[英]Sort matrix according to first column in R
I have a matrix with two columns of the following form:我有一个包含以下形式的两列的矩阵:
1 349
1 393
1 392
4 459
3 49
3 32
2 94
I would like to sort this matrix in increasing order based on the first column but I would like to keep the corresponding values in the second column.我想根据第一列按递增顺序对这个矩阵进行排序,但我想将相应的值保留在第二列中。
The output would look like this:输出将如下所示:
1 349
1 393
1 392
2 94
3 49
3 32
4 459
Read the data:读取数据:
foo <- read.table(text="1 349
1 393
1 392
4 459
3 49
3 32
2 94")
And sort:并排序:
foo[order(foo$V1),]
This relies on the fact that order
keeps ties in their original order.这依赖于order
保持其原始顺序的事实。 See ?order
.参见?order
。
Creating a data.table
with key=V1
automatically does this for you.创建一个带有key=V1
的data.table
自动为你做这件事。 Using Stephan's data foo
使用 Stephan 的数据foo
> require(data.table)
> foo.dt <- data.table(foo, key="V1")
> foo.dt
V1 V2
1: 1 349
2: 1 393
3: 1 392
4: 2 94
5: 3 49
6: 3 32
7: 4 459
Be aware that if you want to have values in the reverse order, you can easily do so:请注意,如果您想以相反的顺序使用值,您可以轻松地这样做:
> example = matrix(c(1,1,1,4,3,3,2,349,393,392,459,49,32,94), ncol = 2)
> example[order(example[,1], decreasing = TRUE),]
[,1] [,2]
[1,] 4 459
[2,] 3 49
[3,] 3 32
[4,] 2 94
[5,] 1 349
[6,] 1 393
[7,] 1 392
如果您的数据位于名为foo
的矩阵中,则您将运行的行是
foo.sorted=foo[order[foo[,1]]
The accepted answer works like a charm unless you're applying it to a vector.除非您将其应用于向量,否则接受的答案就像一个魅力。 Since a vector is non-recursive, you'll get an error like this由于向量是非递归的,你会得到这样的错误
$ operator is invalid for atomic vectors
You can use [
in that case在这种情况下,您可以使用[
foo[order(foo["V1"]),]
You do not need data.table
.您不需要data.table
。
This is what you need A[order(A[,1]), ]
, where A
is the matrix of your data.这就是您需要的A[order(A[,1]), ]
,其中A
是您的数据矩阵。
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