CODEWITHSUNDEEP
rplyr
user1320502
user1320502

Reputation: 2570

represent all levels of grouping factor with ddply

tnt <- data.frame( g = rep(letters[1:2], each = 5), x = c(1:5,2:6), y = "things" ) 

library(plyr)
ddply(tnt, .(g,x), summarise, count = length(y) )

Considering the above code that is counting the number of rows for each x value in each g group:

What is the simplest way to have the level 6 represented in a and 1 represented in b? i.e. all levels of x represented in all levels of g.

My solution would be using as.data.frame(table(x)) after resetting the levels for each ddply chunk to 1:6 but that seems cumbersome.

(I am not tied to using plyr if the simplest way involves base)

EDIT

The by @rrs answer is great (+1), however, it is my fault that in trying to create a easy example I over simplified the problem. There are many grouping factors (not just g) and When I try .drop = FALSE I get the following error:

Error: length(rows) == 1 is not TRUE

I think this might be because of many grouping factors that nevor occur together and never should. How can I do the following:

c4 <- data.frame( g = rep(letters[1:2], each = 5),
                  f = c(sample(letters[24:25], 5, replace = TRUE),
                        sample(letters[25:26], 5, replace = TRUE)),
                  x = c(1:5,2:6), 
                  y = "things" )


ddply(c4, .(g,f,x), summarise, count = length(y), .drop = FALSE)

but the .drop = FALSE should only apply to x not g and f i.e. there should be no z in a.

Upvotes: 0

Views: 629

Answers (1)

rrs
rrs

Reputation: 9893

Use .drop = FALSE:

ddply(tnt, .(g,x), summarise, count = length(y), .drop = FALSE)

Upvotes: 2

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