Hello R Community,
I don't post to these things often so excuse me if I stumble on my forum
etiquette. This is a complex problem for me, which may require two forum
entries, but I will try my best to be concise. Also, I am a self taught
coder, so if my code is not to convention, your constructive criticism is
always welcome.
I need to split up a data frame by participant (gpsarc - factor), status
(testpo - factor), and by date (LocalDate), then sum the distance
(Dist_Bef_m - numeric) records for each date and average them across each
status state for each participant. Each participant has several records
for each date and for at least 1 of 3 different possible status types (max
3). In the end, I want a table with participant number and the status
state as my column headings with means for each status state under their
appropriate heading (see example below). I am confident I made this way
more complicated than it needs to be. I really appreciate any help you can
offer.
Here is my relevant coding so far:
s1 <- split(data[,c(4,10,20,42)], data$gpsarc)
for(i in 1:length(s1))
s1[[i]] <- split(s1[[i]],s1[[i]]$testpo)
s2 <- vector("list", length(s1))
for(i in 1:length(s2))
s2[[i]] <- ldply(s1[[i]],
function(x)
{
if(nrow(x) == 0) #### if one status state does not exist, but still
accounted in the split sublist because its a factor, I would get an error,
so I added this If/Else portion to remove those entries with no records
{
remove(x)
}
else
{
by(x, x[["LocalDate"]],
function(x1)
{
sum(x1[["Dist_Bef_m"]])
})
}
})
s3 <- vector("list", length(s2))
for(i in 1:length(s3))
s3[[i]] <- data.frame(mean = apply(s2[[i]][,-1],1,mean,na.rm=TRUE),
row.names = as.character(s2[[i]][,1]))
here is a sample of the s3 result:
[[1]]
mean
2 12533.2
[[2]]
mean
2 26300.96
3 25313.93
[[3]]
mean
1 48489.15
3 27398.23
[[4]]
mean
1 34783.97
[[5]]
mean
1 21293.19
2 21962.41
3 18272.67
##### I really want it to look like this:
ppt 1 2 3
1 NA 12533.2 NA
2 NA 26300.96 25313.93
3 48489.15 NA 27398.23
4 34783.97 NA NA
5 21293.19 21962.41 18272.67
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