英文:
Insert matrix values based on exact row/column names
问题
I want to insert the information from the cells in a given subsets into the blank matrix based on the combination of row/column name.
以下是要翻译的部分。
英文:
I have a blank matrix with row and column names that is relatively large (6000 times 200). Then, I have multiple matrices which represent subsets of this matrix (typically around 1500 times 120) and also contain row and column names. I want to insert the information from the cells in a given subsets into the blank matrix based on the combination of row/column name.
Here is an example of the data and the desired output:
adj1 <- matrix(0, 3, 4)
colnames(adj1) <- c("g1", "g2", "g3", "g4")
rownames(adj1) <- c("Tim", "John", "Sarah")
adj2 <- matrix(0, 2, 2)
colnames(adj2) <- c("g1", "g4")
rownames(adj2) <- c("Tim", "John")
adj2[1,2] <- 1
adj2
combined_adj <- matrix(0,3,4)
colnames(combined_adj) <- c("g1","g2","g3","g4")
rownames(combined_adj) <- c("Tim","John", "Sarah")
combined_adj[1,4] <- 1
combined_adj
I have somewhat managed to intersect by column names, but not by the exact combination of row/col names.
答案1
得分: 2
以下是您要求的代码部分的翻译：
这是一个允许您按行和列名称替换值的函数。
它的参数包括：
x较大的矩阵;y具有新值的矩阵;z目标矩阵; 如果缺少，则将创建一个具有x的维度和行/列名称的零矩阵。
matreplace <- function(x, y, z){
if(missing(z)) {
z <- x
z[] <- 0
}
ij <- Map(intersect, dimnames(z), dimnames(y))
i <- ij[[1L]]
j <- ij[[2L]]
z[i, j] <- y[i, j]
z
}
adj1 <- matrix(0, 3, 4)
colnames(adj1) <- c("g1", "g2", "g3", "g4")
rownames(adj1) <- c("Tim", "John", "Sarah")
adj2 <- matrix(0, 2, 2)
colnames(adj2) <- c("g1", "g4")
rownames(adj2) <- c("Tim", "John")
adj2[1,2] <- 1
combined_adj <- matreplace(adj1, adj2)
adj2[2, 1] <- pi
matreplace(adj1, adj2, combined_adj)
#> g1 g2 g3 g4
#> Tim 0.000000 0 0 1
#> John 3.141593 0 0 0
#> Sarah 0.000000 0 0 0
在2023-05-07使用reprex v2.0.2创建
英文:
Here is a function that allows you to replace values by row and column names.
Its arguments are
xthe larger matrix;ythe matrix with new values;zthe target matrix; if missing a new matrix of zeros will be created with the dimensions and row/col names ofx.
matreplace <- function(x, y, z){
if(missing(z)) {
z <- x
z[] <- 0
}
ij <- Map(intersect, dimnames(z), dimnames(y))
i <- ij[[1L]]
j <- ij[[2L]]
z[i, j] <- y[i, j]
z
}
adj1 <- matrix(0, 3, 4)
colnames(adj1) <- c("g1", "g2", "g3", "g4")
rownames(adj1) <- c("Tim", "John", "Sarah")
adj2 <- matrix(0, 2, 2)
colnames(adj2) <- c("g1", "g4")
rownames(adj2) <- c("Tim", "John")
adj2[1,2] <- 1
combined_adj <- matreplace(adj1, adj2)
adj2[2, 1] <- pi
matreplace(adj1, adj2, combined_adj)
#> g1 g2 g3 g4
#> Tim 0.000000 0 0 1
#> John 3.141593 0 0 0
#> Sarah 0.000000 0 0 0
<sup>Created on 2023-05-07 with reprex v2.0.2</sup>
答案2
得分: 2
以下是您要翻译的内容：
这里有另外两种替代方法：
# 创建空矩阵
adj1 <- matrix(0, 3, 4)
colnames(adj1) <- c("g1", "g2", "g3", "g4")
rownames(adj1) <- c("Tim", "John", "Sarah")
# 创建子集矩阵
adj2 <- matrix(0, 2, 2)
colnames(adj2) <- c("g1", "g4")
rownames(adj2) <- c("Tim", "John")
# 更改和测试
adj2[1,2] <- 1
adj2[2, 1] <- pi
for 循环
# 解决方案 1 使用 for 循环
# 根据行和列名称在空白矩阵中填入值
for (i in rownames(adj2)) {
for (j in colnames(adj2)) {
adj1[i,j] <- adj2[i,j]
}
}
adj1
match()
# 解决方案 2 使用 `match()`
row_idx <- match(rownames(adj2), rownames(adj1))
col_idx <- match(colnames(adj2), colnames(adj1))
adj1[row_idx, col_idx] <- adj2
adj1
输出
g1 g2 g3 g4
Tim 0.000000 0.000000 0 1.000000
John 3.141593 0.000000 0 0.000000
Sarah 0.000000 0.000000 0 0.000000
<details>
<summary>英文:</summary>
Here are two more alternatives:
Create blank matrix
adj1 <- matrix(0, 3, 4)
colnames(adj1) <- c("g1", "g2", "g3", "g4")
rownames(adj1) <- c("Tim", "John", "Sarah")
Create subset matrix
adj2 <- matrix(0, 2, 2)
colnames(adj2) <- c("g1", "g4")
rownames(adj2) <- c("Tim", "John")
change and test
adj2[1,2] <- 1
adj2[2, 1] <- pi
### for loop
solution 1 for loop
Fill in values in blank matrix based on row and column names
for (i in rownames(adj2)) {
for (j in colnames(adj2)) {
adj1[i,j] <- adj2[i,j]
}
}
adj1
### match()
solution 2 using match()
row_idx <- match(rownames(adj2), rownames(adj1))
col_idx <- match(colnames(adj2), colnames(adj1))
adj1[row_idx, col_idx] <- adj2
adj1
### output
g1 g2 g3 g4
Tim 0.000000 0 0 1
John 3.141593 0 0 0
Sarah 0.000000 0 0 0
</details>
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