结合在多个行中的相同载体

Question

我有这样的载体：

Vec=c("a" , "b", "c ", "d") 

欲将此作为数据帧：

 [,1] [,2] [,3] [,4] 
[1,] a b c d 
[2,] a b c d 
[3,] a b c d 
[4,] a b c d 
[5,] a b c d 

Answer 1

另一种选择：

t(replicate(5, Vec)) 
#  [,1] [,2] [,3] [,4] 
#[1,] "a" "b" "c " "d" 
#[2,] "a" "b" "c " "d" 
#[3,] "a" "b" "c " "d" 
#[4,] "a" "b" "c " "d" 
#[5,] "a" "b" "c " "d" 

Answer 2

一个使用rbind和do.call将方式：

do.call(rbind, replicate(5, Vec, simplify = FALSE)) 

    [,1] [,2] [,3] [,4] 
[1,] "a" "b" "c " "d" 
[2,] "a" "b" "c " "d" 
[3,] "a" "b" "c " "d" 
[4,] "a" "b" "c " "d" 
[5,] "a" "b" "c " "d" 

你可以用你喜欢的任何数字替换5。

replicate返回列表中的Vec 5次（simplify = FALSE创建列表）。这些元素是rbind -ed使用do.call。

更新：

实际使用matrix可能是最好的：

> matrix(Vec, nrow=5, ncol=length(Vec), byrow=TRUE) 
    [,1] [,2] [,3] [,4] 
[1,] "a" "b" "c " "d" 
[2,] "a" "b" "c " "d" 
[3,] "a" "b" "c " "d" 
[4,] "a" "b" "c " "d" 
[5,] "a" "b" "c " "d" 

更改nrow参数任何你想要的号码，你就准备好它。

全部3个回答将需要使用as.data.frame转换为data.frame所以我从微基准排除这样的：

微基准

> microbenchmark::microbenchmark(t(replicate(5, Vec)), 
+        do.call(rbind, replicate(5, Vec, simplify = FALSE)), 
+        matrix(Vec, nrow=5, ncol=4, byrow=TRUE), 
+        times=1000) 
Unit: microseconds 
               expr min  lq  mean median  uq  max neval 
           t(replicate(5, Vec)) 52.854 59.013 68.393740 63.374 70.815 1749.326 1000 
do.call(rbind, replicate(5, Vec, simplify = FALSE)) 18.986 23.092 27.325856 25.144 27.710 105.708 1000 
     matrix(Vec, nrow = 5, ncol = 4, byrow = TRUE) 1.539 2.566 3.474166 3.079 3.593 29.763 1000 

正如你可以看到matrix解决方案迄今为止最好的。