R—选择满足多个条件的矩阵行的最快方法

这是对R中问题的扩展。假设我有矩阵：

       one two three four
 [1,]   1   6    11   16
 [2,]   2   7    12   17
 [3,]   3   8    11   18
 [4,]   4   9    11   19
 [5,]   5  10    15   20
 [6,]   1   6    15   20
 [7,]   5   7    12   20

我想返回所有行，其中

矩阵$two==7

和

矩阵$two==12

。这就是我知道的方法：

 out <- mat[mat$two == 7,]
 final_out <- out[out$three == 12, ]

---

out使用
which
和
arr.ind=TRUE
如中所示：

> mat[which(mat[,"two"]==7 & mat[,"three"] == 12, arr.ind = TRUE),]
  one two three four
2   2   7    12   17
7   5   7    12   20

只需使用带有逻辑比较的
[
子集设置

#  Reproducible data
set.seed(1)
m <- matrix( sample(12,28,repl=T) , 7 , 4 )
     [,1] [,2] [,3] [,4]
[1,]    4    8   10    3
[2,]    5    8    6    8
[3,]    7    1    9    2
[4,]   11    3   12    4
[5,]    3    3    5    5
[6,]   11    9   10    1
[7,]   12    5   12    5


#  Subset according to condition
m[ m[,2] == 3 & m[,3] == 12 , ]
[1] 11  3 12  4

#可复制数据
种子（1）
mR中绝对最快的方法是
ifelse
，这与
if
不同，if
允许向量化条件。您还可以缓存条件向量（例如
isSeven使用微基准更新：

使用基准测试给出了相反的答案。@SimonO101给出的答案似乎提供了一个稍微快一点的实现

require(microbenchmark)
set.seed(1)
m <- matrix( sample(12,100,repl=T) , 25 , 4 )
colnames(m) <- c("one","two","three","four")

bench1 <- microbenchmark(m[which(m[,'two']==7 & m[,'three'] == 12, arr.ind = TRUE),])
summary(bench1$time)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   7700    8750    9449    9688    9800   22400

bench2 <- microbenchmark(m[ m[,2] == 3 & m[,3] == 12 , ])
summary(bench2$time)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
   6300    7350    7351    7599    8050   15400

require（微基准）
种子（1）
m如果有很多行，那么最好还是先进行子集，如下面的代码所示

set.seed(1)
m <- matrix( sample(12,28,repl=T) , 12e6 , 4 )

#  Subset according to condition
microbenchmark(sample0=m[ m[,2] == 3 & m[,3] == 12 , ],times = 10L)

microbenchmark(sample1=m[ m[,2] == 3, ],
           sample2= sample1[sample1[,3] == 12, ],times = 10L)

或
mat[mat[，“two”]==7&mat[，“two”]==12，]
我没有完全阅读这个问题，也没有意识到OP已经使用了
哪个
是适合矩阵的$sub设置？我得到了一个错误。我认为
ifelse
不会比逻辑subsetting快。毕竟，这正是
ifelse
内部所做的。你应该使用
库（microbenchmark）
用于计算时间……它更可靠。@Thomas谢谢你的提示。这是我第一次尝试基准测试。我将在安装软件包后更新答案。这可能需要一点时间……环顾四周，你会看到人们如何使用它的其他示例。
set.seed(1)
m <- matrix( sample(12,1000000000,repl=T) , 1e8 , 10 )
colnames(m) <- c("one","two","three","four","five","six","seven","eight","nine","ten")

system.time(m[which(m[,'two']==7 & m[,'three'] == 12, arr.ind = TRUE),])
   user  system elapsed 
   6.49    1.58    8.06 
system.time(m[ m[,2] == 3 & m[,3] == 12 , ])
   user  system elapsed 
   8.23    1.29    9.52 

set.seed(1)
m <- matrix( sample(12,28,repl=T) , 12e6 , 4 )

#  Subset according to condition
microbenchmark(sample0=m[ m[,2] == 3 & m[,3] == 12 , ],times = 10L)

microbenchmark(sample1=m[ m[,2] == 3, ],
           sample2= sample1[sample1[,3] == 12, ],times = 10L)

microbenchmark(sample0=m[ m[,2] == 3 & m[,3] == 12 , ],times = 10L)
Unit: milliseconds
expr        min         lq        mean     median         uq        max neval
sample0 342.085212 347.333083 381.6039635 349.920741 375.383425 584.068743    10
microbenchmark(sample1=m[ m[,2] == 3, ],
              sample2= sample1[sample1[,3] == 12, ],times = 10L)
Unit: milliseconds
expr        min         lq        mean      median         uq        max neval cld
 sample1 188.647995 189.832552 215.9355769 194.2375715 199.118962 404.631420    10   b
 sample2   5.097811   5.262028   5.3260160   5.2868025   5.401471   5.571351    10  a