我想按行降序排列此矩阵
> 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
> do.call(order,as.list(a[1,],a[2,]))
[1] 1
如何使用do.call和订单对矩阵进行排序？
编辑。修正了以上矩阵以符合上述代码。
答案 0 :(得分：5)
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)]
给出：
[,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
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")])
给出：
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
但应注意，基准测试仅对较大的数据集有意义，因此：
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)
给出：
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。
使用order2 - 包的sort2和grr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：
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)
，并提供：
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
答案 1 :(得分：0)
t(apply(a, 1, sort, decreasing = TRUE))给出：
# [,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
答案 2 :(得分：0)
我也进行了微基准测试，看起来订单解决方案赢了：）
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