do.call和命令以矩阵的降序对每一行进行排序? [英] do.call and order to sort each row to descending order of a matrix?

问题描述

我想将此矩阵按行顺序降序排列

I want to sort this matrix row-wise to descending order

 > set.seed(123); a <- matrix(rbinom(100,10,0.3),ncol=10)

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    2    6    5    6    1    1    4    4    2     1
 [2,]    4    3    4    5    3    3    1    3    4     4
 [3,]    3    4    3    4    3    4    3    4    3     2
 [4,]    5    3    7    4    2    1    2    0    4     4
 [5,]    5    1    4    0    2    3    4    3    1     2
 [6,]    1    5    4    3    1    2    3    2    3     2
 [7,]    3    2    3    4    2    1    4    2    6     4
 [8,]    5    1    3    2    3    4    4    3    5     1
 [9,]    3    2    2    2    2    5    4    2    5     3
[10,]    3    6    1    2    5    2    3    1    2     3

但是

> do.call(order,as.list(a[1,],a[2,]))
[1] 1

如何用do.call和order对矩阵进行排序?

编辑.固定上述矩阵以符合上述代码.

Edit. Fixed above matrix to conform with the above code.

推荐答案

两种选择:

# Jaap
do.call(rbind, lapply(split(a, row(a)), sort, decreasing = TRUE))

# adaption of lmo's solution in the comments
for(i in 1:nrow(a)) a[i,] <- a[i,][order(a[i,], decreasing = TRUE)]

给予:

   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
1     6    6    5    4    4    2    2    1    1     1
2     5    4    4    4    4    3    3    3    3     1
3     4    4    4    4    3    3    3    3    3     2
4     7    5    4    4    4    3    2    2    1     0
5     5    4    4    3    3    2    2    1    1     0
6     5    4    3    3    3    2    2    2    1     1
7     6    4    4    4    3    3    2    2    2     1
8     5    5    4    4    3    3    3    2    1     1
9     5    5    4    3    3    2    2    2    2     2
10    6    5    3    3    3    2    2    2    1     1

具有以下条件的基准:

library(microbenchmark)
microbenchmark(dc.lapply.sort = do.call(rbind, lapply(split(a, row(a)), sort, decreasing = TRUE)),
               t.apply.sort = t(apply(a, 1, sort, decreasing = TRUE)),
               for.order = for(i in 1:nrow(a)) a[i,] <- a[i,][order(a[i,], decreasing = TRUE)],
               for.sort = for(i in 1:nrow(a)) a[i,] <- sort(a[i,], decreasing = TRUE),
               for.sort.list = for(x in seq_len(nrow(a))) a[x,] <- a[x,][sort.list(a[x,], decreasing = TRUE, method="radix")])

给予:

Unit: microseconds
           expr     min       lq      mean   median       uq      max neval cld
 dc.lapply.sort 189.811 206.5890 222.52223 217.8070 228.0905  332.034   100   c
   t.apply.sort 185.474 200.4515 212.59608 210.4930 220.0025  286.288   100  bc
      for.order  82.631  91.1860  98.66552  97.8475 102.9680  176.666   100 a  
       for.sort 167.939 187.5025 192.90728 192.1195 198.8690  256.494   100  b 
  for.sort.list 187.617 206.4475 230.82960 215.7060 221.6115 1541.343   100   c

---

但是应该注意，基准测试仅对较大的数据集有意义，因此:

---

It should be noted however that benchmarks are only meaningful on larger datasets, so:

set.seed(123)
a <- matrix(rbinom(10e5, 10, 0.3), ncol = 10)

microbenchmark(dc.lapply.sort = do.call(rbind, lapply(split(a, row(a)), sort, decreasing = TRUE)),
               t.apply.sort = t(apply(a, 1, sort, decreasing = TRUE)),
               for.order = for(i in 1:nrow(a)) a[i,] <- a[i,][order(a[i,], decreasing = TRUE)],
               for.sort = for(i in 1:nrow(a)) a[i,] <- sort(a[i,], decreasing = TRUE),
               for.sort.list = for(x in seq_len(nrow(a))) a[x,] <- a[x,][sort.list(a[x,], decreasing = TRUE, method="radix")],
               times = 10)

给予:

Unit: seconds
           expr      min       lq     mean   median       uq      max neval  cld
 dc.lapply.sort 6.790179 6.924036 7.036330 7.013996 7.121343 7.351729    10    d
   t.apply.sort 5.032052 5.057022 5.151560 5.081459 5.177159 5.538416    10   c 
      for.order 1.368351 1.463285 1.514652 1.471467 1.583873 1.736544    10 a   
       for.sort 5.028314 5.102993 5.317597 5.154104 5.348614 6.123278    10   c 
  for.sort.list 2.417857 2.464817 2.573294 2.519408 2.726118 2.815964    10  b  

结论:for循环与order组合仍然是最快的解决方案.

Conclusion: the for-loop in combination with order is still the fastest solution.

使用grr -package的order2和sort2函数可以进一步提高速度.将它们与上面最快的解决方案进行比较:

Using the order2 and sort2 functions of the grr-package can give a further improvement in speed. Comparing them with the fastest solution from above:

set.seed(123)
a <- matrix(rbinom(10e5, 10, 0.3), ncol = 10)

microbenchmark(for.order = for(i in 1:nrow(a)) a[i,] <- a[i,][order(a[i,], decreasing = TRUE)],
               for.order2 = for(i in 1:nrow(a)) a[i,] <- a[i,][rev(grr::order2(a[i,]))],
               for.sort2 = for(i in 1:nrow(a)) a[i,] <- rev(grr::sort2(a[i,])),
               times = 10)

给予:

Unit: milliseconds
       expr       min        lq      mean    median        uq      max neval cld
  for.order 1243.8140 1263.4423 1316.4662 1305.1823 1378.5836 1404.251    10   c
 for.order2  956.1536  962.8226 1110.1778 1090.9984 1233.4241 1368.416    10  b 
  for.sort2  830.1887  843.6765  920.5668  847.1601  972.8703 1144.135    10 a  

这篇关于do.call和命令以矩阵的降序对每一行进行排序?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！