如何在没有循环或mapply的R中的向量的不同间隔上使用相同的函数？

Question

Suppose I have a data frame such as

         Date    Value
1  2014-04-14   830.61
2  2014-04-11   815.69
3  2014-04-10   833.08
4  2014-04-09   872.18
5  2014-04-08   851.96
6  2014-04-07   845.04
7  2014-04-04   865.09
8  2014-04-03   888.77
9  2014-04-02   890.90
10 2014-04-01   885.52

Let's name it DF. And suppose I have defined min and max values of the index number.

minvals<-c(1,2,3)
maxvals<-c(5,7,10)

I want to process a function (ie mean value or standard deviation of Value column) for each interval. For example, take the mean of the first interval.

DF[minvals[1]:maxvals[1],"Value"]

         Date    Value
1  2014-04-14   830.61
2  2014-04-11   815.69
3  2014-04-10   833.08
4  2014-04-09   872.18
5  2014-04-08   851.96

mean(DF[minvals[1]:maxvals[1],"Value"])
#840.704

also for other minvals and maxvals. The first thing that comes to mind is mapply. But as my data has minvals and maxvals with thousands of this. Is it possible to do it in an efficient way?

ps In fact, it is quite similar to rolling mean but my date column include only workdays so I am not sure if rollmean function of zoo package can take care of this. Anyway suppose my time intervals are not regular also.

Answer 1

Try data.table

DFvec <- DF$Value
Ints <- data.frame(MIN = c(1,2,3), MAX = c(5,7,10))
library(data.table)
setDT(Ints)[, MEAN := mean(DFvec[MIN:MAX]), by = c("MIN", "MAX")]
Ints
##    MIN MAX     MEAN
## 1:   1   5 840.7040
## 2:   2   7 847.1733
## 3:   3  10 866.5675

---

Another way:

minvals = as.integer(minvals)
maxvals = as.integer(maxvals)
lenvals = maxvals - minvals + 1L
ix  = data.table:::vecseq(minvals, lenvals, sum(lenvals))
grp = rep(seq_along(lenvals), lenvals)

setDT(DF[ix, ])[, list(Value=mean(Value)), by=grp]
#    grp    Value
# 1:   1 840.7040
# 2:   2 847.1733
# 3:   3 866.5675

Answer 2

Here is the mapply solution. If that is too slow (give a reproducible example of you problem size), you could probably do something clever with data.table or use Rcpp.

x <- DF[["Value"]] #avoid data.frame subsetting in a loop
mapply(function(i1, i2) mean.default(x[i1:i2]), minvals, maxvals)

Benchmarks with 1e5 intervals:

library(microbenchmark)
set.seed(42)
i <- sample(1:3, 1e5, TRUE)
minvals<-c(1,2,3)[i]
maxvals<-c(5,7,10)[i]
microbenchmark(mapply(function(i1, i2) mean.default(x[i1:i2]), minvals, maxvals), times=10)

Unit: milliseconds
                                                             expr      min       lq   median       uq      max neval
mapply(function(i1, i2) mean.default(x[i1:i2]), minvals, maxvals) 446.0529 473.4267 489.2375 523.2335 595.5536    10

Answer 3

Here are a several approaches. Its not clear from the description that efficiency is really important here and readability might be more important:

# they all use this:
DF.Value <- DF$Value

# 1
sapply(seq_along(minvals), function(i) mean(DF.Value[minvals[i]:maxvals[i]]))

# 2
f <- function(minvals, maxvals) mean(DF.Value[minvals:maxvals])
mapply(f, minvals, maxvals)

# 3 - this one assumes that minvals equals seq_along(minvals) which is true in example
library(zoo)
w <- maxvals - minvals + 1
rollapply(DF.Value, w, mean, align = "left")[minvals]