R 选择一行矩阵作为矩阵
这一直困扰着我。考虑以下事项:R 选择一行矩阵作为矩阵,r,matrix,R,Matrix,这一直困扰着我。考虑以下事项: # Part A # # Make a silly simple matrix with column names x = matrix(1:4, ncol = 2) colnames(x) = c("a","b") # Part B # # Pick out the first row of the matrix. This is not a matrix, # and the column names are no longer accessible
# Part A #
# Make a silly simple matrix with column names
x = matrix(1:4, ncol = 2)
colnames(x) = c("a","b")
# Part B #
# Pick out the first row of the matrix. This is not a matrix,
# and the column names are no longer accessible by colnames()
y = x[1,]
y
is.matrix(y)
colnames(y)
# Part C #
# But here is what I want:
y = matrix(x[1,], nrow = 1, dimnames = list(c(), colnames(x)))
有没有办法用更少的处理步骤或更少的代码来实现C部分?似乎应该有一个几乎和
x[1,]
一样短的命令来做同样的事情。只需将drop=FALSE设置为:
> y = x[1,, drop=FALSE]
> y
a b
[1,] 1 3
怎么样
x[1,,drop=FALSE]
a b
[1,] 1 3
你不是唯一被窃听的人。数据分析软件的一个gem:“默认值是,而且一直都是,drop=TRUE;很久以前我们可能做出了一个不明智的决定,但现在是不太可能改变的向后兼容性负担之一。”