Question

简单地说，我有378742个观测值（每个观测值都有发布日期和截止日期），我想检查每个观测值的持续时间与其他所有观测值（378741）的重叠情况，并将其汇总。

由于嵌套循环，我正在运行下面的代码，这需要花费很多时间（我估计是205天）。有没有办法加快计算速度？（我将DescTools软件包用于Overlap命令。）

a <- c(1:378742)

for (i in 1:378742) {
  mydata$competition[i] <- sum(a, na.rm = T)
  for (j in 1:378742) {
    a[j] <- Overlap(c(mydata$Launched[i], mydata$Deadline[i]), c(mydata$Launched[j], mydata$Deadline[j]))
  }
}

Answer 1

通过向量化内部循环可以节省大量时间（然后我将apply()用于外部循环）：

# We'll need both DescTools and microbenchmark
library(DescTools)
library(microbenchmark)
# Make example data
set.seed(123) # setting seed for reproducibility
n <- 10
x <- sample(seq(as.Date("2008/10/20"), as.Date("2018/10/20"), "day"), n)
y <- sample(seq(as.Date("2008/10/20"), as.Date("2018/10/20"), "day"), n)
(mat <- cbind(x, y))
#>           x     y
#>  [1,] 15222 17667
#>  [2,] 17050 15827
#>  [3,] 15665 16645
#>  [4,] 17395 16262
#>  [5,] 17603 14547
#>  [6,] 14338 17454
#>  [7,] 16098 15069
#>  [8,] 17425 14325
#>  [9,] 16181 15367
#> [10,] 15835 17650
# First get the answer using nested loops
a <- z <- 1:n
for (i in 1:n) {
    for (j in 1:n) {
        a[j] <- Overlap(mat[i, ],mat[j, ])
    }
    # Noticed I've moved this sum to the bottom,
    # so that our first element isn't just a sum from one to n
    z[i] <- sum(a, na.rm = T)
}
z
#>  [1] 16102  9561  7860  7969 18169 18140  6690 18037  6017 12374
apply(mat, 1, function(r) sum(Overlap(r, mat)))
#>  [1] 16102  9561  7860  7969 18169 18140  6690 18037  6017 12374
microbenchmark(apply = apply(mat, 1, function(r) sum(Overlap(r, mat))),
               loop = for (i in 1:n) {
                   for (j in 1:n) {
                       a[j] <- Overlap(mat[i, ],mat[j, ])
                   }
                   # Noticed I've moved this sum to the bottom,
                   # so that our first element isn't just a sum from one to n
                   z[i] <- sum(a, na.rm = T)
               })
#> Unit: milliseconds
#>   expr       min        lq      mean    median        uq       max neval
#>  apply  7.538967  7.688929  7.894379  7.767989  7.891177  13.57523   100
#>   loop 76.051011 77.203810 80.045325 78.158369 79.206538 114.68139   100
#>  cld
#>   a 
#>    b

^{由reprex package（v0.2.1）于2018-10-20创建}

现在，让我们尝试了解如何（稍微）使用更大的示例数据进行缩放（如果数据太大，则基准测试将永远花费）：

# 
n <- 100
x <- sample(seq(as.Date("2008/10/20"), as.Date("2018/10/20"), "day"), n, r = T)
y <- sample(seq(as.Date("2008/10/20"), as.Date("2018/10/20"), "day"), n, r = T)
mat <- cbind(x, y)
a <- z <- 1:n
for (i in 1:n) {
    for (j in 1:n) {
        a[j] <- Overlap(mat[i, ],mat[j, ])
    }
    z[i] <- sum(a, na.rm = T)
}
# In case you're concerned it still works:
all.equal(z, apply(mat, 1, function(r) sum(Overlap(r, mat))))
#> [1] TRUE
microbenchmark(apply = apply(mat, 1, function(r) sum(Overlap(r, mat))),
               loop = for (i in 1:n) {
                   for (j in 1:n) {
                       a[j] <- Overlap(mat[i, ],mat[j, ])
                   }
                   # Noticed I've moved this sum to the bottom,
                   # so that our first element isn't just a sum from one to n
                   z[i] <- sum(a, na.rm = T)
               })
#> Unit: milliseconds
#>   expr       min        lq      mean    median        uq       max neval
#>  apply  258.1151  262.8007  269.8172  265.9643  276.8799  296.2167   100
#>   loop 5806.9834 5841.3362 5890.4988 5863.7317 5884.2308 6222.1670   100
#>  cld
#>   a 
#>    b

^{由reprex package（v0.2.1）于2018-10-20创建}

Answer 2

在生物信息学中，我们使用GenomicRanges包来查找重叠范围。

我曾经也使用计算机通常计算的5天的for循环和lapply函数进行计算。但是后来我发现了GenomicRanges软件包-它在几秒钟内就完成了！

（可惜，我仍然不知道它是如何工作的……将与有序的树结构和有效的相交有关……部分可能还涉及C++代码？。）

无论如何，结果是闪电般快速。你会惊讶的！

GenomicRanges软件包，可简化快速范围的计算

############################
# Install GenomicRanges package
############################

# since this year introduced: `BiocManager`
# Bioconductor is main code repository for Bionformaticians.
# It is kind of `CRAN` for Bioinformaticians programming with R
install.packages("BiocManager")
require(BiocManager) 
BiocManager::install("GenomicRanges")

# In older systems, you have to do:
install.packages("BiocInstaller")
require(BiocInstaller)
biocLite("GenomicRanges")

############################
# Load the GenomicRanges package
############################

require(GenomicRanges)

############################
# create dates as positive intervals
############################

set.seed(123) # for reproducibility of random stuff

n <-  1000 # later: 378742
x <- sample(seq(as.Date("2008/10/20"), as.Date("2038/10/20"), "day"), replace=TRUE, n)
# y <- sapply(x, function(date) date + sample(1:1000, 1)) # too slow!
deltas <- sample(1:10000, replace=TRUE, n) # immediate response `sapply` needs very long
y <- x + deltas

df <- data.frame(seqnames="1", start=x, end=y)
gr <- GRanges(df)
gr <- sort(gr)

############################
# Be careful, GRanges obj is 1-based system and not 0-based!
############################

# each row is one index - gr behaves when indexing like a vector
gr[5]   # selects fifth row
gr[4:7] # selects 4th to 7th row

############################
# which range overlaps with which range?
############################

system.time({hits <- findOverlaps(gr, gr)})   
# system.time({ your-R-expression }) - very convenient speed measuring!
# the numbers in the table are the index (i-th row) in each of the tables
# query and subject table - which are in this case identical tables - gr

############################
# what is the amount of overlap?
############################

overlaps <- pintersect(gr[queryHits(hits)], gr[subjectHits(hits)])
amount.overlaps <- width(overlaps) - 1 # - 1 because 1-based systems do +1 when ranges

# 1-base versus 0-based coordinate systems: https://www.biostars.org/p/84686/

如何在R中加快循环中的循环计算

2 个答案: