r - R中按列名排列矩阵组列表的矩阵

Question

我有一个包含 6000 列的矩阵，每列属于我需要的 100 个“组”之一。我需要将此矩阵转换为包含 100 个较小矩阵的列表。这是我所拥有的玩具示例:

  mat = cbind(c(2,2,2),c(3,3,3),c(4,4,4),c(1,1,1))
  colnames(mat) = c("2018.3 1","2018.3 2","2019.1 1","2019.2 2")

因此“组”由每个列名的姓氏标识，这里有 2 个组。我需要的结果如下所示:

list(cbind(c(2,2,2),c(4,4,4)),cbind(c(3,3,3),c(1,1,1)))

我一直在想，我觉得应该是这样的:

lapply(do.call(cbind,sapply(something here to find the columns in each group)))

但我还没弄清楚具体怎么做。

Answer 1

#Obtain the last part of each column names
groups = sapply(strsplit(x = colnames(mat), split = " "), function(x) x[2])

#Go through each unique column name and extract the corresponding columns
lapply(unique(groups), function(x) mat[,which(groups == x)])
#[[1]]
#     2018.3 1 2019.1 1
#[1,]        2        4
#[2,]        2        4
#[3,]        2        4

#[[2]]
#     2018.3 2 2019.2 2
#[1,]        3        1
#[2,]        3        1
#[3,]        3        1

或

lapply(split(1:NCOL(mat), sapply(strsplit(x = colnames(mat), split = " "),
                                 function(x) x[2])), function(i) mat[,i])
#$`1`
#     2018.3 1 2019.1 1
#[1,]        2        4
#[2,]        2        4
#[3,]        2        4

#$`2`
#     2018.3 2 2019.2 2
#[1,]        3        1
#[2,]        3        1
#[3,]        3        1