直到向量/矩阵结束的优雅索引

有没有可能在 R 中说-我想要从位置 i到向量/矩阵末尾的所有索引？ 假设我想要一个从第三列开始的子矩阵，我目前只知道这样:

A = matrix(rep(1:8, each = 5), nrow = 5) # just generate some example matrix...


A[,3:ncol(A)] # get submatrix from 3rd column onwards

但是我真的需要写 ncol(A)吗？怎么说“从第三纵队开始”呢？像是 A[,3:]？(或 A[,3:...]) ？

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最佳答案

Sometimes it's easier to tell R what you don't want. In other words, exclude columns from the matrix using negative indexing:

Here are two alternative ways that both produce the same results:

A[, -(1:2)]
A[, -seq_len(2)]

Results:

     [,1] [,2] [,3] [,4] [,5] [,6]
[1,]    3    4    5    6    7    8
[2,]    3    4    5    6    7    8
[3,]    3    4    5    6    7    8
[4,]    3    4    5    6    7    8
[5,]    3    4    5    6    7    8

But to answer your question as asked: Use ncol to find the number of columns. (Similarly there is nrow to find the number of rows.)

A[, 3:ncol(A)]


[,1] [,2] [,3] [,4] [,5] [,6]
[1,]    3    4    5    6    7    8
[2,]    3    4    5    6    7    8
[3,]    3    4    5    6    7    8
[4,]    3    4    5    6    7    8
[5,]    3    4    5    6    7    8
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For rows (not columns as per your example) then head() and tail() could be utilised.

A <- matrix(rep(1:8, each = 5), nrow = 5)
tail(A, 3)

is almost the same as

A[3:dim(A)[1],]

(the rownames/indices printed are different is all).

Those work for vectors and data frames too:

> tail(1:10, 4)
[1]  7  8  9 10
> tail(data.frame(A = 1:5, B = 1:5), 3)
A B
3 3 3
4 4 4
5 5 5

For the column versions, you could adapt tail(), but it is a bit trickier. I wonder if NROW() and NCOL() might be useful here, rather than dim()?:

> A[, 3:NCOL(A)]
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]    3    4    5    6    7    8
[2,]    3    4    5    6    7    8
[3,]    3    4    5    6    7    8
[4,]    3    4    5    6    7    8
[5,]    3    4    5    6    7    8

Or flip this on its head and instead of asking R for things, ask it to drop things instead. Here is a function that encapsulates this:

give <- function(x, i, dimen = 1L) {
ind <- seq_len(i-1)
if(isTRUE(all.equal(dimen, 1L))) { ## rows
out <- x[-ind, ]
} else if(isTRUE(all.equal(dimen, 2L))) { ## cols
out <- x[, -ind]
} else {
stop("Only for 2d objects")
}
out
}


> give(A, 3)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]    1    2    3    4    5    6    7    8
[2,]    1    2    3    4    5    6    7    8
[3,]    1    2    3    4    5    6    7    8
> give(A, 3, dimen = 2)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]    3    4    5    6    7    8
[2,]    3    4    5    6    7    8
[3,]    3    4    5    6    7    8
[4,]    3    4    5    6    7    8
[5,]    3    4    5    6    7    8
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You can use the following instruction:

A[, 3:length(A[, 1])]
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A dplyr readable renewed approach for the same thing:

A %>% as_tibble() %>%
select(-c(V1,V2))
  

A %>% as_tibble() %>%
select(V3:ncol(A))

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