Question

给定一个包含一行，一列或一个单元格的矩阵，我需要在保持矩阵结构的同时重新排序行。我尝试添加drop=F，但它不起作用！我做了什么？

test = matrix(letters[1:5]) # is a matrix
test[5:1,,drop=F]           # not a matrix

test2 = matrix(letters[1:5],nrow=1) # is a matrix
test2[1:1,,drop=F]                  # not a matrix

test3 = matrix(1)  # is a matrix
test3[1:1,,drop=F] # not a matrix

Answer 1

我猜它是被覆盖的F; F可以设置为变量，在这种情况下它不再是假的。始终完全写出FALSE，不能将其设置为变量。

请参阅Is there anything wrong with using T & F instead of TRUE & FALSE?

同样，R Inferno，第8.1.32节，是一个很好的参考。

> F <- 1
> test = matrix(letters[1:5]) # is a matrix
> test[5:1,,drop=F]           # not a matrix
[1] "e" "d" "c" "b" "a"
> test[5:1,,drop=FALSE]       # but this is a matrix
     [,1]
[1,] "e" 
[2,] "d" 
[3,] "c" 
[4,] "b" 
[5,] "a" 
> rm(F)
> test[5:1,,drop=F]           # now a matrix again
     [,1]
[1,] "e" 
[2,] "d" 
[3,] "c" 
[4,] "b" 
[5,] "a"

Answer 2

您的问题中的代码在新的R会话中正常工作：

test = matrix(letters[1:5]) # is a matrix
result = test[5:1,,drop=F]  
result
#      [,1]
# [1,] "e" 
# [2,] "d" 
# [3,] "c" 
# [4,] "b" 
# [5,] "a" 
class(result)  # still a matrix
# [1] "matrix"
dim(result)
# [1] 5 1

即使在1x1矩阵上：

test3 = matrix(1)  # is a matrix
result3 = test3[1:1,,drop=F]
class(result3)
# [1] "matrix"
dim(result3)
# [1] 1 1

也许您已经加载了覆盖默认行为的其他软件包？是什么让你认为你不会以矩阵结束？

Answer 3

以下作品：

test <- matrix(test[5:1,, drop = F], nrow = 5, ncol = 1)

使用is.matrix进行测试时，输出为矩阵。同时，您指定行数（nrow）和列数（ncol）以将其强制为您需要的行数和列数。

R：保留1行/列矩阵

3 个答案: