Ответ 1
Я бы преобразовал в data.table а затем data.table необходимые вычисления:
ld <- lengths(data)
D <- data.table(val = unlist(data),
id = rep(1:length(data), times = ld),
Ntotal = rep(ld, times = ld))
D <- D[, .N, keyby = .(id, val, Ntotal)]
D[, freq := N/Ntotal]
ii <- data.table(val = nms, ind = seq_along(nms))
D <- ii[D, on = 'val']
sp <- with(D, sparseMatrix(i = id, j = ind, x = freq,
dims = c(max(id), length(nms))))
Контрольные показатели для n = 100
data2 <- data[1:100]
Unit: milliseconds
expr min lq mean median uq max neval cld
OP 102.150200 106.235148 113.117848 109.98310 116.79734 142.859832 10 b
F. Privé 122.314496 123.804442 149.999595 126.76936 164.97166 233.034447 10 c
minem 5.617658 5.827209 6.307891 6.10946 6.15137 9.199257 10 a
user20650 11.012509 11.752350 13.580099 12.59034 14.31870 21.961725 10 a
Контрольные показатели по всем данным
Позволяет сравнить 3 самых быстрых функций, потому что остальные из них (OP, user20650_v1 и F.Privé) будут замедлять работу по всем данным.
user20650_v2 <- function(x) {
dt2 = data.table(lst = rep(1:length(x), lengths(x)),
V1 = unlist(x))
dt2[, V1 := factor(V1, levels = nms)]
x3 = xtabs(~ lst + V1, data = dt2, sparse = TRUE)
x3/rowSums(x3)
}
user20650_v3 <- function(x) {
x3 = xtabs(~ rep(1:length(x), lengths(x)) + factor(unlist(x), levels = nms),
sparse = TRUE)
x3/rowSums(x3)
}
minem <- function(x) {
ld <- lengths(x)
D <- data.table(val = unlist(x), id = rep(1:length(x), times = ld),
Ntotal = rep(ld, times = ld))
D <- D[, .N, keyby = .(id, val, Ntotal)]
D[, freq := N/Ntotal]
ii <- data.table(val = nms, ind = seq_along(nms))
D <- ii[D, on = 'val']
sparseMatrix(i = D$id, j = D$ind, x = D$freq,
dims = c(max(D$id), length(nms)))
}
Сравните результаты minem и user20650_v3:
x1 <- minem(data)
x2 <- user20650_v3(data)
all.equal(x1, x2)
# [1] "Component "Dimnames": names for current but not for target"
# [2] "Component "Dimnames": Component 1: target is NULL, current is character"
# [3] "Component "Dimnames": Component 2: target is NULL, current is character"
# [4] "names for target but not for current"
x2 имеет дополнительные имена. удалить их:
dimnames(x2) <- names([email protected]) <- NULL
all.equal(x1, x2)
# [1] TRUE # all equal
Тайминги:
x <- bench::mark(minem(data),
user20650_v2(data),
user20650_v3(data),
iterations = 5, check = F)
as.data.table(x)[, 1:10]
# expression min mean median max itr/sec mem_alloc n_gc n_itr total_time
# 1: minem(data) 324ms 345ms 352ms 371ms 2.896187 141MB 7 5 1.73s
# 2: user20650_v2(data) 604ms 648ms 624ms 759ms 1.544380 222MB 10 5 3.24s
# 3: user20650_v3(data) 587ms 607ms 605ms 633ms 1.646977 209MB 10 5 3.04s
связанная память:
OPdirty <- function(x) {
indxs <- lapply(x, function(x) sapply(x, function(x) which(x == nms),
USE.NAMES = FALSE))
relFreq <- lapply(indxs, function(idx) table(idx)/length(idx))
dims <- c(length(indxs), length(nms))
mm <- Matrix(0, nrow = dims[1], ncol = dims[2])
for (idx in 1:dims[1]) {
mm[idx, as.numeric(names(relFreq[[idx]]))] <- as.numeric(relFreq[[idx]])
}
mm
}
xx <- data[1:1000]
all.equal(OPdirty(xx), minem(xx))
# true
x <- bench::mark(minem(xx),
FPrive(xx),
OPdirty(xx),
iterations = 3, check = T)
as.data.table(x)[, 1:10]
expression min mean median max itr/sec mem_alloc n_gc n_itr total_time
1: minem(xx) 12.69ms 14.11ms 12.71ms 16.93ms 70.8788647 3.04MB 0 3 42.33ms
2: FPrive(xx) 1.46s 1.48s 1.47s 1.52s 0.6740317 214.95MB 4 3 4.45s
3: OPdirty(xx) 2.12s 2.14s 2.15s 2.16s 0.4666106 914.91MB 9 3 6.43s
См. mem_alloc...