Hi John,
Just printing the result gives a good indication where the problem lies:
frm %% rowwise() %% do(MM=max(as.numeric(.)))
Source: local data frame [6 x 1]
Groups: by row
MM
1 dbl[1]
2 dbl[1]
3 dbl[1]
4 dbl[1]
5 dbl[1]
6 dbl[1]
do() is designed to produce scalars (e.g. a linear model), not
vectors, so it doesn't join the results back into a single vector. You
can either fix this yourself with unlist(), or use tidyr::unnest()
which will also handle vectors with length 1.
Hadley
On Mon, Feb 23, 2015 at 2:54 PM, John Posner john.pos...@mjbiostat.com wrote:
I'm using the dplyr package to perform one-row-at-a-time processing of a data
frame:
rnd6 = function() sample(1:300, 6)
frm = data.frame(AA=rnd6(), BB=rnd6(), CC=rnd6())
frm
AA BB CC
1 123 50 45
2 12 30 231
3 127 147 100
4 133 32 129
5 66 235 71
6 38 264 261
The interface is nice and straightforward:
library(dplyr)
dplyr_result = frm %% rowwise() %% do(MM=max(as.numeric(.)))
I've gotten used to the fact that dplyr_result is not a good old vanilla
data frame. The as.data.frame() function *seems* to do the trick:
dplyr_result_2 = as.data.frame(dplyr_result)
dplyr_result_2
MM
1 123
2 231
3 147
4 133
5 235
6 264
... but there's trouble ahead:
mean(dplyr_result_2$MM)
[1] NA
Warning message:
In mean.default(dplyr_result_2$MM) :
argument is not numeric or logical: returning NA
I need to enlist unlist() to get me to my destination:
mean(unlist(dplyr_result_2$MM))
[1] 188.8333
[NOTE: dplyr's as_data_frame() function does a better job than
as.data.frame() of indicating that I was headed for trouble. ]
By contrast, the plyr package's adply() function *does* produce a vanilla
data frame:
library(plyr)
plyr_result = adply(frm, .margins=1, function(onerowfrm)
max(as.numeric(onerowfrm[1,])))
mean(plyr_result$V1)
[1] 188.8333
Is there a good reason for dplyr to require the extra processing? My (naïve
?) recommendation would be to have as_data_frame() produce a vanilla data
frame.
Tx,
John
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