I have a list of matrices each with the same number of columns. I'd like to create a single data frame with the first column from matrix1 followed by the first column from matrix2 followed by the first column from matrix3...then the second column from matrix1 followed by the second column from matrix2, etc.
我有一个矩阵列表，每个矩阵的列数相同。我想创建一个单独的数据框架，第一列来自matrix1，第一列来自matrix2，第一列来自matrix3…然后是矩阵x1的第二列，然后是矩阵x2的第二列，等等。
counts <- data.frame(apples=round(rnorm(10, mean=5), digits=0),
oranges=round(rnorm(10, mean=5), digits=0),
pears=round(rnorm(10, mean=5), digits=0))
weights <- data.frame(apples=round(rnorm(10, mean=3), digits=1),
oranges=round(rnorm(10, mean=3), digits=1),
pears=round(rnorm(10, mean=3), digits=1))
diameters <- data.frame(apples=round(rnorm(10, mean=10), digits=1),
oranges=round(rnorm(10, mean=10), digits=1),
pears=round(rnorm(10, mean=10), digits=1))
fruitdata <- list(counts, weights, diameters)
> fruitdata
[[1]]
apples oranges pears
1 4 4 4
2 6 5 5
3 6 4 6
4 6 5 4
5 7 4 5
6 6 7 5
7 4 6 5
8 4 6 5
9 4 5 5
10 5 7 5
[[2]]
apples oranges pears
1 2.5 3.1 1.1
2 4.3 2.4 4.2
3 2.8 3.5 1.3
4 2.8 1.5 2.5
5 2.9 3.3 1.9
6 3.7 1.5 2.2
7 2.9 2.7 5.1
8 3.0 2.5 3.0
9 2.3 2.3 1.7
10 2.7 2.9 1.4
[[3]]
apples oranges pears
1 10.5 10.4 12.3
2 10.0 9.8 10.1
3 9.7 11.1 10.5
4 9.1 10.9 9.9
5 8.5 9.4 9.7
6 8.9 12.2 10.0
7 11.0 9.7 10.8
8 9.4 8.6 12.1
9 8.6 9.9 11.0
10 11.9 10.2 11.2
What I want is something like this...
我想要的是这样的东西……
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 4 2.5 10.5 4 3.1 10.4
[2,] 6 4.3 10.0 5 2.4 9.8
[3,] 6 2.8 9.7 4 3.5 11.1
[4,] 6 2.8 9.1 5 1.5 10.9
[5,] 7 2.9 8.5 4 3.3 9.4
[6,] 6 3.7 8.9 7 1.5 12.2
[7,] 4 2.9 11.0 6 2.7 9.7
[8,] 4 3.0 9.4 6 2.5 8.6
[9,] 4 2.3 8.6 5 2.3 9.9
[10,] 5 2.7 11.9 7 2.9 10.2
I can use sapply to pull the first column from each matrix:
我可以用sapply从每个矩阵中拉出第一列:
test1 <- sapply(fruitdata, function(x) x[,1])
> test1
[,1] [,2] [,3]
[1,] 4 2.5 10.5
[2,] 6 4.3 10.0
[3,] 6 2.8 9.7
[4,] 6 2.8 9.1
[5,] 7 2.9 8.5
[6,] 6 3.7 8.9
[7,] 4 2.9 11.0
[8,] 4 3.0 9.4
[9,] 4 2.3 8.6
[10,] 5 2.7 11.9
So I tried:
所以我试着:
test2 <- sapply(fruitdata, function(x) x[,1:ncol(fruitdata[[1]])])
But it did not get me what I wanted.
但这并没有得到我想要的。
>test2
[,1] [,2] [,3]
apples Numeric,10 Numeric,10 Numeric,10
oranges Numeric,10 Numeric,10 Numeric,10
pears Numeric,10 Numeric,10 Numeric,10
I can, of course, write a for loop to do this task, but it seems like I should be able to do it with the apply or plyr suites instead, but I can't quite figure out how to do it.
当然，我可以编写一个for循环来完成这个任务，但是看起来我应该能够使用应用程序或plyr套件来完成它，但是我不知道怎么做。
Thanks!
谢谢!
1 个解决方案
#1
2
EDIT: incorporated rawr's clever suggestion.
编辑:结合罗尔斯的聪明建议。
Is this what you want?
这就是你想要的吗?
do.call(cbind, fruitdata)[order(rep(1:3, 3))]
# apples apples.1 apples.2 oranges oranges.1 oranges.2 pears pears.1 pears.2
# 1 4 3.2 10.1 6 4.9 10.0 4 2.6 10.8
# 2 6 4.3 10.0 4 4.2 8.4 4 3.7 9.4
# 3 5 2.9 9.4 4 4.4 8.7 5 5.5 9.2
# 4 6 2.8 12.6 6 1.9 10.4 5 1.9 11.0
# 5 8 3.6 9.5 6 3.8 9.7 4 3.7 8.9
# 6 4 4.3 10.1 4 2.3 9.7 5 4.1 10.5
# 7 5 2.1 10.2 4 3.2 8.7 6 0.9 10.6
# 8 4 4.2 9.8 7 5.0 10.0 5 2.0 9.2
# 9 4 4.1 10.5 6 1.8 8.2 6 3.2 9.1
# 10 6 2.4 9.3 5 3.1 8.6 4 3.6 10.5
Here, we just stick all the matrices together, and then re-order the columns so that they end up how you want.
在这里，我们把所有的矩阵粘在一起，然后重新排序列，这样它们就可以按照你想要的方式结束。
#1
2
EDIT: incorporated rawr's clever suggestion.
编辑:结合罗尔斯的聪明建议。
Is this what you want?
这就是你想要的吗?
do.call(cbind, fruitdata)[order(rep(1:3, 3))]
# apples apples.1 apples.2 oranges oranges.1 oranges.2 pears pears.1 pears.2
# 1 4 3.2 10.1 6 4.9 10.0 4 2.6 10.8
# 2 6 4.3 10.0 4 4.2 8.4 4 3.7 9.4
# 3 5 2.9 9.4 4 4.4 8.7 5 5.5 9.2
# 4 6 2.8 12.6 6 1.9 10.4 5 1.9 11.0
# 5 8 3.6 9.5 6 3.8 9.7 4 3.7 8.9
# 6 4 4.3 10.1 4 2.3 9.7 5 4.1 10.5
# 7 5 2.1 10.2 4 3.2 8.7 6 0.9 10.6
# 8 4 4.2 9.8 7 5.0 10.0 5 2.0 9.2
# 9 4 4.1 10.5 6 1.8 8.2 6 3.2 9.1
# 10 6 2.4 9.3 5 3.1 8.6 4 3.6 10.5
Here, we just stick all the matrices together, and then re-order the columns so that they end up how you want.
在这里，我们把所有的矩阵粘在一起，然后重新排序列，这样它们就可以按照你想要的方式结束。