#!/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
# 感覺計(jì)算數(shù)組這里應(yīng)該還有辦法簡化的吧;
# 我一時(shí)沒想到，請(qǐng)大家提一下建議，多謝;
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

# 找到了解決數(shù)組計(jì)算的辦法了，不過感覺for循環(huán)計(jì)算的方式還是有些繁瑣；
# 不知道有沒有那種對(duì)數(shù)組內(nèi)所有值一并計(jì)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)
}