Goal: from a list of vectors of equal length, create a matrix where each vector becomes a row.
目的:从长度相等的向量列表中,创建一个矩阵,使每个向量成为一行。
Example:
例子:
> a <- list()
> for (i in 1:10) a[[i]] <- c(i,1:5)
> a
[[1]]
[1] 1 1 2 3 4 5
[[2]]
[1] 2 1 2 3 4 5
[[3]]
[1] 3 1 2 3 4 5
[[4]]
[1] 4 1 2 3 4 5
[[5]]
[1] 5 1 2 3 4 5
[[6]]
[1] 6 1 2 3 4 5
[[7]]
[1] 7 1 2 3 4 5
[[8]]
[1] 8 1 2 3 4 5
[[9]]
[1] 9 1 2 3 4 5
[[10]]
[1] 10 1 2 3 4 5
I want:
我想要:
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
6 个解决方案
#1
114
One option is to use do.call()
:
一个选项是使用do.call():
> do.call(rbind, a)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#2
11
Not straightforward, but it works:
这并不简单,但它确实有效:
> t(sapply(a, unlist))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#3
11
simplify2array
is a base function that is fairly intuitive. However, since R's default is to fill in data by columns first, you will need to transpose the output. (sapply
uses simplify2array
, as documented in help(sapply)
.)
simplify2array是一个非常直观的基函数。但是,由于R的默认值是先用列填充数据,所以需要转置输出。(sapply使用了在help(sapply)中记录的simplify2array。)
> t(simplify2array(a))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#4
7
The built-in matrix
function has the nice option to enter data byrow
. Combine that with an unlist
on your source list will give you a matrix. We also need to specify the number of rows so it can break up the unlisted data. That is:
内置的矩阵函数有一个很好的选项来输入数据。将它与源列表上的unlist结合起来会得到一个矩阵。我们还需要指定行数,这样它就可以分解未列出的数据。那就是:
> matrix(unlist(a), byrow=TRUE, nrow=length(a) )
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#5
5
t(sapply(a, '[', 1:max(sapply(a, length))))
where 'a' is a list. Would work for unequal row size
其中“a”是一个列表。对于不相等的行大小是否可行
#6
3
> library(plyr)
> as.matrix(ldply(a))
V1 V2 V3 V4 V5 V6
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#1
114
One option is to use do.call()
:
一个选项是使用do.call():
> do.call(rbind, a)
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#2
11
Not straightforward, but it works:
这并不简单,但它确实有效:
> t(sapply(a, unlist))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#3
11
simplify2array
is a base function that is fairly intuitive. However, since R's default is to fill in data by columns first, you will need to transpose the output. (sapply
uses simplify2array
, as documented in help(sapply)
.)
simplify2array是一个非常直观的基函数。但是,由于R的默认值是先用列填充数据,所以需要转置输出。(sapply使用了在help(sapply)中记录的simplify2array。)
> t(simplify2array(a))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#4
7
The built-in matrix
function has the nice option to enter data byrow
. Combine that with an unlist
on your source list will give you a matrix. We also need to specify the number of rows so it can break up the unlisted data. That is:
内置的矩阵函数有一个很好的选项来输入数据。将它与源列表上的unlist结合起来会得到一个矩阵。我们还需要指定行数,这样它就可以分解未列出的数据。那就是:
> matrix(unlist(a), byrow=TRUE, nrow=length(a) )
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5
#5
5
t(sapply(a, '[', 1:max(sapply(a, length))))
where 'a' is a list. Would work for unequal row size
其中“a”是一个列表。对于不相等的行大小是否可行
#6
3
> library(plyr)
> as.matrix(ldply(a))
V1 V2 V3 V4 V5 V6
[1,] 1 1 2 3 4 5
[2,] 2 1 2 3 4 5
[3,] 3 1 2 3 4 5
[4,] 4 1 2 3 4 5
[5,] 5 1 2 3 4 5
[6,] 6 1 2 3 4 5
[7,] 7 1 2 3 4 5
[8,] 8 1 2 3 4 5
[9,] 9 1 2 3 4 5
[10,] 10 1 2 3 4 5