#--------------------------------------------------
# Replace all instances of value x in a data frame
#	with value y
#--------------------------------------------------


# The basic approach is to apply a replacement function
#	to each column of the data frame.  There are a
#	couple of different variations on this theme.

# Setup some data

df <- data.frame(x=rnorm(25), y=runif(25, min=-2, max=2))

#--------------------------------------------------
# Version 1, write a function and lapply it.
#     The replace() function nearly does what we want.
#	Then lapply() repeats the modified replace() for
#	each column. lapply() returns a list, so we
#	convert it back to a data frame. 
#--------------------------------------------------

replace.neg <- function(z) {replace(z, z<0, NA)}

df.na <- as.data.frame(lapply(df, replace.neg))
data.frame(df,df.na)

#--------------------------------------------------
# Version 2, use an anonymous function
#	Harder to read, so its harder to debug,
#	but it leaves fewer objects in the workspace.
#--------------------------------------------------

df.na2 <- as.data.frame(lapply(df, function(z){replace(z, z<0, NA)}))
data.frame(df,df.na2)

#--------------------------------------------------
# Version 3, replace using extractors, see ?replace
#	Just to illustrate how you write a function
#	using "[]", which is what replace() does.
#--------------------------------------------------

replace.na <- function(z) {
	z[is.na(z)]<-0
	z     
	# the last expression evaluated within a function
	# is what gets returned
	}

df.zero <- as.data.frame(lapply(df.na, replace.na))
data.frame(df.na, df.zero)

#--------------------------------------------------
# Version 4, use an anonymous function again
#--------------------------------------------------

df.zero2 <- as.data.frame(lapply(df.na, function(z){z[is.na(z)]<-0; z}))
data.frame(df.na, df.zero2)