R 从字符矩阵计算值

我有一个7358444行110列的矩阵。矩阵由字符向量组成，如下所示：

      FORMAT     eQTL188                      eQTL193                   eQTL178                   eQTL179                       eQTL238                      
[1,] "GT:DS:GP" "0/1:0.79:0.221,0.767,0.011" "0/0:0.031:0.97,0.03,0"   "0/0:0.033:0.967,0.033,0" "0/0:0.079:0.922,0.077,0.001" "0/0:0.344:0.664,0.329,0.007"
[2,] "GT:DS:GP" "0/0:0.047:0.953,0.047,0"    "0/0:0.007:0.993,0.007,0" "0/0:0.006:0.994,0.006,0" "0/0:0.008:0.992,0.008,0"     "0/1:0.525:0.477,0.52,0.002" 
[3,] "GT:DS:GP" "0/0:0.047:0.953,0.047,0"    "0/0:0.007:0.993,0.007,0" "0/0:0.006:0.994,0.006,0" "0/0:0.008:0.992,0.008,0"     "0/1:0.527:0.476,0.521,0.003"
[4,] "GT:DS:GP" "0/0:0.048:0.952,0.048,0"    "0/0:0.007:0.993,0.007,0" "0/0:0.006:0.994,0.006,0" "0/0:0.008:0.992,0.008,0"     "0/1:0.518:0.485,0.512,0.003"

我需要为我的每个样本（带有模式eQTL的列）计算等位基因1的剂量。这可以使用每列中第二个

：

后的GP值进行计算。我需要应用的公式是

P（A1）=2*P（A1/A1）+P（A1/A2）

，其中P1是第二个

：

之后的第一个元素，A2是第二个元素

我要查找的结果（数字矩阵）如下所示

     eQTL188  eQTL193 eQTL178 eQTL179 eQTL238                      
[1,] 1.209    1.970   1.967   1.921   1.657
[2,] 1.953    1.903   1.994   1.992   1.474 
[3,] 1.953    1.993   1.994   1.992   1.473
[4,] 1.952    1.993   1.994   1.99    1.482

---

由于矩阵非常庞大，速度可能是一个问题

一种方法是首先检索第二个

：

之后的数字，然后通过逗号检索strsplit。公式可以应用于<代码> LePix<代码>，例如考虑这个

df <- matrix(c("GT:DS:GP" ,"0/1:0.79:0.221,0.767,0.011" ,"0/0:0.031:0.97,0.03,0" ,  "0/0:0.033:0.967,0.033,0", "0/0:0.079:0.922,0.077,0.001", "0/0:0.344:0.664,0.329,0.007",
            "GT:DS:GP" ,"0/0:0.047:0.953,0.047,0" ,   "0/0:0.007:0.993,0.007,0", "0/0:0.006:0.994,0.006,0" ,"0/0:0.008:0.992,0.008,0"  ,   "0/1:0.525:0.477,0.52,0.002",
            "GT:DS:GP", "0/0:0.047:0.953,0.047,0"  ,  "0/0:0.007:0.993,0.007,0", "0/0:0.006:0.994,0.006,0" ,"0/0:0.008:0.992,0.008,0"  ,   "0/1:0.527:0.476,0.521,0.003",
            "GT:DS:GP" ,"0/0:0.048:0.952,0.048,0" ,   "0/0:0.007:0.993,0.007,0", "0/0:0.006:0.994,0.006,0" ,"0/0:0.008:0.992,0.008,0"   ,  "0/1:0.518:0.485,0.512,0.003"), 
            ncol=6, byrow=TRUE)


df <- df[, -1]
df <- gsub(".+:.+:(.*)", "\\1", df)
out <- lapply(strsplit(df, ","), function(x) {
  x <- as.numeric(x)
  return(2 * x[1] / x[1] + x[1] / x[2])
})

out <- do.call(rbind, out)
dim(out) <- dim(df)

          [,1]      [,2]      [,3]      [,4]     [,5]
[1,]  2.288136  34.33333  31.30303  13.97403 4.018237
[2,] 22.276596 143.85714 167.66667 126.00000 2.917308
[3,] 22.276596 143.85714 167.66667 126.00000 2.913628
[4,] 21.833333 143.85714 167.66667 126.00000 2.947266

---

df不是A1/A1总是1.0吗？什么是
P（）
-函数？如果包含一个。查看矩阵的内容很有帮助，但给出生成矩阵的代码更有用。然后向我们展示你所尝试的。你可以考虑<代码> GSUB/COD>在第一个/第二个代码>之前和之后拆分数字：。这可以包含在
函数中，该函数还将计算GP值，并使用
sapply
应用于所有列。