每次从R中的矩阵中检索3行（以及列的子集）

我试图在矩阵上应用一个循环（matrixExample，如下所述），每次检索3行上的V1到V6列

matrixExample
      ID      CHR SEGNUM GENOSEG  V1  V2  V3  V4  V5  V6 
 [1,] "CHAR8" "1" "1"    "102505" "1" "0" "2" "5" "0" "5"
 [2,] "LIMO9" "1" "1"    "012505" "0" "1" "2" "5" "0" "5"
 [3,] "SIM10" "1" "1"    "122505" "1" "2" "2" "5" "0" "5"
 [4,] "CHAR8" "1" "2"    "111520" "1" "1" "1" "5" "2" "0"
 [5,] "LIMO9" "1" "2"    "221520" "2" "2" "1" "5" "2" "0"
 [6,] "SIM10" "1" "2"    "222520" "2" "2" "2" "5" "2" "0"
 [7,] "CHAR8" "1" "3"    "501111" "5" "0" "1" "1" "1" "1"
 [8,] "LIMO9" "1" "3"    "501100" "5" "0" "1" "1" "0" "0"
 [9,] "SIM10" "1" "3"    "502011" "5" "0" "2" "0" "1" "1"
[10,] "CHAR8" "2" "1"    "102505" "1" "0" "2" "5" "0" "5"
[11,] "LIMO9" "2" "1"    "012505" "0" "1" "2" "5" "0" "5"
[12,] "SIM10" "2" "1"    "122505" "1" "2" "2" "5" "0" "5"
[13,] "CHAR8" "2" "2"    "111520" "1" "1" "1" "5" "2" "0"
[14,] "LIMO9" "2" "2"    "221520" "2" "2" "1" "5" "2" "0"
[15,] "SIM10" "2" "2"    "222520" "2" "2" "2" "5" "2" "0"
[16,] "CHAR8" "2" "3"    "501111" "5" "0" "1" "1" "1" "1"
[17,] "LIMO9" "2" "3"    "501100" "5" "0" "1" "1" "0" "0"
[18,] "SIM10" "2" "3"    "502011" "5" "0" "2" "0" "1" "1"

例如，从前3行开始，我想使用子矩阵进行一些矩阵计算：

"1" "0" "2" "5" "0" "5"
"0" "1" "2" "5" "0" "5"
"1" "2" "2" "5" "0" "5"

在这个计算之后，我需要转到第4到第6行

我尝试了以下代码：

 for(i in seq(1, dim(exampleDoubleSort)[1], 3))
{
  print(matrixExample[(i:i+2),c(4:10)]) # using print only as an example  
  # do some matrix calculations using the subset matrix
}

它只打印一行中的元素，而不打印三行的组合

如果我尝试跳出循环，我可以获得预期的结果

      print(matrixExample[(1:3),c(5:10)])

     V1  V2  V3  V4  V5  V6 
[1,] "1" "0" "2" "5" "0" "5"
[2,] "0" "1" "2" "5" "0" "5"
[3,] "1" "2" "2" "5" "0" "5"

请告诉我如何每次读取3行并检索矩阵子集以进行进一步计算，好吗？ 如果我有30个how，我需要重试10个矩阵作为子集并执行10个计算。。。以函数为例进行计算

提前谢谢！ 干杯

---

RV

您的问题在于行的索引，以及行索引向量的生成。查看

3:3+2

和

3：（3+2）

之间的输出差异，作为旁注，您可能最好使用

data.frame

而不是列表，因为您有不同类的值（字符/因子和数字）但是一个矩阵只能包含一个类，因此所有值都被强制为charactersHanks很多Heroka！解决了这个问题！干杯谢谢docendo discimus！很好的提示…这只是另一个使用

split

lappy（split（data，list（data$CHR，data$SEGNUM）），函数（x）作为.matrix（x[，-c（1:4）]）的选项）

将数据作为

data.frame