准确计算 Linux 中 CPU 使用率的百分比?
- 2024-11-07 08:55:00
- admin 原创
- 32
问题描述:
这个问题已被问过很多次,但是我找不到有力支持的答案。
很多人建议使用 top 命令,但是如果你运行一次 top(因为你有一个脚本,例如每 1 秒收集一次 CPU 使用情况),它将始终给出相同的 CPU 使用情况结果(示例 1、示例 2)。
计算 CPU 使用率的更准确方法是通过读取中的值/proc/stat
,但大多数答案仅使用中的前 4 个字段/proc/stat
来计算(此处有一个例子)。
/proc/stat/
从 Linux 内核 2.6.33 开始,每个 CPU 核心有 10 个字段!
我还发现这个使用 /proc/stat 准确计算 Linux 中的 CPU 利用率问题指出了同样的问题,即大多数其他问题仅考虑了许多字段中的 4 个 - 但这里给出的答案仍然以“我认为”(不确定)开头,除此之外,它只关注前 7 个字段(共 10 个/proc/stat/
)
这个perl 脚本使用所有字段来计算 CPU 使用率,经过进一步调查后我再次认为这是不正确的。
快速浏览一下此处的内核代码后,它看起来像,例如,guest_nice
和总是与和guest fields
一起增加(因此它们不应包括在 CPU 使用率计算中,因为它们已经包含在和字段中)nice
`usernice
user`
/*
* Account guest cpu time to a process.
* @p: the process that the cpu time gets accounted to
* @cputime: the cpu time spent in virtual machine since the last update
* @cputime_scaled: cputime scaled by cpu frequency
*/
static void account_guest_time(struct task_struct *p, cputime_t cputime,
cputime_t cputime_scaled)
{
u64 *cpustat = kcpustat_this_cpu->cpustat;
/* Add guest time to process. */
p->utime += cputime;
p->utimescaled += cputime_scaled;
account_group_user_time(p, cputime);
p->gtime += cputime;
/* Add guest time to cpustat. */
if (task_nice(p) > 0) {
cpustat[CPUTIME_NICE] += (__force u64) cputime;
cpustat[CPUTIME_GUEST_NICE] += (__force u64) cputime;
} else {
cpustat[CPUTIME_USER] += (__force u64) cputime;
cpustat[CPUTIME_GUEST] += (__force u64) cputime;
}
}
那么总结一下,在 Linux 中,如何准确计算 CPU 使用率,以及在计算中应该考虑哪些字段以及如何计算(哪些字段属于空闲时间,哪些字段属于非空闲时间)?
解决方案 1:
根据htop源代码,我的假设看起来是正确的:
(参见LinuxProcessList.cstatic inline double LinuxProcessList_scanCPUTime(LinuxProcessList* this)
中的函数)
// Guest time is already accounted in usertime
usertime = usertime - guest; # As you see here, it subtracts guest from user time
nicetime = nicetime - guestnice; # and guest_nice from nice time
// Fields existing on kernels >= 2.6
// (and RHEL's patched kernel 2.4...)
unsigned long long int idlealltime = idletime + ioWait; # ioWait is added in the idleTime
unsigned long long int systemalltime = systemtime + irq + softIrq;
unsigned long long int virtalltime = guest + guestnice;
unsigned long long int totaltime = usertime + nicetime + systemalltime + idlealltime + steal + virtalltime;
因此,从第一行列出的字段:(请参阅文档/proc/stat
第 1.8 节)
user nice system idle iowait irq softirq steal guest guest_nice
cpu 74608 2520 24433 1117073 6176 4054 0 0 0 0
从算法上来说,我们可以计算出 CPU 使用率百分比,如下所示:
PrevIdle = previdle + previowait
Idle = idle + iowait
PrevNonIdle = prevuser + prevnice + prevsystem + previrq + prevsoftirq + prevsteal
NonIdle = user + nice + system + irq + softirq + steal
PrevTotal = PrevIdle + PrevNonIdle
Total = Idle + NonIdle
# differentiate: actual value minus the previous one
totald = Total - PrevTotal
idled = Idle - PrevIdle
CPU_Percentage = (totald - idled)/totald
解决方案 2:
以下是基于 Vangelis 的回答的 bash 脚本。它产生如下输出:
total 49.1803
cpu0 14.2857
cpu1 100
cpu2 28.5714
cpu3 100
cpu4 30
cpu5 25
创建一个名为get_cpu_usage.sh
使用以下命令运行它:bash get_cpu_usage.sh 0.2
参数是要测量的秒数。在本例中为 200 毫秒。
内容如下:
#!/bin/sh
sleepDurationSeconds=$1
previousDate=$(date +%s%N | cut -b1-13)
previousStats=$(cat /proc/stat)
sleep $sleepDurationSeconds
currentDate=$(date +%s%N | cut -b1-13)
currentStats=$(cat /proc/stat)
cpus=$(echo "$currentStats" | grep -P 'cpu' | awk -F " " '{print $1}')
for cpu in $cpus
do
currentLine=$(echo "$currentStats" | grep "$cpu ")
user=$(echo "$currentLine" | awk -F " " '{print $2}')
nice=$(echo "$currentLine" | awk -F " " '{print $3}')
