R-将df或矩阵的每一行乘以一个向量

b*rep(unlist(a),each=nrow(b))
#      [,1] [,2] [,3] [,4]
# [1,]    2    4    6    8
# [2,]    2    4    6    8
# [3,]    2    4    6    8

或者只是 b*rep(a,each=nrow(b)) 如果您定义 a <- 1:4

这只是一个矢量化的元素乘法,没有任何变换 rep 。

编辑:

似乎是rep拖慢了我的解决方案。这里是一个基准测试,我在其中包含了一个带有预计算rep的选项,以及对sweep选项的一些改进(仅从源代码中获取相关部分)。

a <- data.table(t(1:200))
b <- matrix(data=2, nrow=100000, ncol=200)

a_vec <- unlist(a)
rep_a <- rep(a_vec,each=nrow(b))
microbenchmark::microbenchmark(
  mkr1 = a[,lapply(.SD,function(x)(x*b[,x]))],
  mkr2 = t(t(b) * (as.matrix(a)[1,])),
  mkr_update = a[,lapply(V1:V4,function(i)(a[[i]]*b[,i]))],
  mm = b*rep(unlist(a),each=nrow(b)),
  mm_cheat = b*rep_a,
  regular_sweep = sweep(b,2,unlist(a),`*`),
  regular_sweep2 = sweep(b,2,a_vec,`*`),
  improved_sweepA1 = b*aperm(array(unlist(a),rev(dim(b)))),
  improved_sweepA2 = b*aperm(array(a_vec,rev(dim(b)))),
  improved_sweepB1 = b*a[rep_len(1,nrow(b)),],
  improved_sweepB2 = b*t(a_vec)[rep_len(1,nrow(b)),],
  unit = "relative",
  times=50)


Unit: relative
             expr       min        lq      mean    median        uq       max neval
             mkr1  42.12228  44.15266  50.23959  46.35240  57.20280  65.07289    50
             mkr2 114.58427 124.19653 125.25660 131.08677 124.17058 114.91137    50
       mkr_update   1.00000   1.00000   1.00000   1.00000   1.00000   1.00000    50
               mm 231.34331 223.74365 217.50145 225.91117 215.90765 165.64814    50
         mm_cheat  13.38838  13.22556  14.94682  13.36649  12.95260  25.15564    50
    regular_sweep  96.15758 124.26746 121.04428 128.67282 129.19407 119.20210    50
   regular_sweep2  97.79001 124.69191 124.74650 134.64249 134.97407 107.47152    50
 improved_sweepA1  96.57837 124.86189 116.93736 127.08909 124.92805 105.83318    50
 improved_sweepA2  96.27737 122.49773 118.45262 128.13369 126.15029 106.58669    50
 improved_sweepB1 214.95773 227.39523 226.04339 248.38553 232.50401 161.45341    50
 improved_sweepB2  31.20967  32.61873  37.74552  33.70969  41.52149  55.93362    50

在我这方面,我会使用内置的R方法进行矩阵乘法 %*% 。

考虑到 矢量 :[注: data.table 是 不 vector ]

a <- c(1:4)

考虑到矩阵:

b <- matrix(data=2, nrow=3, ncol=4)

您的输出由以下内容给出:

output <- b %*% diag(a)

     [,1] [,2] [,3] [,4]
[1,]    2    4    6    8
[2,]    2    4    6    8
[3,]    2    4    6    8

如果您认为此解决方案对您的需求效率很低,那么我建议使用内置功能 sweep :

sweep(b, 2, a, FUN = "*")

     [,1] [,2] [,3] [,4]
[1,]    2    4    6    8
[2,]    2    4    6    8
[3,]    2    4    6    8

选项#1: 使用 data.table 特点:

注意:它之所以有效,是因为列编号和值与 a

a[,lapply(.SD,function(x)(x*b[,x]))]
#   V1 V2 V3 V4
#1:  2  4  6  8
#2:  2  4  6  8
#3:  2  4  6  8

选项2: 可能是:

t(t(b) * (as.matrix(a)[1,]))
     [,1] [,2] [,3] [,4]
[1,]    2    4    6    8
[2,]    2    4    6    8
[3,]    2    4    6    8

更新

选项3: 在中处理十进制/实际值的步骤 一

#Cases when `a` contains decimal values can be handled as
a <- data.table(t(c(1, 0.24, 3, 4)))
b <- matrix(data=2, nrow=3, ncol=4)

a[,lapply(V1:V4,function(i)(a[[i]]*b[,i]))]
#   V1   V2 V3 V4
#1:  2 0.48  6  8
#2:  2 0.48  6  8
#3:  2 0.48  6  8

感谢您的回复。我已经在速度(向量和矩阵的实际大小)上测试了上述建议的解决方案,以使用最有效的解决方案:

a <- data.table(t(1:200))
b <- matrix(data=2, nrow=100000, ncol=200)

system.time(sweep(b, MARGIN=2, t(a), "*"))
#   user  system elapsed 
#   0.31    0.06    0.39 

system.time(a[,lapply(.SD,function(x)(x*b[,x]))])
#   user  system elapsed 
#    0.2     0.0     0.2 

#system.time(bind_rows(apply(b,1,`*`,a)))     
#took 100+ so stopped it manually

system.time(t(t(b)*(as.matrix(a)[1,])))
#   user  system elapsed 
#   0.31    0.05    0.36 

system.time(apply(b, 1, `*`, 1:200))
#   user  system elapsed 
#   1.20    0.11    1.31 

system.time(b*rep(unlist(a),each=nrow(b)))
#   user  system elapsed 
#   0.83    0.05    0.89 

system.time(b*rep((1:200),each=nrow(b)))
#   user  system elapsed 
#   0.36    0.06    0.42

dplyr::bind_rows(apply(b, 1, `*`, a))
   V1 V2 V3 V4
1:  2  4  6  8
2:  2  4  6  8
3:  2  4  6  8

棘手的是你的 a 是一个数据。桌子如果它实际上是一个向量,那么它要简单得多:

apply(b, 1, `*`, 1:4)
     [,1] [,2] [,3]
[1,]    2    2    2
[2,]    4    4    4
[3,]    6    6    6
[4,]    8    8    8