kp.pwcurv {kernplus} | R Documentation |
Predict Wind Power Output by Using a Multivariate Power Curve
Description
Takes multiple environmental variable inputs measured on an operating wind farm and predicts the wind power output under the given environmental condition.
Usage
kp.pwcurv(y, x, x.new = x, id.spd = 1, id.dir = NA)
Arguments
y |
An |
x |
An |
x.new |
A matrix or a data frame containing new input conditions of the
|
id.spd |
The column number of |
id.dir |
The column number of |
Value
A vector representing the predicted power output for the new
wind/weather condition specified in x.new
. If x.new
is not
supplied, this function returns the fitted power output for the given
x
.
Note
This function is developed for wind power prediction. As such, the response
y
represents wind power output and the covariatesx
include multiple wind and weather variables that potentially affect the power output.The data matrix
x
is expected to include at least wind speed and wind direction data. As measurements of other environmental variables become available, they can be added to thex
. Typically, the first column ofx
corresponds to wind speed data and the second column to wind direction data and, as such,id.spd = 1
andid.dir = 2
.If
x
has a single variable of wind speed, i.e.,p = 1
andid.spd = 1
, this function returns an estimate (or prediction) of the Nadaraya-Watson estimator with a Gaussian kernel by using theksmooth
function in the stats package.
References
Lee, G., Ding, Y., Genton, M.G., and Xie, L. (2015) Power Curve Estimation with Multivariate Environmental Factors for Inland and Offshore Wind Farms, Journal of the American Statistical Association 110(509):56-67.
See Also
Examples
head(windpw)
### Power curve estimation.
# By using a single input of wind speed.
pwcurv.est <- kp.pwcurv(windpw$y, windpw$V)
# By using wind speed and direction: id.dir needs to be set.
pwcurv.est <- kp.pwcurv(windpw$y, windpw[, c('V', 'D')], id.dir = 2)
# By using full covariates: confirm whether id.spd and id.dir are correctly specified.
pwcurv.est <- kp.pwcurv(windpw$y, windpw[, c('V', 'D', 'rho', 'I', 'Sb')], id.spd = 1, id.dir = 2)
### Wind power prediction.
# Suppose only 90% of data are available and use the rest 10% for prediction.
df.tr <- windpw[1:900, ]
df.ts <- windpw[901:1000, ]
id.cov <- c('V', 'D', 'rho', 'I', 'Sb')
pred <- kp.pwcurv(df.tr$y, df.tr[, id.cov], df.ts[, id.cov], id.dir = 2)
### Evaluation of wind power prediction based on 10-fold cross validation.
# Partition the given dataset into 10 folds.
index <- sample(1:nrow(windpw), nrow(windpw))
n.fold <- round(nrow(windpw) / 10)
ls.fold <- rep(list(c()), 10)
for(fold in 1:9) {
ls.fold[[fold]] <- index[((fold-1)*n.fold+1):(fold*n.fold)]
}
ls.fold[[10]] <- index[(9*n.fold+1):nrow(windpw)]
# Predict wind power output.
pred.res <- rep(list(c()), 10)
id.cov <- c('V', 'D', 'rho', 'I', 'Sb')
for(k in 1:10) {
id.fold <- ls.fold[[k]]
df.tr <- windpw[-id.fold, ]
df.ts <- windpw[id.fold, ]
pred <- kp.pwcurv(df.tr$y, df.tr[, id.cov], df.ts[, id.cov], id.dir = 2)
pred.res[[k]] <- list(obs = df.ts$y, pred)
}
# Calculate rmse and its mean and standard deviation.
rmse <- sapply(pred.res, function(res) with(res, sqrt(mean((obs - pred)^2))))
mean(rmse)
sd(rmse)