如何识别哪些列不是“NA”？矩阵中的每行？

发布于 2024-12-05 04:03:50 字数 570 浏览 0 评论 0原文

我有一个 12 行和 77 列的矩阵，但简单地让我们使用：

p <- matrix(NA,5,7)  
p[1,2]<-0.3  
p[1,3]<-0.5  
p[2,4]<-0.9  
p[2,7]<-0.4  
p[4,5]<-0.6

我想知道每行哪些列不是“NA”，所以我想要得到的是这样的：

[1] 2,3  
[2] 4  
[3] 0  
[4] 5  
[5] 0

但如果我这样做 > which(p[]!="NA") 我得到 [1] 6 11 17 24 32

我尝试使用循环：

aux <- matrix(NA,5,7)  
for(i in 1:5) {  
    aux[i,]<-which(p[i,]!="NA")  
}

但我得到一个错误： number of要替换的项目不是替换长度的倍数

有没有办法做到这一点？提前致谢

原文

I have a matrix with 12 rows and 77 columns, but to simply lets use:

p <- matrix(NA,5,7)  
p[1,2]<-0.3  
p[1,3]<-0.5  
p[2,4]<-0.9  
p[2,7]<-0.4  
p[4,5]<-0.6

I want to know which columns are not "NA" per row, so what I would like to get would be something like:

[1] 2,3  
[2] 4  
[3] 0  
[4] 5  
[5] 0

but if I do > which(p[]!="NA") I get [1] 6 11 17 24 32

I tried using a loop:

aux <- matrix(NA,5,7)  
for(i in 1:5) {  
    aux[i,]<-which(p[i,]!="NA")  
}

but I just get an error: number of items to replace is not a multiple of replacement length

Is there a way of doing this? Thanks in advance

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才能让你更想念 2024-12-12 04:03:50

尝试：

which( !is.na(p), arr.ind=TRUE)

我认为这与您指定的输出一样信息丰富，并且可能比您指定的输出更有用，但如果您确实想要列表版本，那么可以使用：

> apply(p, 1, function(x) which(!is.na(x)) )
[[1]]
[1] 2 3

[[2]]
[1] 4 7

[[3]]
integer(0)

[[4]]
[1] 5

[[5]]
integer(0)

或者甚至与粘贴一起涂抹：

lapply(apply(p, 1, function(x) which(!is.na(x)) ) , paste, collapse=", ")

which< 的输出/code> 函数建议的方法提供逻辑测试的非零 (TRUE) 位置的行和列：

> which( !is.na(p), arr.ind=TRUE)
     row col
[1,]   1   2
[2,]   1   3
[3,]   2   4
[4,]   4   5
[5,]   2   7

如果没有将 arr.ind 参数设置为非默认 TRUE，您只能得到“向量位置”使用 R 的主要排序列确定作为其惯例。 R 矩阵只是“折叠向量”。

> which( !is.na(p) )
[1]  6 11 17 24 32

Try:

which( !is.na(p), arr.ind=TRUE)

Which I think is just as informative and probably more useful than the output you specified, But if you really wanted the list version, then this could be used:

> apply(p, 1, function(x) which(!is.na(x)) )
[[1]]
[1] 2 3

[[2]]
[1] 4 7

[[3]]
integer(0)

[[4]]
[1] 5

[[5]]
integer(0)

Or even with smushing together with paste:

lapply(apply(p, 1, function(x) which(!is.na(x)) ) , paste, collapse=", ")

The output from which function the suggested method delivers the row and column of non-zero (TRUE) locations of logical tests:

> which( !is.na(p), arr.ind=TRUE)
     row col
[1,]   1   2
[2,]   1   3
[3,]   2   4
[4,]   4   5
[5,]   2   7

Without the arr.ind parameter set to non-default TRUE, you only get the "vector location" determined using the column major ordering the R has as its convention. R-matrices are just "folded vectors".