system=$(echo "$currentLine" | awk -F " " '{print $4}')
idle=$(echo "$currentLine" | awk -F " " '{print $5}')
iowait=$(echo "$currentLine" | awk -F " " '{print $6}')
irq=$(echo "$currentLine" | awk -F " " '{print $7}')
softirq=$(echo "$currentLine" | awk -F " " '{print $8}')
steal=$(echo "$currentLine" | awk -F " " '{print $9}')
guest=$(echo "$currentLine" | awk -F " " '{print $10}')
guest_nice=$(echo "$currentLine" | awk -F " " '{print $11}')
previousLine=$(echo "$previousStats" | grep "$cpu ")
prevuser=$(echo "$previousLine" | awk -F " " '{print $2}')
prevnice=$(echo "$previousLine" | awk -F " " '{print $3}')
prevsystem=$(echo "$previousLine" | awk -F " " '{print $4}')
previdle=$(echo "$previousLine" | awk -F " " '{print $5}')
previowait=$(echo "$previousLine" | awk -F " " '{print $6}')
previrq=$(echo "$previousLine" | awk -F " " '{print $7}')
prevsoftirq=$(echo "$previousLine" | awk -F " " '{print $8}')
prevsteal=$(echo "$previousLine" | awk -F " " '{print $9}')
prevguest=$(echo "$previousLine" | awk -F " " '{print $10}')
prevguest_nice=$(echo "$previousLine" | awk -F " " '{print $11}')
PrevIdle=$((previdle + previowait))
Idle=$((idle + iowait))
PrevNonIdle=$((prevuser + prevnice + prevsystem + previrq + prevsoftirq + prevsteal))
NonIdle=$((user + nice + system + irq + softirq + steal))
PrevTotal=$((PrevIdle + PrevNonIdle))
Total=$((Idle + NonIdle))
totald=$((Total - PrevTotal))
idled=$((Idle - PrevIdle))
CPU_Percentage=$(awk "BEGIN {print ($totald - $idled)/$totald*100}")
if [[ "$cpu" == "cpu" ]]; then
echo "total "$CPU_Percentage
else
echo $cpu" "$CPU_Percentage
fi
done
解决方案 3:
嘿,我也在研究这个主题,发现这个帖子真的很有帮助。我使用 Vangelis Tasoulas 公式为此编写了一个小型 Python 脚本。附件是我针对该问题编写的 Python 代码。它每秒加载每个 cpu_id 的 CPU 使用情况。也许它也对其他人有帮助。也欢迎发表评论/建议 :-)
#!/usr/bin/python
# -*- coding: utf-8 -*-
'''
Created on 04.12.2014
@author: plagtag
'''
from time import sleep
import sys
class GetCpuLoad(object):
'''
classdocs
'''
def __init__(self, percentage=True, sleeptime = 1):
'''
@parent class: GetCpuLoad
@date: 04.12.2014
@author: plagtag
@info:
@param:
@return: CPU load in percentage
'''
self.percentage = percentage
self.cpustat = '/proc/stat'
self.sep = ' '
self.sleeptime = sleeptime
def getcputime(self):
'''
http://stackoverflow.com/questions/23367857/accurate-calculation-of-cpu-usage-given-in-percentage-in-linux
read in cpu information from file
The meanings of the columns are as follows, from left to right:
0cpuid: number of cpu
1user: normal processes executing in user mode
2nice: niced processes executing in user mode
3system: processes executing in kernel mode
4idle: twiddling thumbs
5iowait: waiting for I/O to complete
6irq: servicing interrupts
7softirq: servicing softirqs
#the formulas from htop
user nice system idle iowait irq softirq steal guest guest_nice
cpu 74608 2520 24433 1117073 6176 4054 0 0 0 0
Idle=idle+iowait
NonIdle=user+nice+system+irq+softirq+steal
Total=Idle+NonIdle # first line of file for all cpus
CPU_Percentage=((Total-PrevTotal)-(Idle-PrevIdle))/(Total-PrevTotal)
'''
cpu_infos = {} #collect here the information
with open(self.cpustat,'r') as f_stat:
lines = [line.split(self.sep) for content in f_stat.readlines() for line in content.split('
') if line.startswith('cpu')]
#compute for every cpu
for cpu_line in lines:
if '' in cpu_line: cpu_line.remove('')#remove empty elements
cpu_line = [cpu_line[0]]+[float(i) for i in cpu_line[1:]]#type casting
