Approximate sunrise and sunset times

This function is not perfect, but it does a reasonable job estimating sunrise and sunset times for my field site. If more accurate data are required, try here. Note, the command to calculate day of year is: strptime(x, "%m/%d/%Y")$yday+1

suncalc<-function(d,Lat=48.1442,Long=-122.7551){
  ## d is the day of year
  ## Lat is latitude in decimal degrees
  ## Long is longitude in decimal degrees (negative == West)
  
  ##This method is copied from:
  ##Teets, D.A. 2003. Predicting sunrise and sunset times.
  ##  The College Mathematics Journal 34(4):317-321.
 
  ## At the default location the estimates of sunrise and sunset are within
  ## seven minutes of the correct times (http://aa.usno.navy.mil/data/docs/RS_OneYear.php)
  ## with a mean of 2.4 minutes error.

  ## Function to convert degrees to radians
  rad<-function(x)pi*x/180
  
  ##Radius of the earth (km)
  R=6378
  
  ##Radians between the xy-plane and the ecliptic plane
  epsilon=rad(23.45)

  ##Convert observer's latitude to radians
  L=rad(Lat)

  ## Calculate offset of sunrise based on longitude (min)
  ## If Long is negative, then the mod represents degrees West of
  ## a standard time meridian, so timing of sunrise and sunset should
  ## be made later.
  timezone = -4*(abs(Long)%%15)*sign(Long)

  ## The earth's mean distance from the sun (km)
  r = 149598000

  theta = 2*pi/365.25*(d-80)

  z.s = r*sin(theta)*sin(epsilon)
  r.p = sqrt(r^2-z.s^2)

  t0 = 1440/(2*pi)*acos((R-z.s*sin(L))/(r.p*cos(L)))
  
  ##a kludge adjustment for the radius of the sun
  that = t0+5 

  ## Adjust "noon" for the fact that the earth's orbit is not circular:
  n = 720-10*sin(4*pi*(d-80)/365.25)+8*sin(2*pi*d/365.25)

  ## now sunrise and sunset are:
  sunrise = (n-that+timezone)/60
  sunset = (n+that+timezone)/60

  return(list("sunrise" = sunrise,"sunset" = sunset))
}

Convert polar coordinates to Cartesian

When I want to calculate the coordinates of a location (e.g., a nest or burrow) based on distance and bearing from a grid point, this function helps me avoid writing down SOH-CAH-TOA every time. Just note that the bearing in this case is from the grid point (known location) to the unknown location.

polar2cart<-function(x,y,dist,bearing,as.deg=FALSE){
  ## Translate Polar coordinates into Cartesian coordinates
  ## based on starting location, distance, and bearing
  ## as.deg indicates if the bearing is in degrees (T) or radians (F)
  
  if(as.deg){
    ##if bearing is in degrees, convert to radians
    bearing=bearing*pi/180
  }
  
  newx<-x+dist*sin(bearing)  ##X
  newy<-y+dist*cos(bearing)  ##Y
  return(list("x"=newx,"y"=newy))
}


##Example

oldloc=c(0,5) 
bearing=200 #degrees
dist=5

newloc<-polar2cart(oldloc[1],oldloc[2],dist,bearing,TRUE)
plot(oldloc[1],oldloc[2],xlim=c(-10,10),ylim=c(-10,10))
points(newloc$x,newloc$y,col="red")

Reorder factor levels

Very often, especially when plotting data, I need to reorder the levels of a factor because the default order is alphabetical. There must be many ways of reordering the levels; however, I always forget which package to look in. A direct way of reordering, using standard syntax is as follows:

## generate data
x = factor(sample(letters[1:5],100, replace=TRUE))

print(levels(x))  ## This will show the levels of x are "Levels: a b c d e"

## To reorder the levels:
## note, if x is not a factor use levels(factor(x))
x = factor(x,levels(x)[c(4,5,1:3)])

print(levels(x))  ## Now "Levels: d e a b c"

In this case, the level order could be set in the first line; however, if there are many levels (and you don't want to type out all of the levels explicitly), the above method is preferred. Again, if there is a better way to do this, please let me know in the comments.