选择矩阵中满足条件的行

发布于 2024-10-24 08:31:37 字数 410 浏览 8 评论 0原文

在带有矩阵的 R 中：

     one two three four
 [1,]   1   6    11   16
 [2,]   2   7    12   17
 [3,]   3   8    11   18
 [4,]   4   9    11   19
 [5,]   5  10    15   20

我想提取其行具有第三列 = 11 的子矩阵。也就是说：

      one two three four
 [1,]   1   6    11   16
 [3,]   3   8    11   18
 [4,]   4   9    11   19

我想在不循环的情况下执行此操作。我是 R 新手，所以这可能非常明显，但是文档通常比较简洁。

原文

In R with a matrix:

     one two three four
 [1,]   1   6    11   16
 [2,]   2   7    12   17
 [3,]   3   8    11   18
 [4,]   4   9    11   19
 [5,]   5  10    15   20

I want to extract the submatrix whose rows have column three = 11. That is:

      one two three four
 [1,]   1   6    11   16
 [3,]   3   8    11   18
 [4,]   4   9    11   19

I want to do this without looping. I am new to R so this is probably very obvious but the
documentation is often somewhat terse.

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月竹挽风 2024-10-31 08:31:38

m <- matrix(1:20, ncol = 4) 
colnames(m) <- letters[1:4]

以下命令将选择上面矩阵的第一行。

subset(m, m[,4] == 16)

这将选择最后三个。

subset(m, m[,4] > 17)

在这两种情况下，结果都是矩阵。
如果您想使用列名称来选择列，那么您最好将其转换为数据框，

mf <- data.frame(m)

然后您可以选择，

mf[ mf$a == 16, ]

或者，您可以使用子集命令。

m <- matrix(1:20, ncol = 4) 
colnames(m) <- letters[1:4]

The following command will select the first row of the matrix above.

subset(m, m[,4] == 16)

And this will select the last three.

subset(m, m[,4] > 17)

The result will be a matrix in both cases.
If you want to use column names to select columns then you would be best off converting it to a dataframe with

mf <- data.frame(m)

Then you can select with

mf[ mf$a == 16, ]

Or, you could use the subset command.

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疯了 2024-10-31 08:31:38

我将选择使用 dplyr 包的简单方法。

如果数据框是数据。

library(dplyr)
result <- filter(data, three == 11)

I will choose a simple approach using the dplyr package.

If the dataframe is data.

library(dplyr)
result <- filter(data, three == 11)

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寒江雪… 2024-10-31 08:31:38

Subset是一个非常慢的函数，我个人觉得它没什么用。

我假设您有一个名为 Mat 的 data.frame、数组、矩阵，其中 A、B、C 作为列姓名；那么您需要做的就是：

在一列上有一个条件的情况下，假设 A 列
```
Mat[which(Mat[,'A'] == 10), ]
```

如果不同列上有多个条件，您可以创建一个虚拟变量。假设条件为 A = 10、B = 5 和 C > 1。 2，那么我们有：

    aux = which(Mat[,'A'] == 10)
    aux = aux[which(Mat[aux,'B'] == 5)]
    aux = aux[which(Mat[aux,'C'] > 2)]
    Mat[aux, ]

通过使用system.time测试速度优势，which方法比subset快10倍方法。

Subset is a very slow function , and I personally find it useless.

I assume you have a data.frame, array, matrix called Mat with A, B, C as column names; then all you need to do is:

In the case of one condition on one column, lets say column A
```
Mat[which(Mat[,'A'] == 10), ]
```

In the case of multiple conditions on different column, you can create a dummy variable. Suppose the conditions are A = 10, B = 5, and C > 2, then we have:

    aux = which(Mat[,'A'] == 10)
    aux = aux[which(Mat[aux,'B'] == 5)]
    aux = aux[which(Mat[aux,'C'] > 2)]
    Mat[aux, ]

By testing the speed advantage with system.time, the which method is 10x faster than the subset method.

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他是夢罘是命 2024-10-31 08:31:38

如果您的矩阵名为 m，只需使用：

R> m[m$three == 11, ]

If your matrix is called m, just use :

R> m[m$three == 11, ]

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心凉怎暖 2024-10-31 08:31:38

如果数据集称为data，则所有行满足列“pm2.5”的值>“pm2.5”的条件。 300 可以通过 -

data[data['pm2.5'] >300,]

If the dataset is called data, then all the rows meeting a condition where value of column 'pm2.5' > 300 can be received by -

data[data['pm2.5'] >300,]

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断念 2024-10-31 08:31:37

如果使用 as.data.frame() 将矩阵转换为数据框，这会更容易做到。在这种情况下，前面的答案（使用子集或 m$三）将起作用，否则将不起作用。

要对矩阵执行操作，您可以按名称定义列：

m[m[, "three"] == 11,]

或按数字定义列：

m[m[,3] == 11,]

请注意，如果只有一行匹配，则结果是整数向量，而不是矩阵。

This is easier to do if you convert your matrix to a data frame using as.data.frame(). In that case the previous answers (using subset or m$three) will work, otherwise they will not.

To perform the operation on a matrix, you can define a column by name: