从坐标列表中分配矩阵中的1

时间：2016-10-12 01:15:08

标签： r vectorization

我想要一个坐标列表。对于坐标中的每个位置，我想为矩阵指定1。最终矩阵应如下所示。我想使用快速矢量化方法而不是for循环。

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示例数据：

    > sample.matrix
      A B C D E F G H I J K L M N O P Q R S T
    A 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    B 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
    C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    D 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
    E 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    F 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
    G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    H 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
    I 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    J 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
    K 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    L 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    M 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
    N 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
    O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
    P 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
    Q 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
    R 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
    S 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
    T 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

我尝试过的是什么。

    set.seed(15)
    dat <- data.frame(x=sample(LETTERS[1:20], 15,set.seed(15)),y=sample(LETTERS[1:20], 15,set.seed(15)), stringsAsFactors = FALSE)
    sample.matrix <- matrix(data=0, nrow = 20, ncol = 20, dimnames = list(LETTERS[1:20],LETTERS[1:20]))

我尝试的不起作用。 

    for ( n in 1:15){
      sample.matrix[dat[n,]$x,dat[n,]$y] =1
    }

3 个答案:

答案 0 :(得分：8)

这是该语言的一个很酷的编程方面。在R中，我们可以使用两列矩阵进行子集化：

sample.matrix[as.matrix(dat)] <- 1

有关工作原理的更多信息

来自?Extract的帮助（粗体文字是我的重点）：

  

第三种索引形式是通过带有一列的数字矩阵   对于每个维度：索引矩阵的每一行然后选择一个   数组的单个元素，结果是一个向量。负   索引矩阵中不允许使用索引。 NA和零值是   允许：忽略包含零的索引矩阵的行，   而包含NA的行在结果中产生NA。

     

通过字符矩阵进行索引，每个维度一列   如果数组具有维度名称，也会受到支持。和数字一样   矩阵索引，索引矩阵的每一行选择一个单独的元素   数组。指数与适当的维度相匹配   名。允许NA并在结果中产生NA。独一无二   索引以及空字符串（“”）是不允许的   导致错误。

在您的情况下，您有一个矩阵，其中的字母代表维度名称。 

     x   y  
[1,] "B" "C"
[2,] "S" "S"
[3,] "L" "B"
[4,] "N" "L"
[5,] "E" "G"
[6,] "M" "A"

当我们将此矩阵插入到一组括号[as.matrix(dat)]中时，求值程序将检测矩阵并使用特殊类型的子集。第一列表示行，第二列表示列。

head(sample.matrix)
  A B C D E F G H I J K L M N O P Q R S T
A 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
B 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
D 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
E 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
F 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

答案 1 :(得分：3)

您可以使用dcast中的data.table：

dat$z <- 1
dcast(dat,x~y,fill=0, value="z")

   x B C D G H I J K L M N O P Q T
1  A 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0
2  B 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
3  C 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
4  D 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
5  F 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
6  H 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
7  J 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0
8  K 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
9  L 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
10 M 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
11 N 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
12 O 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0
13 Q 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
14 R 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
15 T 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

答案 2 :(得分：1)

这也有效：

sample.matrix[t(apply(dat, 1, function(x) 
             c(which(dimnames(sample.matrix)[[1]]==x[1]),
               which(dimnames(sample.matrix)[[2]]==x[2]))))] <- 1

sample.matrix
  A B C D E F G H I J K L M N O P Q R S T
A 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
B 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
C 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
D 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0
E 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
F 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
G 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
H 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0
I 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
J 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
K 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
L 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
M 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
N 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
O 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
P 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
Q 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
R 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
S 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
T 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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