子集的稀疏矩阵data.table R中

我试图解决与data.table包以下问题： Is there a faster way to subset a sparse Matrix than '['?子集的稀疏矩阵data.table R中

，但我得到了这个错误：

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")]

这里我的解决方案：

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')]

问题是关于试图复制上面保存为“res”的函数的代码的最后一行。我是否在data.table中做了错误，或者这是不可能的？谢谢你的帮助！

来源

2016-09-24 Daniel Winkler

不，我不认为你可以加快使用data.table访问矩阵对象。但是，如果你愿意使用data.table代替矩阵...

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 
)

它给像

 gene_name   V1 
    1:  g1 3.720619 
    2:  g2 35.727923 
    3:  g3 -3.949385 
    4:  g4 -18.253456 
    5:  g5 5.970879 
---      
496:  g496 -20.979669 
497:  g497 63.880925 
498:  g498 16.498587 
499:  g499 -17.417110 
500:  g500 45.169608

当然的结果，您可能需要保持在一个数据稀疏矩阵出于其他原因，但这是我的电脑上快得多，并具有更简单的输入和输出。

来源

2016-09-24 01:54:53 Frank

我认为sum()太简单了，无法估计时间，当您显示更真实的function时，您会得到更合适的答案。（I走近不data.table()）

例如，这function看上去比data.table()方法相等或更快的（当然，这种方法不能与复杂function使用）;

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)

这是比data.table()方法更通用但明显更慢的方法。

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)

来源

2016-09-24 07:37:02 cuttlefish44

子集的稀疏矩阵data.table R中

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