Vectorize my thinking: Vector Operations in R

问题:

So earlier I answered my own question on thinking in vectors in R. But now I have another problem which I can't 'vectorize.' I know vectors are faster and loops slower, but I can't figure out how to do this in a vector method:

I have a data frame (which for sentimental reasons I like to call my.data) which I want to do a full marginal analysis on. I need to remove certain elements one at a time and 'value' the data frame then I need to do the iterating again by removing only the next element. Then do again... and again... The idea is to do a full marginal analysis on a subset of my data. Anyhow, I can't conceive of how to do this in a vector efficient way.

I've shortened the looping part of the code down and it looks something like this:

for (j in my.data$item[my.data$fixed==0]) { # <-- selects the items I want to loop 
                                            #     through
    my.data.it <- my.data[my.data$item!= j,] # <-- this kicks item j out of the list
    sum.data <-aggregate(my.data.it, by=list(year), FUN=sum, na.rm=TRUE) #<-- do an
                                                                         # aggregation

    do(a.little.dance) && make(a.little.love) -> get.down(tonight) # <-- a little
                                                                   #  song and dance

    delta <- (get.love)                                         # <-- get some love
    delta.list<-append(delta.list, delta, after=length(delta.list)) #<-- put my love
                                                                    #    in a vector 
}

So obviously I hacked out a bunch of stuff in the middle, just to make it less clumsy. The goal would be to remove the j loop using something more vector efficient. Any ideas?

回答1:

Here's what seems like another very R-type way to generate the sums. Generate a vector that is as long as your input vector, containing nothing but the repeated sum of n elements. Then, subtract your original vector from the sums vector. The result: a vector (isums) where each entry is your original vector less the ith element.

> (my.data$item[my.data$fixed==0])
[1] 1 1 3 5 7
> sums <- rep(sum(my.data$item[my.data$fixed==0]),length(my.data$item[my.data$fixed==0]))
> sums
[1] 17 17 17 17 17
> isums <- sums - (my.data$item[my.data$fixed==0])
> isums
[1] 16 16 14 12 10

回答2:

Strangely enough, learning to vectorize in R is what helped me get used to basic functional programming. A basic technique would be to define your operations inside the loop as a function:

data = ...;
items = ...;

leave_one_out = function(i) {
   data1 = data[items != i];
   delta = ...;  # some operation on data1
   return delta;
}


for (j in items) {
   delta.list = cbind(delta.list, leave_one_out(j));
}

To vectorize, all you do is replace the for loop with the sapply mapping function:

delta.list = sapply(items, leave_one_out);

回答3:

This is no answer, but I wonder if any insight lies in this direction:

> tapply((my.data$item[my.data$fixed==0])[-1], my.data$year[my.data$fixed==0][-1], sum)

tapply produces a table of statistics (sums, in this case; the third argument) grouped by the parameter given as the second argument. For example

2001 2003 2005 2007
1    3    5    7

The [-1] notation drops observation (row) one from the selected rows. So, you could loop and use [-i] on each loop

for (i in 1:length(my.data$item)) {
  tapply((my.data$item[my.data$fixed==0])[-i], my.data$year[my.data$fixed==0][-i], sum)
}

keeping in mind that if you have any years with only 1 observation, then the tables returned by the successive tapply calls won't have the same number of columns. (i.e., if you drop out the only observation for 2001, then 2003, 2005, and 2007 would be te only columns returned).