Question

让我们看一下示例矩阵并计算相关性：

some.data <- data.frame(
  A1.1 = c(1,3,4,5,6),
  A1.2 = c(4,5,6,2,3),
  A1.3 = c(3,3,4,2,1),
  A2.1 = c(3,4,5,2,4),
  A2.2 = c(4,5,5,4,2),
  A2.3 = c(1,1,2,2,3),
  A3.1 = c(1,3,4,5,6),
  A3.2 = c(1,4,3,3,4),
  A3.3 = c(4,4,4,4,5)
)
cor.mat <- cor(some.data)

哪个给：

            A1.1       A1.2       A1.3       A2.1       A2.2       A2.3        A3.1       A3.2       A3.3
A1.1  1.00000000 -0.4109975 -0.6155470 0.06839411 -0.5305954  0.9009862  1.00000000  0.7428336  0.6393620
A1.2 -0.41099747  1.0000000  0.8320503 0.83205029  0.6454972 -0.3779645 -0.41099747  0.0000000 -0.3535534
A1.3 -0.61554702  0.8320503  1.0000000 0.42307692  0.8951436 -0.6289709 -0.61554702 -0.3580574 -0.7844645
A2.1  0.06839411  0.8320503  0.4230769 1.00000000  0.1790287  0.1572427  0.06839411  0.3580574  0.1961161
A2.2 -0.53059545  0.6454972  0.8951436 0.17902872  1.0000000 -0.7319251 -0.53059545 -0.1666667 -0.9128709
A2.3  0.90098616 -0.3779645 -0.6289709 0.15724273 -0.7319251  1.0000000  0.90098616  0.4879500  0.8017837
A3.1  1.00000000 -0.4109975 -0.6155470 0.06839411 -0.5305954  0.9009862  1.00000000  0.7428336  0.6393620
A3.2  0.74283363  0.0000000 -0.3580574 0.35805744 -0.1666667  0.4879500  0.74283363  1.0000000  0.4564355
A3.3  0.63936201 -0.3535534 -0.7844645 0.19611614 -0.9128709  0.8017837  0.63936201  0.4564355  1.0000000

在我的原始数据中，某些列是相关的，这里用前缀（A1，A2，A3）表示。由于这些对我而言不感兴趣，因此我想将具有相同前缀的相关性设置为零，如下所示：

           A1.1       A1.2       A1.3       A2.1       A2.2       A2.3        A3.1       A3.2       A3.3
A1.1          0         0          0   0.06839411 -0.5305954  0.9009862  1.00000000  0.7428336  0.6393620
A1.2          0         0          0   0.83205029  0.6454972 -0.3779645 -0.41099747  0.0000000 -0.3535534
A1.3          0         0          0   0.42307692  0.8951436 -0.6289709 -0.61554702 -0.3580574 -0.7844645
A2.1  0.06839411  0.8320503  0.4230769          0          0          0  0.06839411  0.3580574  0.1961161
A2.2 -0.53059545  0.6454972  0.8951436          0          0          0 -0.53059545 -0.1666667 -0.9128709
A2.3  0.90098616 -0.3779645 -0.6289709          0          0          0  0.90098616  0.4879500  0.8017837
A3.1  1.00000000 -0.4109975 -0.6155470 0.06839411 -0.5305954  0.9009862           0          0          0
A3.2  0.74283363  0.0000000 -0.3580574 0.35805744 -0.1666667  0.4879500           0          0          0 
A3.3  0.63936201 -0.3535534 -0.7844645 0.19611614 -0.9128709  0.8017837           0          0          0

我可以使用for循环来做到这一点，但我想这样做可以比这容易得多吗？

Answer 1

一种选择是将数据从宽到长整形，使其包含三列

cor.mat_long <- reshape2::melt(cor.mat)
cor.mat_long
#   Var1 Var2       value
#1  A1.1 A1.1  1.00000000
#2  A1.2 A1.1 -0.41099747
#3  A1.3 A1.1 -0.61554702
#4  A2.1 A1.1  0.06839411
#5  A2.2 A1.1 -0.53059545
#6  A2.3 A1.1  0.90098616
#...

