我正在尝试使python脚本每分钟通过cpu /内存使用情况来打印前5个进程。但是,cpu结果在循环时似乎并没有改变。
当cpu循环时如何获得新的测量值?
我的代码在下面。
谢谢您的帮助!
addfiglab <- function(lab, xl = par()$mar[2], yl = par()$mar[3], cex = 1) {
text(x = line2user(xl, 2), y = line2user(yl, 3),
lab, xpd = NA, font = 2, cex = cex, adj = c(0, 1))
}
line2user <- function(line, side, outer = FALSE) {
unit <- if (outer) "nic" else "npc"
lh <- par('cin')[2] * par('cex') * par('lheight')
x_off <- diff(grconvertX(c(0, lh), 'inches', unit))
y_off <- diff(grconvertY(c(0, lh), 'inches', unit))
switch(side,
`1` = grconvertY(-line * y_off, unit, 'user'),
`2` = grconvertX(-line * x_off, unit, 'user'),
`3` = grconvertY(1 + line * y_off, unit, 'user'),
`4` = grconvertX(1 + line * x_off, unit, 'user'),
stop("Side must be 1, 2, 3, or 4", call.=FALSE))
}
layout(matrix(c(rep(1, 9), 2:10), ncol = 3, byrow = TRUE))
plot(1:10); addfiglab("A")
for (i in 2:10) {plot(i); addfiglab(LETTERS[i])}
答案 0 :(得分:0)
似乎psutil.process_iter就是答案。下面的代码有效。
import psutil
import time;
from functools import cmp_to_key
def log(line):
print(line)
with open("log.txt", "a") as f:
f.write("{}\n".format(line))
def cmpCpu(a, b):
a = a['cpu']
b = b['cpu']
if a > b:
return -1
elif a == b:
return 0
else:
return 1
def cmpMemory(a, b):
a = a['memory']
b = b['memory']
if a > b:
return -1
elif a == b:
return 0
else:
return 1
while True:
localtime = time.localtime(time.time())
timestamp = str(localtime.tm_hour)+":"+str(localtime.tm_min)
log(timestamp)
#Collect information for each process
processes = []
for proc in psutil.process_iter(attrs=['name', 'cpu_percent', 'memory_info']):
processes.append({'name': proc.info['name'], 'cpu': proc.info['cpu_percent'], 'memory': int(proc.info['memory_info'].rss/1024/1024)})
#Sort by cpu usage
log("CPU:")
processes.sort(key=cmp_to_key(cmpCpu))
for i in range(5):
info = processes[i]
info = info['name']+", "+str(info['cpu'])+"%"
log(info)
#Sort by memory usage
log("Memory:")
processes.sort(key=cmp_to_key(cmpMemory))
for i in range(5):
info = processes[i]
info = info['name']+", "+str(info['memory'])+"MB"
log(info)
time.sleep(60)