我有一个矩阵“a”,如下所示:
a<-rbind(c("a1","ost1;ost2;ost3","utr;body;pro"),
c("a2","idh1;idh2","pro;body"),
c("a3","dnm1","body"))
>a
[,1] [,2] [,3]
[1,] "a1" "ost1;ost2;ost3" "utr;body;pro"
[2,] "a2" "idh1;idh2" "pro;body"
[3,] "a3" "dnm1" "body"
我想得到像这样的矩阵“b”
[,1] [,2] [,3]
[1,] "a1" "ost1" "utr"
[2,] "a1" "ost2" "body"
[3,] "a1" "ost3" "pro"
[4,] "a2" "idh1" "pro"
[5,] "a2" "idh2" "body"
[6,] "a3" "dnm1" "body"
好的,明白了:
b<-do.call(rbind, (apply(a, 1, function(x) {do.call(cbind, strsplit(x,";"))})))
答案 0 :(得分:1)
您的解决方案,没有不必要的括号:
do.call(rbind, apply(a, 1, function(x) do.call(cbind, strsplit(x, ";"))))
这也有效:
do.call(rbind, lapply(apply(a, 1, strsplit, ';'), do.call, what = cbind))
并非使用匿名函数(function(x){...}
)有任何问题,但有些人发现它更“优雅”而没有任何问题。