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R *应用向量作为输入;矩阵作为输出

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为了拥有一个矩阵，我想'd like to apply over each element of a vector, a function that outputs a vector. After applying the function to each element of that vector, I should have many vectors, which I' .

代码应该等同于以下内容：

my_function <- function(x) x:(x+10)

my_vec <- 1:10
x <- vector()
for(i in seq_along(vec)){
  x <- rbind(x,my_function(my_vec[i]))
}

当然， my_function 和 my_vec 就是例子 .

r matrix vector apply

2 回答

  • 1

    这是一个替代方案：

    matrix( unlist(lapply(my_vec,my_function)), length(my_vec), byrow=TRUE )

    速度几乎相同：

    library(microbenchmark)

my_function <- function(x) sin(x:(x+10))

for ( n in 1:4 )
{
  my_vec <- 1:10^n

  print(
    microbenchmark( mra68 = matrix( unlist(lapply(my_vec,my_function)), length(my_vec), byrow=TRUE ),
                    stas.g = do.call(rbind, lapply(my_vec, my_function)),
                    times = 1000 )
  )

  print("identical?")
  print( identical( matrix( unlist(lapply(my_vec,my_function)), length(my_vec), byrow=TRUE ),
                    do.call(rbind, lapply(my_vec, my_function)) ) )  
}

    .

    Unit: microseconds
   expr    min     lq     mean median      uq     max neval
  mra68 38.496 40.307 68.00539 41.213 110.052 282.148  1000
 stas.g 41.213 42.572 72.86443 43.930 115.939 445.186  1000
[1] "identical?"
[1] TRUE
Unit: microseconds
   expr     min      lq     mean   median       uq      max neval
  mra68 793.002 810.212 850.4857 818.3640 865.2375 7231.669  1000
 stas.g 876.786 894.901 946.8165 906.2235 966.9100 7051.873  1000
[1] "identical?"
[1] TRUE
Unit: milliseconds
   expr      min       lq     mean   median       uq      max neval
  mra68 2.605448 3.028442 5.269003 4.020940 7.807512 14.51225  1000
 stas.g 2.959604 3.390071 5.823661 4.500546 8.800462 92.54977  1000
[1] "identical?"
[1] TRUE
Unit: milliseconds
   expr      min       lq     mean   median       uq      max neval
  mra68 27.29810 30.99387 51.44223 41.20167 79.46185 559.0059  1000
 stas.g 33.63622 37.22420 60.10224 49.07643 92.94333 395.3315  1000
[1] "identical?"
[1] TRUE
>
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  • 2

    尝试：

    tmp <- lapply(my_vec, my_function)
 do.call(rbind, tmp)

    或者像Heroka建议的那样，使用 sapply . 我更喜欢 lapply ，然后以我喜欢的方式绑定我的输出（ rbind / cbind ）而不是潜在的转置 .

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