如何用R中的向量元素划分矩阵的每一行

Question

我想用固定的向量划分矩阵的每一行.例如

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

给mat/dev分割各列由dev.

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

但是,我希望得到这个结果,即按行进行操作:

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

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

到那儿有明确的命令吗？

Answer 1

以下是增加代码长度的几种方法:

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

数据框架

mat /如果mat是数据帧并且产生数据帧结果,则所有开始的解决方案也起作用.sweep解决方案和最后一个解决方案也是如此mat2.该mapply解决方案可与data.frames但产生的矩阵.

向量

如果mat是普通矢量而不是矩阵,则其中任何一个都返回一个列矩阵

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

这一个返回一个向量:

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

其他人给出错误,警告或不是所需的答案.

基准

代码的简洁和清晰可能比速度更重要,但为了完整性,这里有一些使用10次重复然后100次重复的基准.

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)])

赠送:

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

重新运行所有那些花费<20毫秒,重复100次的测试:

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)])

赠送:

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)都是最快的.原因可能在于它几乎直接呼吁BLAS,而#1依赖于更昂贵的t.

Answer 2

您正在寻找apply应用于行的函数：

t(apply(mat, 1, function(x) x/dev))