cpu_id,user,nice,system,idle,iowait,irq,softrig,steal,guest,guest_nice = cpu_line
Idle=idle+iowait
NonIdle=user+nice+system+irq+softrig+steal
Total=Idle+NonIdle
#update dictionionary
cpu_infos.update({cpu_id:{'total':Total,'idle':Idle}})
return cpu_infos
def getcpuload(self):
'''
CPU_Percentage=((Total-PrevTotal)-(Idle-PrevIdle))/(Total-PrevTotal)
'''
start = self.getcputime()
#wait a second
sleep(self.sleeptime)
stop = self.getcputime()
cpu_load = {}
for cpu in start:
Total = stop[cpu]['total']
PrevTotal = start[cpu]['total']
Idle = stop[cpu]['idle']
PrevIdle = start[cpu]['idle']
CPU_Percentage=((Total-PrevTotal)-(Idle-PrevIdle))/(Total-PrevTotal)*100
cpu_load.update({cpu: CPU_Percentage})
return cpu_load
if __name__=='__main__':
x = GetCpuLoad()
while True:
try:
data = x.getcpuload()
print data
except KeyboardInterrupt:
sys.exit("Finished")
解决方案 4:
idnt.net对如何使用 /proc/stat cpu 数据进行了很好的描述,包括用于提取 cpu 的 bash 脚本和行描述。我只是想在这里链接它,因为我发现它很有价值。
解决方案 5:
我也在寻找同样的东西。这是我基于 Vangelis Tasoulas 的答案编写的 ruby 程序:
#!/usr/bin/env ruby
$VERBOSE = true
prev_file = IO.readlines(::File.join('', 'proc', 'stat')).select { |line| line.start_with?('cpu') }
Kernel.sleep(0.05)
file = IO.readlines(::File.join('', 'proc', 'stat')).select { |line| line.start_with?('cpu') }
file.size.times do |i|
data, prev_data = file[i].split.map(&:to_f), prev_file[i].split.map(&:to_f)
%w(user nice sys idle iowait irq softirq steal).each_with_index do |el, index|
eval "@#{el}, @prev_#{el} = #{data[index + 1]}, #{prev_data[index + 1]}"
end
previdle, idle = @prev_idle + @prev_iowait, @idle + @iowait
totald = idle + (@user + @nice + @sys + @irq + @softirq + @steal) -
(previdle + (@prev_user + @prev_nice + @prev_sys + @prev_irq + @prev_softirq + @prev_steal))
puts "CPU #{i}: #{((totald - (idle - previdle)) / totald * 100).round(2)} %"
end
解决方案 6:
以下是基于 Fidel 的回答和 arberg 的链接构建的 bash 脚本。
我想降低 cat awk grep 和 date 调用的使用率,并减少 CPU 使用率,以找出 CPU 使用率。
输出:
total: 4%
cpu0: 10%
cpu1: 5%
cpu2: 1%
cpu3: 1%
使用以下命令创建一个 bash 脚本:
#!/bin/bash
# Paramiter one used to set time in sec between reads
sleepDurationSeconds=$1
# read cpu stats to arrays
readarray -t previousStats < <( awk '/^cpu /{flag=1}/^intr/{flag=0}flag' /proc/stat )
sleep $sleepDurationSeconds
readarray -t currentStats < <( awk '/^cpu /{flag=1}/^intr/{flag=0}flag' /proc/stat )
# loop through the arrays
for i in "${!previousStats[@]}"; do
# Break up arrays 1 line sting into an array element for each item in string
previousStat_elemant_array=(${previousStats[i]})
currentStat_elemant_array=(${currentStats[i]})
# Get all columns from user to steal
previousStat_colums="${previousStat_elemant_array[@]:1:7}"
currentStat_colums="${currentStat_elemant_array[@]:1:7}"
# Replace the column seperator (space) with +
previous_cpu_sum=$((${previousStat_colums// /+}))
current_cpu_sum=$((${currentStat_colums// /+}))
# Get the delta between two reads
cpu_delta=$((current_cpu_sum - previous_cpu_sum))
# Get the idle time Delta
cpu_idle=$((currentStat_elemant_array[4]- previousStat_elemant_array[4]))
# Calc time spent working
cpu_used=$((cpu_delta - cpu_idle))
# Calc percentage
cpu_usage=$((100 * cpu_used / cpu_delta))
# Get cpu used for calc cpu percentage used
cpu_used_for_calc="${currentStat_elemant_array[0]}"
if [[ "$cpu_used_for_calc" == "cpu" ]]; then
echo "total: "$cpu_usage"%"
else
echo $cpu_used_for_calc": "$cpu_usage"%"
fi
done
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