Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

How to ddply() without sorting?

Tags:

sorting

r

plyr

I use the following code to summarize my data, grouped by Compound, Replicate and Mass.

summaryDataFrame <- ddply(reviewDataFrame, .(Compound, Replicate, Mass), 
  .fun = calculate_T60_Over_T0_Ratio)

An unfortunate side effect is that the resulting data frame is sorted by those fields. I would like to do this and keep Compound, Replicate and Mass in the same order as in the original data frame. Any ideas? I tried adding a "Sorting" column of sequential integers to the original data, but of course I can't include that in the .variables since I don't want to 'group by' that, and so it is not returned in the summaryDataFrame.

Thanks for the help.

like image 721
James Avatar asked Aug 29 '11 20:08

James


People also ask

How does ddply work?

Today we will emphasize ddply() which accepts a data. frame, splits it into pieces based on one or more factors, computes on the pieces, then returns the results as a data.

What is the Ddply function in R?

ddply: Split data frame, apply function, and return results in a data frame.


2 Answers

This came up on the plyr mailing list a while back (raised by @kohske no less) and this is a solution offered by Peter Meilstrup for limited cases:

#Peter's version used a function gensym to
# create the col name, but I couldn't track down
# what package it was in.
keeping.order <- function(data, fn, ...) { 
  col <- ".sortColumn"
  data[,col] <- 1:nrow(data) 
  out <- fn(data, ...) 
  if (!col %in% colnames(out)) stop("Ordering column not preserved by function") 
  out <- out[order(out[,col]),] 
  out[,col] <- NULL 
  out 
} 

#Some sample data 
d <- structure(list(g = c(2L, 2L, 1L, 1L, 2L, 2L), v = c(-1.90127112738315, 
-1.20862680183042, -1.13913266070505, 0.14899803094742, -0.69427656843677, 
0.872558638137971)), .Names = c("g", "v"), row.names = c(NA, 
-6L), class = "data.frame") 

#This one resorts
ddply(d, .(g), mutate, v=scale(v)) #does not preserve order of d 

#This one does not
keeping.order(d, ddply, .(g), mutate, v=scale(v)) #preserves order of d 

Please do read the thread for Hadley's notes about why this functionality may not be general enough to roll into ddply, particularly as it probably applies in your case as you are likely returning fewer rows with each piece.

Edited to include a strategy for more general cases

If ddply is outputting something that is sorted in an order you do not like you basically have two options: specify the desired ordering on the splitting variables beforehand using ordered factors, or manually sort the output after the fact.

For instance, consider the following data:

d <- data.frame(x1 = rep(letters[1:3],each = 5), 
                x2 = rep(letters[4:6],5),
                x3 = 1:15,stringsAsFactors = FALSE)

using strings, for now. ddply will sort the output, which in this case will entail the default lexical ordering:

> ddply(d,.(x1,x2),summarise, val = sum(x3))
  x1 x2 val
1  a  d   5
2  a  e   7
3  a  f   3
4  b  d  17
5  b  e   8
6  b  f  15
7  c  d  13
8  c  e  25
9  c  f  27


> ddply(d[sample(1:15,15),],.(x1,x2),summarise, val = sum(x3))
  x1 x2 val
1  a  d   5
2  a  e   7
3  a  f   3
4  b  d  17
5  b  e   8
6  b  f  15
7  c  d  13
8  c  e  25
9  c  f  27

If the resulting data frame isn't ending up in the "right" order, it's probably because you really want some of those variables to be ordered factors. Suppose that we really wanted x1 and x2 ordered like so:

d$x1 <- factor(d$x1, levels = c('b','a','c'),ordered = TRUE)
d$x2 <- factor(d$x2, levels = c('d','f','e'), ordered = TRUE)

Now when we use ddply, the resulting sort will be as we intend:

> ddply(d,.(x1,x2),summarise, val = sum(x3))
  x1 x2 val
1  b  d  17
2  b  f  15
3  b  e   8
4  a  d   5
5  a  f   3
6  a  e   7
7  c  d  13
8  c  f  27
9  c  e  25

The moral of the story here is that if ddply is outputting something in an order you didn't intend, it's a good sign that you should be using ordered factors for the variables you're splitting on.

like image 181
joran Avatar answered Sep 20 '22 02:09

joran


I eventually ended up adding an 'indexing' column to the original data frame. It consisted of two columns pasted with sep="_". Then I made another data frame made of only unique members of the 'indexing' column and a counter 1:length(df). I did my ddply() on the data which returned a sorted data frame. Then to get things back in the original order I did merge() the results data frame and the index data frame (making sure the columns are named the same thing makes this easier). Finally, I did order and removed the extraneous columns.

Not an elegant solution, but one that works.

Thanks for the assist. It got me thinking in the right direction.

like image 29
James Avatar answered Sep 21 '22 02:09

James