R help - Jan 2014 - How to "apply" correctly.

Folks,

# I have the following function:

breakByFreq<-function(freq, defData) {
  breakUpFun<-function(freq, defs) {
    if(freq != 1) {
      defs<-diff(c(0, defs))
      defs<-cumsum(rep(defs/freq, each = freq))
    }
    defs
  }
  defMat<-sapply(defData[,-1], function(x) breakUpFun(freq, x))
  data.frame(Year = 1:nrow(defMat), defMat)
}

# And this data (year column and then 2 columns of data):

dum<-structure(list(Year = 1:10, 
            c1 = c(0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 0.1), 
            c2 = c(0.02, 0.04, 0.06, 0.08, 0.1, 0.12, 0.14, 0.16, 0.18, 0.2)), 
.Names = c("Year", "c1", "c2"), row.names = c(NA,
-10L), class = "data.frame")

# This works:
breakByFreq(1, dum)

# This doesn't:
breakByFreq(1, dum[,-3])


# How do I use and choose the appropriate apply function to make this work when
there is one and or more columns to be "processed".

Thanks for your time,
KW



--

I think you will be okay if you change
one line to:

defMat<-sapply(defData[,-1, drop=FALSE], function(x) breakUpFun(freq, x))

In your example that doesn't work you are
ending up with a vector rather than a one
column data frame.

Pat


On 07/01/2014 17:44, Keith S Weintraub wrote:
> Folks,
>
> # I have the following function:
>
> breakByFreq<-function(freq, defData) {
>    breakUpFun<-function(freq, defs) {
>      if(freq != 1) {
>        defs<-diff(c(0, defs))
>        defs<-cumsum(rep(defs/freq, each = freq))
>      }
>      defs
>    }
>    defMat<-sapply(defData[,-1], function(x) breakUpFun(freq, x))
>    data.frame(Year = 1:nrow(defMat), defMat)
> }
>
> # And this data (year column and then 2 columns of data):
>
> dum<-structure(list(Year = 1:10,
>              c1 = c(0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09,
0.1),
>              c2 = c(0.02, 0.04, 0.06, 0.08, 0.1, 0.12, 0.14, 0.16, 0.18,
0.2)),
> .Names = c("Year", "c1", "c2"), row.names =
c(NA, -10L), class = "data.frame")
>
> # This works:
> breakByFreq(1, dum)
>
> # This doesn't:
> breakByFreq(1, dum[,-3])
>
>
> # How do I use and choose the appropriate apply function to make this work
when there is one and or more columns to be "processed".
>
> Thanks for your time,
> KW
>
>
>
> --
>
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-- 
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