python和bash统计CPU利用率的方法

#!/usr/bin/python
# -*- coding:utf8 -*-
__author__ = 'chenwx'
def cpu_rate():
 import time
 def cpu_r():
 f = open("/proc/stat","r")
 for f_line in f:
 break
 f.close()
 f_line = f_line.split(" ")
 f_line_a=[]
 for i in f_line:
 if i.isdigit():
 i=int(i)
 f_line_a.append(i)
 total = sum(f_line_a)
 idle = f_line_a[3]
 return total,idle
 total_a,idle_a=cpu_r()
 time.sleep(2)
 total_b,idle_b=cpu_r()
 sys_idle = idle_b - idle_a
 sys_total = total_b - total_a
 sys_us = sys_total - sys_idle
 cpu_a = (float(sys_us)/sys_total)*100
 return cpu_a
# print cpu_rate()

#!/bin/bash
# 感觉计算数组这里应该还有办法简化的吧;
# 我一时没想到，请大家提一下建议，多谢;
cpu_a=(`grep 'cpu ' /proc/stat`)
total_a=$((${cpu_a[1]}+${cpu_a[2]}+${cpu_a[3]}+${cpu_a[4]}+${cpu_a[5]}+${cpu_a[6]}+${cpu_a[7]}+${cpu_a[8]}+${cpu_a[9]}))
idle_a=${cpu_a[4]}
sleep 5
cpu_b=(`grep 'cpu ' /proc/stat`)
total_b=$((${cpu_b[1]}+${cpu_b[2]}+${cpu_b[3]}+${cpu_b[4]}+${cpu_b[5]}+${cpu_b[6]}+${cpu_b[7]}+${cpu_b[8]}+${cpu_b[9]}))
idle_b=${cpu_b[4]}
sys_idle=$(($idle_b-$idle_a))
sys_total=$(($total_b-$total_a))
sys_us=$(($sys_total-$sys_idle))
echo "scale=2;$sys_us/$sys_total*100" | bc

# 找到了解决数组计算的办法了，不过感觉for循环计算的方式还是有些繁琐；
# 不知道有没有那种对数组内所有值一并计算的方法；
cpu_rate_a () {
cpu_a=(`grep 'cpu ' /proc/stat`)
for i in ${cpu_a[@]:1}
do
 total_a=$(($total_a+$i))
done
idle_a=${cpu_a[4]}
sleep 5
cpu_b=(`grep 'cpu ' /proc/stat`)
for i in ${cpu_b[@]:1}
do
 total_b=$(($total_b+$i))
done
idle_b=${cpu_b[4]}
sys_idle=$(($idle_b-$idle_a))
sys_total=$(($total_b-$total_a))
sys_us=$(($sys_total-$sys_idle))
local_cpu_rate=$(echo "scale=2;$sys_us/$sys_total*100" | bc)
}