22

所以我试图对矩阵的行求和，其中有 inf。我如何对行求和，省略inf？

4

5 回答 5

36

将您的矩阵乘以 的结果is.finite(m)并使用 调用rowSums乘积na.rm=TRUE。这有效，因为Inf*0is NaN。

m <- matrix(c(1:3,Inf,4,Inf,5:6),4,2)
rowSums(m*is.finite(m),na.rm=TRUE)
于 2013-03-13T18:26:54.063 回答
21 
A[is.infinite(A)]<-NA
rowSums(A,na.rm=TRUE)

一些基准比较：

library(microbenchmark)


rowSumsMethod<-function(A){
 A[is.infinite(A)]<-NA
 rowSums(A,na.rm=TRUE)
}
applyMethod<-function(A){
 apply( A , 1 , function(x){ sum(x[!is.infinite(x)])})
}

rowSumsMethod2<-function(m){
  rowSums(m*is.finite(m),na.rm=TRUE) 
}

rowSumsMethod0<-function(A){
 A[is.infinite(A)]<-0
 rowSums(A)
}

A1 <- matrix(sample(c(1:5, Inf), 50, TRUE), ncol=5)
A2 <- matrix(sample(c(1:5, Inf), 5000, TRUE), ncol=5)
microbenchmark(rowSumsMethod(A1),rowSumsMethod(A2),
               rowSumsMethod0(A1),rowSumsMethod0(A2),
               rowSumsMethod2(A1),rowSumsMethod2(A2),
               applyMethod(A1),applyMethod(A2))

Unit: microseconds
               expr      min        lq    median        uq      max neval
  rowSumsMethod(A1)   13.063   14.9285   16.7950   19.3605 1198.450   100
  rowSumsMethod(A2)  212.726  220.8905  226.7220  240.7165  307.427   100
 rowSumsMethod0(A1)   11.663   13.9960   15.3950   18.1940  112.894   100
 rowSumsMethod0(A2)  103.098  109.6290  114.0610  122.9240  159.545   100
 rowSumsMethod2(A1)    8.864   11.6630   12.5960   14.6955   49.450   100
 rowSumsMethod2(A2)   57.380   60.1790   63.4450   67.4100   81.172   100
    applyMethod(A1)   78.839   84.4380   92.1355   99.8330  181.005   100
    applyMethod(A2) 3996.543 4221.8645 4338.0235 4552.3825 6124.735   100

所以约书亚的方法赢了！而且 apply 方法显然比其他两种方法慢（当然相对而言）。

于 2013-03-13T18:15:10.730 回答
11

我会使用applyandis.infinite来避免用@Hemmo的答案替换Inf值。NA

> set.seed(1)
> Mat <- matrix(sample(c(1:5, Inf), 50, TRUE), ncol=5)
> Mat # this is an example
      [,1] [,2] [,3] [,4] [,5]
 [1,]    2    2  Inf    3    5
 [2,]    3    2    2    4    4
 [3,]    4    5    4    3    5
 [4,]  Inf    3    1    2    4
 [5,]    2    5    2    5    4
 [6,]  Inf    3    3    5    5
 [7,]  Inf    5    1    5    1
 [8,]    4  Inf    3    1    3
 [9,]    4    3  Inf    5    5
[10,]    1    5    3    3    5
> apply(Mat, 1, function(x) sum(x[!is.infinite(x)]))
 [1] 12 15 21 10 18 16 12 11 17 17
于 2013-03-13T18:22:46.727 回答
8

尝试这个...

m <- c( 1 ,2 , 3 , Inf , 4 , Inf ,5 )
sum(m[!is.infinite(m)])

或者

m <- matrix( sample( c(1:10 , Inf) , 100 , rep = TRUE ) , nrow = 10 )
sums <- apply( m , 1 , FUN = function(x){ sum(x[!is.infinite(x)])})

> m
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    8    9    7  Inf    9    2    2    6    1   Inf
 [2,]    8    7    4    5    9    5    8    4    7    10
 [3,]    7    9    3    4    7    3    3    6    9     4
 [4,]    7  Inf    2    6    4    8    3    1    9     9
 [5,]    4  Inf    7    5    9    5    3    5    9     9
 [6,]    7    3    7  Inf    7    3    7    3    7     1
 [7,]    5    7    2    1  Inf    1    9    8    1     5
 [8,]    4  Inf   10  Inf    8   10    4    9    7     2
 [9,]   10    7    9    7    2  Inf    4  Inf    4     6
[10,]    9    4    6    3    9    6    6    5    1     8

> sums
 [1] 44 67 55 49 56 45 39 54 49 57
于 2013-03-13T18:14:51.180 回答
3

这是一种“非应用”且非破坏性的方法：

rowSums( matrix(match(A, A[is.finite(A)]), nrow(A)), na.rm=TRUE)
[1] 2 4

尽管它相当有效，但它不如 Johsua 的乘法方法快。

于 2013-03-13T18:48:10.597 回答