Tha*_*uys 17 memory r sparse-matrix
我有一个50000字符串向量的列表,包含6000个独特字符串的各种组合.
目标:我想在"相对频率"(table(x)/length(x))中对它们进行转换,并将它们存储在稀疏矩阵中.低内存消耗比速度更重要.目前记忆是瓶颈.(即使源数据大约有~50 mb,目标格式的数据~10mb - >转换似乎效率低下,......)
生成样本数据
dims <- c(50000, 6000)
nms <- paste0("A", 1:dims[2])
lengths <- sample(5:30, dims[1], replace = T)
data <- lapply(lengths, sample, x = nms, replace = T)
可能的尝试:
1)sapply()简化为稀疏矩阵？
library(Matrix)
sparseRow <- function(stringVec){
relFreq <- c(table(factor(stringVec, levels = nms)) / length(stringVec))
Matrix(relFreq, 1, dims[2], sparse = TRUE)
}
sparseRows <- sapply(data[1:5], sparseRow)
sparseMat <- do.call(rbind, sparseRows)
问题:我的瓶颈似乎是sparseRows因为行没有直接组合成稀疏矩阵.(如果我在完整样本上运行上面的代码,我得到一个Error: cannot allocate vector of size 194 Kb
Error during wrapup: memory exhausted (limit reached?)- 我的硬件有8 GB RAM.)
显然,在组合行之前,有更多的内存消耗用于创建行列表,而不是直接填充稀疏矩阵. - >所以在我的情况下使用(s/l)apply对内存不友好？
object.size(sparseRows)
object.size(sparseMat)
2)肮脏的解决方法(？)
我的目标似乎是创建一个空的稀疏矩阵并按行填充.下面是一个脏的方法(它适用于我的硬件).
indxs <- lapply(data, function(data) sapply(data, function(x) which(x == nms),
USE.NAMES = FALSE))
relFreq <- lapply(indxs, function(idx) table(idx)/length(idx))
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]])
}
#sapply(1:dims[1], function(idx) mm[idx,
# as.numeric(names(relFreq[[idx]]))] <<- as.numeric(relFreq[[idx]]))
我想问一下,是否有一种更优雅/更有效的方法来实现最低RAM量.
min*_*nem 12
我会转换为data.table然后进行必要的计算:
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))))
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
让最快的函数基准测试3,因为其余的(OP,user20650_v1和F.Privé)会减慢所有数据.
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)))
}
比较的结果minem和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有其他名称.删除它们:
dimnames(x2) <- names(x2@x) <- NULL
all.equal(x1, x2)
# [1] TRUE # all equal
时序:
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
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
见专栏mem_alloc......
使用循环以列方式填充预先分配的稀疏矩阵(然后转置它):
res <- Matrix(0, dims[2], length(data), sparse = TRUE)
for (i in seq_along(data)) {
ind.match <- match(data[[i]], nms)
tab.match <- table(ind.match)
res[as.integer(names(tab.match)), i] <- as.vector(tab.match) / length(data[[i]])
}
# Verif
stopifnot(identical(t(res), sparseMat))
基准测试:
data2 <- data[1:50]
microbenchmark::microbenchmark(
OP = {
sparseMat <- do.call(rbind, sapply(data2, sparseRow))
},
ME = {
res <- Matrix(0, dims[2], length(data2), sparse = TRUE)
for (i in seq_along(data2)) {
ind.match <- match(data2[[i]], nms)
tab.match <- table(ind.match)
res[as.integer(names(tab.match)), i] <- as.vector(tab.match) / length(data2[[i]])
}
res2 <- t(res)
}
)
stopifnot(identical(res2, sparseMat))
Unit: milliseconds
expr min lq mean median uq max neval cld
OP 56.28020 59.61689 63.24816 61.16986 62.80294 206.18689 100 b
ME 46.60318 48.27268 49.77190 49.50714 50.92287 55.23727 100 a
所以,它的内存效率并不慢.