extremes: Calculate Quarter with Extremes for Temperature and...
In jjvanderwal/climates: Methods for working with weather & climate

Description Usage Arguments Details Value Author(s) See Also Examples

extremes calculates quarter (3 month or 13 week) in wich highest and lowest temperature and highest and lowest precipitation occurred given input of temperature max and min and precipitation. Calculated the same was as in bioclim.Rd.

1 2	extremes(tmin = NULL, tmax = NULL, prec = NULL, tmean = NULL, period = "month", tiebreak = "first")

`tmin`	a data.frame or matrix with 12 or 52 columns representing monthly or weekly minimum temperature data; rows represent different locations. (required)
`tmax`	a data.frame or matrix as with `tmin` representing maximum temperature data. (required)
`prec`	a data.frame or matrix as with `tmin` representing precipitation data. (required)
`tmean`	a data.frame or matrix as with `tmin` representing mean temperature data. (optional; will be calculated as (tmax+tmin)/2)
`period`	can be either "month" or "week" representing the temporal period for which values are calculated; see details for further description.
`tiebreak`	determines how calls to max.col will decide ties (multiple periods with same maximum). Options are "random", "first", "last" but there is no min.col and which.min only finds first min.

If "month" is specified for period, 12 columns of data are expected, if "week" is specified, 52 columns are expected.

a matrix with columns representing the number of the first month (or week) of the "Warmest","Coldest","Wettest","Driest" quarter. The number of rows (and order of them) is the same as the input tmin, tmax, prec or tmean.

Patrick D. Lorch plorch@kent.edu

bioclim.Rd

## Not run: 
# Need to fill in to match bioclim example

##---- Should be DIRECTLY executable !! ----
##-- ==>  Define data, use random,
##--	or do  help(data=index)  for the standard data sets.

## The function is currently defined as
function(tmin = NULL, tmax = NULL, prec = NULL, tmean = NULL, period = "month",tiebreak="first")
{
# 	This function finds the warmest, wettest, coldest, driest periods for each location

	tsize = NULL
	m.per.indx=function(x){c(x,(x:(x+1))%%12+1)} # index for 3 month quarter
	w.per.indx=function(x){c(x,(x:(x+11))%%52+1)} # index for 13 week quarter

	# Function to check for various input errors, stop program and print an error message
	error.check=function(datum,datum.name,dsize=tsize){
		if (is.null(datum))
			stop(paste(datum.name,"is needed for the variables selected"))
		else if (is.data.frame(datum) | is.matrix(tmin)) {
			if (!(dim(datum)[2] %in% c(12, 52))) # check for correct number of columns
				stop(paste(datum.name,"must have 12 or 52 columns --
  			one for each month or week"))
			dsize = c(dsize, dim(datum)[1])
		}
		else
			stop(paste(datum.name,"must be a data.frame or matrix"))
		return(dsize)
	}

	# Check for valid input data; tmin, tmax, prec are always needed
	tsize=error.check(tmin,"tmin")
	tsize=error.check(tmax,"tmax")
	tsize=error.check(prec,"prec")

	if (is.null(tmean)) {
		tmean=(tmax+tmin)/2
		print("Calculated tmean as (tmax+tmin)/2")
	}
	else
		tsize=error.check(tmean,"tmean")

	if (!all(tsize == mean(tsize))) # Check all input vars are the same length
		stop("all input data must be of the same length") # redundant?

	tsize=mean(tsize)
	out = matrix(NA, nr = tsize, nc = 4)  # Set up output matrix

	if (period == "month") { # Warmest/coldest and wettest/driest quarter sums and means
		tt1 = matrix(NA, nr = tsize, nc = 12)
		tt2 = matrix(NA, nr = tsize, nc = 12)
		for (ii in 1:12) {  # Find temperature means for 3 month quarters
			tt1[, ii] = rowMeans(tmean[, m.per.indx(ii)], na.rm = T)
			tt2[, ii] = rowSums(prec[, m.per.indx(ii)], na.rm = T)
		}
	}
	else {
		tt1 = matrix(NA, nr = tsize, nc = 52)
		tt2 = matrix(NA, nr = tsize, nc = 52)
		for (ii in 1:52) {  # Find temperature means for 13 week quarters
			tt1[, ii] = rowMeans(tmean[, w.per.indx(ii)], na.rm = T)
			tt2[, ii] = rowSums(prec[, w.per.indx(ii)], na.rm = T)
		}
	}
	# 1 Warmest quarter
	out[, 1] = max.col(tt1,tiebreak)
	# 2 Coldest quarter; there is no min.col!; this can only find first match
	out[, 2] = unname(apply(tt1, 1, which.min))
	# 3 Wettest quarter
	out[, 3] = max.col(tt2,tiebreak)
	# 4 Driest quarter; this can only find first match
	out[, 4] = unname(apply(tt2, 1, which.min))

#	out=data.frame(out)
	colnames(out) = c("Warmest","Coldest","Wettest","Driest")
	return(out)
  }

## End(Not run)