kernplus (version 0.1.2)

kp.pwcurv: 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 \(n\)-dimensional vector or a matrix of size \(n\) by 1 containing wind power output data. This along with x trains the multidimensional power curve model.

x

An \(n\) by \(p\) matrix or a data frame containing the input data for \(p\) predictor variables (wind and weather variables). This x must have the same number of rows as y, i.e., \(n\).

x.new

A matrix or a data frame containing new input conditions of the \(p\) predictor variables for which a prediction of wind power output will be made. This is an optional parameter and will be set to x by default, if it is not supplied.

id.spd

The column number of x (and of x.new, if supplied) indicating wind speed data . Default to 1.

id.dir

The column number of x (and of x.new, if supplied) indicating wind direction data. Default to NA, but this parameter needs to be set if x includes wind direction data.

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.

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

windpw, ksmooth

Examples

# NOT RUN {
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)
# }