在大矩阵中使用strsplit应用函数

时间:2018-03-08 20:13:11

标签: r

我有一个8Million x 10列的矩阵。该矩阵的子集如下所示:

df <- matrix(c("0.221,0.767,0.011" ,"0.97,0.03,0",       "0.967,0.033,0",
               "0.922,0.077,0.001", "0.664,0.329,0.007", "0.953,0.047,0", 
               "0.993,0.007,0",     "0.994,0.006,0",     "0.992,0.008,0",
               "0.477,0.52,0.002",  "0.953,0.047,0",     "0.993,0.007,0", 
               "0.994,0.006,0",     "0.992,0.008,0",     "0.476,0.521,0.003",
               "0.952,0.048,0" ,    "0.993,0.007,0",     "0.994,0.006,0",
               "0.992,0.008,0", "0.485,0.512,0.003"), ncol=5, byrow=TRUE)

我需要在此矩阵中进行一系列计算。例如,如果x <- df[1,1],那么我需要将第一个元素0.221乘以两倍,并将结果加到第二个元素0.767

结果如下:

       [,1]  [,2]  [,3]  [,4]  [,5]
  [1,] 1.209 1.970 1.967 1.921 1.657
  [2,] 1.953 1.993 1.994 1.992 1.474
  [3,] 1.953 1.993 1.994 1.992 1.473
  [4,] 1.952 1.993 1.994 1.992 1.482

我使用@erocoar提供的这个解决方案,适用于小型数据集:

out <- lapply(strsplit(df, ","), function(x) {
x <- as.numeric(x)
return((2 * x[1]) + x[2])
})
out <- do.call(rbind, out)
dim(out) <- dim(df)

然而,这个解决方案需要大量内存,并且在我的真实数据集中需要花费大量时间。我也试过这个:

y = function(x) {a <- strsplit(x, ",")
    z <- as.numeric(a)
    return((2 * z[1]) + z[2])
    }
 m <- matrix(-9, nrow = nrow(df), ncol = ncol(df))
 m[] <- vapply(df, y, numeric(1))

但是这给我一个关于格式的错误。

更新

原始文件来自存储遗传信息的vcf文件(遗传变异文件),

首先我用read.vcfR读取了文件,这里是代码:

library(vcfR)
vcf <- read.vcfR("/mnt/lustre/scratch/home/proximal.vcf.gz")
df <- vcf@gt
df <- dosages[, -1]
df <- gsub(".+:.+:(.*)", "\\1", dosages, perl = TRUE)

然后我想在帖子的第一部分进行计算

1 个答案:

答案 0 :(得分:1)

您不需要任何循环,只需再次读取数据即可。这次sep字符定义了两次。对于粘贴sep=\nread.table sep=\n

首先将矩阵设为dataframe。将列逐列粘贴在一起,然后再次将其读回。

 dd=do.call(paste,c(data.frame(df),sep="\n"))
 m=read.table(text=dd,sep=",")
 matrix(2*m[,1]+m[,2],4,byrow=T)
      [,1]  [,2]  [,3]  [,4]  [,5]
[1,] 1.209 1.970 1.967 1.921 1.657
[2,] 1.953 1.993 1.994 1.992 1.474
[3,] 1.953 1.993 1.994 1.992 1.473
[4,] 1.952 1.993 1.994 1.992 1.482