如何将矩阵的每一行除以 R 中向量的元素 [英] How to divide each row of a matrix by elements of a vector in R

问题描述

我想将矩阵的每一行除以一个固定向量.例如

I would like to divide each row of a matrix by a fixed vector. For example

mat<-matrix(1,ncol=2,nrow=2,TRUE)
dev<-c(5,10)

Giving mat/dev 将每个列除以 dev.

Giving mat/dev divides each column by dev.

     [,1] [,2]
[1,]  0.2  0.2
[2,]  0.1  0.1

但是，我希望得到这样的结果，即按行进行操作:

However, I would like to have this as a result, i.e. do the operation row-wise :

rbind(mat[1,]/dev, mat[2,]/dev)

     [,1] [,2]
[1,]  0.2  0.1
[2,]  0.2  0.1

是否有明确的命令可以到达那里?

Is there an explicit command to get there?

推荐答案

以下是几种增加代码长度的方法:

Here are a few ways in order of increasing code length:

t(t(mat) / dev)

mat / dev[col(mat)] #  @DavidArenburg & @akrun

mat %*% diag(1 / dev)

sweep(mat, 2, dev, "/")

t(apply(mat, 1, "/", dev))

plyr::aaply(mat, 1, "/", dev)

mat / rep(dev, each = nrow(mat))

mat / t(replace(t(mat), TRUE, dev))

mapply("/", as.data.frame(mat), dev)  # added later

mat / matrix(dev, nrow(mat), ncol(mat), byrow = TRUE)  # added later

do.call(rbind, lapply(as.data.frame(t(mat)), "/", dev))

mat2 <- mat; for(i in seq_len(nrow(mat2))) mat2[i, ] <- mat2[i, ] / dev

数据帧

所有以 mat/ 开头的解决方案也适用于如果 mat 是一个数据框并产生一个数据框结果.sweep 解决方案和最后一个，即 mat2 解决方案的情况也是如此.mapply 解决方案适用于 data.frames，但会生成一个矩阵.

Data Frames

All the solutions that begin with mat / also work if mat is a data frame and produce a data frame result. The same is also the case for the sweep solution and the last, i.e. mat2, solution. The mapply solutions works with data.frames but produces a matrix.

如果 mat 是一个普通向量而不是一个矩阵，那么它们中的任何一个都返回一个单列矩阵

If mat is a plain vector rather than a matrix then either of these return a one column matrix

t(t(mat) / dev)
mat / t(replace(t(mat), TRUE, dev))

这个返回一个向量:

plyr::aaply(mat, 1, "/", dev)

其他人给出错误、警告或不是想要的答案.

The others give an error, warning or not the desired answer.

代码的简洁性和清晰度可能比速度更重要，但为了完整起见，这里有一些基准测试，分别使用 10 次重复和 100 次重复.

The brevity and clarity of the code may be more important than speed but for purposes of completeness here are some benchmarks using 10 repetitions and then 100 repetitions.

library(microbenchmark)
library(plyr)

set.seed(84789)

mat<-matrix(runif(1e6),nrow=1e5)
dev<-runif(10)

microbenchmark(times=10L,
  "1" = t(t(mat) / dev),
  "2" = mat %*% diag(1/dev),
  "3" = sweep(mat, 2, dev, "/"),
  "4" = t(apply(mat, 1, "/", dev)),
  "5" = mat / rep(dev, each = nrow(mat)),
  "6" = mat / t(replace(t(mat), TRUE, dev)),
  "7" = aaply(mat, 1, "/", dev),
  "8" = do.call(rbind, lapply(as.data.frame(t(mat)), "/", dev)),
  "9" = {mat2 <- mat; for(i in seq_len(nrow(mat2))) mat2[i, ] <- mat2[i, ] / dev},
 "10" = mat/dev[col(mat)])

给予:

Unit: milliseconds
 expr         min          lq       mean      median          uq        max neval
    1    7.957253    8.136799   44.13317    8.370418    8.597972  366.24246    10
    2    4.678240    4.693771   10.11320    4.708153    4.720309   58.79537    10
    3   15.594488   15.691104   16.38740   15.843637   16.559956   19.98246    10
    4   96.616547  104.743737  124.94650  117.272493  134.852009  177.96882    10
    5   17.631848   17.654821   18.98646   18.295586   20.120382   21.30338    10
    6   19.097557   19.365944   27.78814   20.126037   43.322090   48.76881    10
    7 8279.428898 8496.131747 8631.02530 8644.798642 8741.748155 9194.66980    10
    8  509.528218  524.251103  570.81573  545.627522  568.929481  821.17562    10
    9  161.240680  177.282664  188.30452  186.235811  193.250346  242.45495    10
   10    7.713448    7.815545   11.86550    7.965811    8.807754   45.87518    10

对所有耗时 <20 毫秒并重复 100 次的测试重新运行:

Re-running the test on all those that took <20 milliseconds with 100 repetitions:

microbenchmark(times=100L,
  "1" = t(t(mat) / dev),
  "2" = mat %*% diag(1/dev),
  "3" = sweep(mat, 2, dev, "/"),
  "5" = mat / rep(dev, each = nrow(mat)),
  "6" = mat / t(replace(t(mat), TRUE, dev)),
 "10" = mat/dev[col(mat)])

给予:

Unit: milliseconds
 expr       min        lq      mean    median        uq       max neval
    1  8.010749  8.188459 13.972445  8.560578 10.197650 299.80328   100
    2  4.672902  4.734321  5.802965  4.769501  4.985402  20.89999   100
    3 15.224121 15.428518 18.707554 15.836116 17.064866  42.54882   100
    5 17.625347 17.678850 21.464804 17.847698 18.209404 303.27342   100
    6 19.158946 19.361413 22.907115 19.772479 21.142961  38.77585   100
   10  7.754911  7.939305  9.971388  8.010871  8.324860  25.65829   100

所以在这两个测试中，#2(使用 diag)是最快的.原因可能在于它几乎直接对 BLAS 有吸引力，而 #1 依赖于更昂贵的 t.

So on both these tests #2 (using diag) is fastest. The reason may lie in its almost direct appeal to the BLAS, whereas #1 relies on the costlier t.

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