根据Var1和Var2的前缀创建逻辑向量，以指示这些前缀何时相同。使用此向量将cor.mat_long$value替换为0，其结果为TRUE

cor.mat_long$value[with(cor.mat_long, sub("\\.\\d+$", "", Var1) == sub("\\.\\d+$", "", Var2))] <- 0

最后再次调整为宽格式。

cor.mat2 <- reshape2::dcast(cor.mat_long, Var1 ~ Var2)
cor.mat2
#  Var1        A1.1       A1.2       A1.3       A2.1       A2.2       A2.3        A3.1       A3.2       A3.3
#1 A1.1  0.00000000  0.0000000  0.0000000 0.06839411 -0.5305954  0.9009862  1.00000000  0.7428336  0.6393620
#2 A1.2  0.00000000  0.0000000  0.0000000 0.83205029  0.6454972 -0.3779645 -0.41099747  0.0000000 -0.3535534
#3 A1.3  0.00000000  0.0000000  0.0000000 0.42307692  0.8951436 -0.6289709 -0.61554702 -0.3580574 -0.7844645
#4 A2.1  0.06839411  0.8320503  0.4230769 0.00000000  0.0000000  0.0000000  0.06839411  0.3580574  0.1961161
#5 A2.2 -0.53059545  0.6454972  0.8951436 0.00000000  0.0000000  0.0000000 -0.53059545 -0.1666667 -0.9128709
#6 A2.3  0.90098616 -0.3779645 -0.6289709 0.00000000  0.0000000  0.0000000  0.90098616  0.4879500  0.8017837
#7 A3.1  1.00000000 -0.4109975 -0.6155470 0.06839411 -0.5305954  0.9009862  0.00000000  0.0000000  0.0000000
#8 A3.2  0.74283363  0.0000000 -0.3580574 0.35805744 -0.1666667  0.4879500  0.00000000  0.0000000  0.0000000
#9 A3.3  0.63936201 -0.3535534 -0.7844645 0.19611614 -0.9128709  0.8017837  0.00000000  0.0000000  0.0000000

如果您不希望将Var1用作显式列，请

rownames(cor.mat2) <- cor.mat2$Var1
cor.mat2 <- cor.mat2[-1]

不知道这是否比循环容易得多。

Answer 2

我们可以乘以1的块对角矩阵

library(Matrix)
as.matrix(cor.mat * !bdiag(replicate(3, matrix(1, 3, 3), simplify = FALSE)))
#        A1.1       A1.2       A1.3       A2.1       A2.2       A2.3        A3.1       A3.2       A3.3
#A1.1  0.00000000  0.0000000  0.0000000 0.06839411 -0.5305954  0.9009862  1.00000000  0.7428336  0.6393620
#A1.2  0.00000000  0.0000000  0.0000000 0.83205029  0.6454972 -0.3779645 -0.41099747  0.0000000 -0.3535534
#A1.3  0.00000000  0.0000000  0.0000000 0.42307692  0.8951436 -0.6289709 -0.61554702 -0.3580574 -0.7844645
#A2.1  0.06839411  0.8320503  0.4230769 0.00000000  0.0000000  0.0000000  0.06839411  0.3580574  0.1961161
#A2.2 -0.53059545  0.6454972  0.8951436 0.00000000  0.0000000  0.0000000 -0.53059545 -0.1666667 -0.9128709
#A2.3  0.90098616 -0.3779645 -0.6289709 0.00000000  0.0000000  0.0000000  0.90098616  0.4879500  0.8017837
#A3.1  1.00000000 -0.4109975 -0.6155470 0.06839411 -0.5305954  0.9009862  0.00000000  0.0000000  0.0000000
#A3.2  0.74283363  0.0000000 -0.3580574 0.35805744 -0.1666667  0.4879500  0.00000000  0.0000000  0.0000000
#A3.3  0.63936201 -0.3535534 -0.7844645 0.19611614 -0.9128709  0.8017837  0.00000000  0.0000000  0.0000000

或者另一种选择是使用row/column索引

replace(cor.mat, cbind(rep(1:9, each = 3),
       c(sapply(list(1:3, 4:6, 7:9), rep, 3))), 0)

或使用outer来构建逻辑矩阵并与cor.mat

相乘

nm1 <- sub("\\.\\d+$", "", colnames(cor.mat))
cor.mat * outer(nm1, nm1, `!=`)

R：根据匹配的行和列名称从相关矩阵中删除项目

2 个答案: