rrd 数据库，求平均值

[freeskyo]

有做过rrd数据库的吗

问一个问题

求AVEARAGE 的PDP时，我知道，当数据递增时是如何计算出来的 (后面的值 - 前面的值）/ 步长(秒)

如: (3-2)/300 = 3.333333e+03 (科学计数法呀)

那么递减呢


如: (2-3)/300 = ? (科学计数法呀)

[healer_kx]

不懂~

[freeskyo]

见过: 下面的吗

1345540140: 1.1848083667e+06 3.9428056667e+05
1345540200: 1.2186805833e+06 4.0250340000e+05
1345540260: 1.1743186000e+06 3.8986008333e+05
1345540320: 1.0629253333e+06 3.5841306667e+05
1345540380: 1.1375081500e+06 3.8327231667e+05
1345540440: 1.1879058000e+06 3.9307803333e+05
1345540500: 1.2410891667e+06 4.1273376667e+05
1345540560: 1.1520188333e+06 3.8479356667e+05
1345540620: 1.0901494500e+06 3.6950656667e+05
1345540680: 1.1499634333e+06 3.9492798333e+05
1345540740: 1.1826454333e+06 3.9661748333e+05
1345540800: 1.1662577833e+06 4.0059935000e+05
1345540860: 1.0425469500e+06 3.4692160000e+05
1345540920: 1.1382243333e+06 3.8745860000e+05
1345540980: 9.5189188333e+05 3.1793993333e+05
1345541040: 9.6472183333e+05 3.1489348333e+05
1345541100: 1.0926876667e+06 3.6478835000e+05
1345541160: 1.0549178167e+06 3.5396311667e+05
1345541220: 9.8150300000e+05 3.2347238333e+05
1345541280: 1.0004072167e+06 3.3736493333e+05
1345541340: 1.0424256333e+06 3.5190113333e+05
1345541400: 9.1698593333e+05 3.0749868333e+05
1345541460: 9.8497828333e+05 3.2352073333e+05
1345541520: 8.9819545000e+05 3.0560705000e+05
1345541580: 9.0673400000e+05 3.0531975000e+05
1345541640: 9.7121075000e+05 3.2026621667e+05

[freeskyo]

rrdtool update 1.rrd 1345541400:7000:1000
rrdtool update 1.rrd 1345541700:8000:1000
rrdtool update 1.rrd 1345542000:9000:1000
rrdtool update 1.rrd 1345542300:8000:1000

rrdtool fetch 1.rrd AVERAGE
1345541400: nan nan
1345541700: 3.3333333333e+00 0.0000000000e+00
1345542000: 3.3333333333e+00 0.0000000000e+00
1345542300: 1.4316554320e+07 0.0000000000e+00

递增好说 (9000-8000)/300 = 3.3333333333e+00
但递减是如何计算出 1.4316554320e+07 的呀