.check_output(commands, shell=True).decode()
end_time = time.time()
print(end_time-start_time)
实际的代码有很多服务,并且此代码运行非常频繁。 这段代码运行得更快的原因是什么?有没有其他方法可以更快地获取此信息
System: Debian-like Linux x64, Python: 3.9
目前没有回答
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