Question

我尝试使用data.table包解决以下问题： Is there a faster way to subset a sparse Matrix than '['?

但是我得到了这个错误：

    Error in Z[, cols] : invalid or not-yet-implemented 'Matrix' subsetting 
10 stop("invalid or not-yet-implemented 'Matrix' subsetting") 
9 Z[, cols] 
8 Z[, cols] 
7 FUN(X[[i]], ...) 
6 lapply(X = ans[index], FUN = FUN, ...) 
5 tapply(.SD, INDEX = "gene_name", FUN = simple_fun, Z = Z, simplify = FALSE) 
4 eval(expr, envir, enclos) 
3 eval(jsub, SDenv, parent.frame()) 
2 `[.data.table`(lkupdt, , tapply(.SD, INDEX = "gene_name", FUN = simple_fun, 
Z = Z, simplify = FALSE), .SDcols = c("snps")) 
1 lkupdt[, tapply(.SD, INDEX = "gene_name", FUN = simple_fun, Z = Z, 
simplify = FALSE), .SDcols = c("snps")]

这是我的解决方案：

library(data.table)
library(Matrix)

seed(1)

n_subjects <- 1e3
n_snps <- 1e5
sparcity <- 0.05


n <- floor(n_subjects*n_snps*sparcity) 

# create our simulated data matrix
Z <- Matrix(0, nrow = n_subjects, ncol = n_snps, sparse = TRUE)
pos <- sample(1:(n_subjects*n_snps), size = n, replace = FALSE)
vals <- rnorm(n)
Z[pos] <- vals

# create the data frame on how to split
# real data set the grouping size is between 1 and ~1500
n_splits <- 500
sizes <- sample(2:20, size = n_splits, replace = TRUE)  
lkup <- data.frame(gene_name=rep(paste0("g", 1:n_splits), times = sizes),
                   snps = sample(n_snps, size = sum(sizes)))

# simple function that gets called on the split
# the real function creates a cols x cols dense upper triangular matrix
# similar to a covariance matrix
simple_fun <- function(Z, cols) {sum(Z[ , cols])}

# split our matrix based look up table
system.time(
  res <- tapply(lkup[ , "snps"], lkup[ , "gene_name"], FUN=simple_fun, Z=Z, simplify = FALSE)
)
lkupdt <- data.table(lkup)
lkupdt[, tapply(.SD, INDEX = 'gene_name' , FUN = simple_fun, Z = Z, simplify = FALSE), .SDcols = c('snps')]

问题是关于尝试将上面保存的函数复制到“res”的最后一行代码。我在使用data.table做错了什么，或者这根本不可能？谢谢你的帮助！

Answer 1

不，我不认为您可以使用data.table加快访问Matrix对象的速度。但是，如果您愿意使用data.table而不是Matrix ...

addEntry

它给出的结果如

ZDT = setDT(summary(Z))
system.time(
  resDT <- ZDT[lkupdt, on = c(j = "snps")][, sum(x), by=gene_name]
)

# verify correctness
all.equal(
  unname(unlist(res))[order(as.numeric(substring(names(res), 2, nchar(names(res)))))],
  resDT$V1
)

当然，出于其他原因，您可能需要将数据保存在稀疏矩阵中，但这在我的计算机上要快得多，并且输入和输出更简单。

Answer 2

我认为sum()过于简单无法估计时间，当您展示更真实的function时，您会得到更合适的答案。（我没有data.table()）

例如，此function看起来等于或快于data.table()方法（当然，此方法不能用于复杂function）;

sum.func <- function(Z, lkup) {
  Zsum <- colSums(Z)[lkup$snps]
  Z2 <- cbind(Zsum, lkup$gene_name)
  res <- c(tapply(Z2[,1], Z2[,2], sum))
  names(res) <- levels(lkup$gene_name)
  return(c(res))
}

system.time(
  test.res <- sum.func(Z, lkup)
)

all.equal(unlist(res), test.res)

这更为一般但明显慢于data.table()方法。

general.fun <- function(Z, lkup) {
  Z2 <- Z[, lkup$snps]
  num.gn <- as.numeric(lkup$gene_name)
  res <- sapply(1:max(num.gn), function(x) sum(Z2[, which(num.gn == x)]))
  names(res) <- levels(lkup$gene_name)
  return(res)
}

system.time(
  test.res2 <- general.fun(Z, lkup)
)

all.equal(unlist(res), test.res2)

使用R中的data.table子集稀疏矩阵

2 个答案: