Suppose I have the following frequency table.
> print(dat)
V1 V2
1 1 11613
2 2 6517
3 3 2442
4 4 687
5 5 159
6 6 29
# V1 = Score
# V2 = Frequency
How can I efficiently compute the Mean and standard deviation? Yielding: SD=0.87 MEAN=1.66. Replicating the score by frequency takes too long to compute.
I might be missing something, but this seems to work very quickly, even substituting millions in the frequency column:
dset <- data.frame(V1=1:6,V2=c(11613,6517,2442,687,159,29))
mean(rep(dset$V1,dset$V2))
#[1] 1.664102
sd(rep(dset$V1,dset$V2))
#[1] 0.8712242
Mean is easy. SD is a little trickier (can't just use fastmean() again because there's an n-1 in the denominator.
> dat <- data.frame(freq=seq(6),value=runif(6)*100)
> fastmean <- function(dat) {
+ with(dat, sum(freq*value)/sum(freq) )
+ }
> fastmean(dat)
[1] 55.78302
>
> fastRMSE <- function(dat) {
+ mu <- fastmean(dat)
+ with(dat, sqrt(sum(freq*(value-mu)^2)/(sum(freq)-1) ) )
+ }
> fastRMSE(dat)
[1] 34.9316
>
> # To test
> expanded <- with(dat, rep(value,freq) )
> mean(expanded)
[1] 55.78302
> sd(expanded)
[1] 34.9316
Note that fastRMSE calculates sum(freq) twice. Eliminating this would probably result in another minor speed boost.
Benchmarking
> microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 12.433 13.5335 14.776 15.398 23.921
2 mean(with(dat, rep(value, freq))) 21.225 22.3990 22.714 23.406 86.434
> dat <- data.frame(freq=seq(60),value=runif(60)*100)
>
> dat <- data.frame(freq=seq(60),value=runif(60)*100)
> microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 13.177 14.544 15.8860 17.2905 54.983
2 mean(with(dat, rep(value, freq))) 42.610 48.659 49.8615 50.6385 151.053
> dat <- data.frame(freq=seq(600),value=runif(600)*100)
> microbenchmark(
+ fastmean(dat),
+ mean( with(dat, rep(value,freq) ) )
+ )
Unit: microseconds
expr min lq median uq max
1 fastmean(dat) 15.706 17.489 25.8825 29.615 79.113
2 mean(with(dat, rep(value, freq))) 1827.146 2283.551 2534.7210 2884.933 26196.923
The replicating solution appears to be O( N^2 ) in the number of entries.
The fastmean solution appears to have a 12ms or so fixed cost after which it scales beautifully.
More benchmarking
Comparison with dot product.
dat <- data.frame(freq=seq(600),value=runif(600)*100)
dbaupp <- function(dat) {
total.count <- sum(dat$freq)
as.vector(dat$freq %*% dat$value) / total.count
}
microbenchmark(
fastmean(dat),
mean( with(dat, rep(value,freq) ) ),
dbaupp(dat)
)
Unit: microseconds
expr min lq median uq max
1 dbaupp(dat) 20.162 21.6875 25.6010 31.3475 104.054
2 fastmean(dat) 14.680 16.7885 20.7490 25.1765 94.423
3 mean(with(dat, rep(value, freq))) 489.434 503.6310 514.3525 583.2790 30130.302
How about:
> m = sum(dat$V1 * dat$V2) / sum(dat$V2)
> m
[1] 1.664102
> sigma = sqrt(sum((dat$V1 - m)**2 * dat$V2) / (sum(dat$V2)-1))
> sigma
[1] 0.8712242
No replication here.
The following code doesn't use replication, and it uses R builtins (for the dot product especially) as much as possible so it is probably more efficient that solutions that use sum(V1 * V2). (Edit: this is possibly false: @gsk3's solution seems to be about 1.5 - 2 times faster from my testing.)
The definition of mean (or expectation) is sum(n * freq(n)) / total.count where n is the "score" and freq(n) is the frequency of n (total.count is just sum(freq(n))). The sum in the numerator is precisely the dot product of the scores with the frequencies.
The dot product in R is %*% (it returns a matrix, but this can basically be treated at a vector for most purposes):
> total.count <- sum(dat$V2)
> mean <- dat$V1 %*% dat$V2 / total.count
> mean
[,1]
[1,] 1.664102
There is a formula at the end of this section of the Wikipedia article, which translates to the following code
> sqrt(dat$V1^2 %*% dat$V2 / total.count - mean^2)
[,1]
[1,] 0.8712039
