Description Usage Format Note Source Examples
Amount of daily rainfall in Melbourne, Australia, 1981-1990, measured in millimeters. The amounts are integers with many zeros and three days of more than 500mm rain.
1 |
A time-series of length 3653 with the amount of daily rainfall in mm.
Because of the two leap years 1984 and '88, we have constructed it
with ts(*, start=1981, frequency=365.25,
end = 1981+ (3653 - 1)/365.25)
.
There must be one extra observation since for the ten years with two leap years, there are only 3652 days. In 61 out of 100 days, there's no rain.
‘rainfall.dat’ in Rob J. Hyndman's Time Series Data Library, http://www-personal.buseco.monash.edu.au/~hyndman/TSDL/
originally, Australian Bureau of Meteorology, http://www.abs.gov.au.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 | data(OZrain)
(n <- length(OZrain)) ## should be 1 more than
ISOdate(1990,12,31) - ISOdate(1981, 1,1)## but it's 2 ..
has.rain <- OZrain > 0
summary(OZrain[has.rain])# Median = 18, Q3 = 50
table(rain01 <- as.integer(has.rain))
table(rain4c <- cut(OZrain, c(-.1, 0.5, 18.5, 50.1, 1000)))
AIC(v1 <- vlmc(rain01))# cutoff = 1.92
AIC(v00 <- vlmc(rain01, cut = 1.4))
AIC(v0 <- vlmc(rain01, cut = 1.5))
hist(OZrain)
hist(OZrain, breaks = c(0,1,5,10,50,1000), xlim = c(0,100))
plot(OZrain, main = "Rainfall 1981-1990 in Melbourne")
plot(OZrain, log="y", main = "Non-0 Rainfall [LOG scale]")
lOZ <- lowess(log10(OZrain[has.rain]), f= .05)
lines(time(OZrain)[has.rain], 10^lOZ$y, col = 2, lwd = 2)
|
[1] 3653
Time difference of 3651 days
Min. 1st Qu. Median Mean 3rd Qu. Max.
2.00 6.00 18.00 43.63 50.00 692.00
0 1
2237 1416
(-0.1,0.5] (0.5,18.5] (18.5,50.1] (50.1,1e+03]
2237 713 356 347
[1] 4547.764
[1] 4560.286
[1] 4554.815
Warning message:
In xy.coords(x, NULL, log = log, setLab = FALSE) :
2237 y values <= 0 omitted from logarithmic plot
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