根据每日最大值子集数据框

Question

I have an excel csv with a date/time column and a value associated with that date/time. I'm trying to write a script that will go through this format (see below), and find 1) the maximum value per day, and 2) the time on that day that the maximum occurs. Preferably R would return both values to me in a new dataframe.

The data looks something like this:

         V1    V2 V3
1  5/1/2012  3:00  1
2  5/1/2012  6:00  2
3  5/1/2012  9:00  5
4  5/1/2012 12:00  3
5  5/1/2012 15:00  6
6  5/1/2012 18:00  2
7  5/1/2012 21:00  1
8  5/2/2012  0:00  2
9  5/2/2012  3:00  3
10 5/2/2012  6:00  6
11 5/2/2012  9:00  4
12 5/2/2012 12:00  6
13 5/2/2012 15:00  7
14 5/2/2012 18:00  9
15 5/2/2012 21:00  1

So the function I'm envisioning would return:

1 5/1/2012 15:00 6
2 5/2/2012 18:00 9

Any ideas?

Answer 1

A solution using the plyr package, which I find very elegant for problems like this.

dat.str <- '         V1    V2 V3
1  5/1/2012  3:00  1
2  5/1/2012  6:00  2
3  5/1/2012  9:00  5
4  5/1/2012 12:00  3
5  5/1/2012 15:00  6
6  5/1/2012 18:00  2
7  5/1/2012 21:00  1
8  5/2/2012  0:00  2
9  5/2/2012  3:00  3
10 5/2/2012  6:00  6
11 5/2/2012  9:00  4
12 5/2/2012 12:00  6
13 5/2/2012 15:00  7
14 5/2/2012 18:00  9
15 5/2/2012 21:00  1'

dat <- read.table(textConnection(dat.str), row.names=1, header=TRUE)

library(plyr)
ddply(dat, .(V1), function(x){
   x[which.max(x$V3), ]
})

Answer 2

If you are dealing with time series data, I suggest you use a time series class like zoo or xts

dat <- read.table(text="         V1    V2 V3
1  5/1/2012  3:00  1
2  5/1/2012  6:00  2
3  5/1/2012  9:00  5
4  5/1/2012 12:00  3
5  5/1/2012 15:00  6
6  5/1/2012 18:00  2
7  5/1/2012 21:00  1
8  5/2/2012  0:00  2
9  5/2/2012  3:00  3
10 5/2/2012  6:00  6
11 5/2/2012  9:00  4
12 5/2/2012 12:00  6
13 5/2/2012 15:00  7
14 5/2/2012 18:00  9
15 5/2/2012 21:00  1", row.names=1, header=TRUE)

require("xts")
# create an xts object
xobj <- xts(dat[, 3], order.by=as.POSIXct(paste(dat[, 1], dat[, 2]), format="%m/%d/%Y %H:%M"))

If you just wanted to get the daily maximums, and you were okay with using the last time of the day as the index, you could use apply.daily

apply.daily(xobj, max)
#                    [,1]
#2012-05-01 21:00:00    6
#2012-05-02 21:00:00    9

To keep the timestamps at which it occurs, you could do this

do.call(rbind, lapply(split(xobj, "days"), function(x) x[which.max(x), ]))
#                    [,1]
2012-05-01 15:00:00    6
2012-05-02 18:00:00    9

split(xobj, "days") creates a list with one day's data in each element.

lapply applies a function to each day; the function, in this case, simply returns the max observation for each day. The lapply call will return a list of xts objects. To turn it back into a single xts object, use do.call .

do.call(rbind, X) constructs a call to rbind using each element of the list. It is equivalent to rbind(X[[1]], X[[2]], ..., X[[n]])

Answer 3

For another alternative, you could use data.table :

dat_table <- data.table(dat)

dat_table [ , list(is_max = V3==max(V3), V2, V3), by= 'V1'][which(is_max),][,is_max :=NULL]

EDIT as per @MattDowle's comment

dat_table[, .SD[which.max(V3)], by=V1]

For an even simpler data.table solution.

Answer 4

here you go:

dat.str <- '         V1    V2 V3
1  5/1/2012  3:00  1
2  5/1/2012  6:00  2
3  5/1/2012  9:00  5
4  5/1/2012 12:00  3
5  5/1/2012 15:00  6
6  5/1/2012 18:00  2
7  5/1/2012 21:00  1
8  5/2/2012  0:00  2
9  5/2/2012  3:00  3
10 5/2/2012  6:00  6
11 5/2/2012  9:00  4
12 5/2/2012 12:00  6
13 5/2/2012 15:00  7
14 5/2/2012 18:00  9
15 5/2/2012 21:00  1'

dat <- read.table(textConnection(dat.str), row.names=1, header=TRUE)

do.call(rbind, 
        by(dat, INDICES=dat$V1, FUN=function(x) tail(x[order(x$V3), ], 1)))