英文:
Create a diagonal matrix from a dataframe in r
问题
我有这个数据框：
data.frame(Name = c("A", "B"),Value = c(1,2))
我想得到一个矩阵，其中Name值作为矩阵的行名和列名，值作为以下减法乘积的结果：
又被称为
英文:
I have this data frame:
data.frame(Name = c("A", "B"),Value = c(1,2))
and I would like to arrive at a matrix with Name values as the matrix's row names and column names and values as the product of a subtraction like the following
AKA
答案1
得分: 3
我假设您想要使用矩阵运算来完成，所以这是我的努力。
dd <- data.frame(Name = c("A", "B"),Value = c(1,2))M <- matrix( dd$Value, 2,2, dimnames=list( dd$Name, dd$Name))M# A B#A 1 1#B 2 2M - t(M)# A B#A 0 -1#B 1 0
如果您需要进一步的帮助，请告诉我。
英文:
I assumed you wanted it done with matrix operations, so here's my effort.
dd <- data.frame(Name = c("A", "B"),Value = c(1,2))M <- matrix( dd$Value, 2,2, dimnames=list( dd$Name, dd$Name))M# A B#A 1 1$B 2 2M - t(M)# A B#A 0 -1#B 1 0
答案2
得分: 2
你可以使用函数 outer -
df <- data.frame(Name = c("A", "B"), Value = c(1,2))mat <- t(outer(df$Value, df$Value, `-`))dimnames(mat) <- list(df$Name, df$Name)mat# A B#A 0 1#B -1 0
英文:
You can use the function outer -
df <- data.frame(Name = c("A", "B"), Value = c(1,2))mat <- t(outer(df$Value, df$Value, `-`))dimnames(mat) <- list(df$Name, df$Name)mat# A B#A 0 1#B -1 0
答案3
得分: 1
A tidyverse solution:
df <- data.frame(name = c("A", "B"), value = c(1,2))df |>bind_cols(df |> pivot_wider()) |>transmute(across(A:B, \(x) x - value))A B1 0 12 -1 0
*注意，此示例中的 name 和 value 是小写。如果它们是大写的，或者存在其他列名，可以在 pivot_wider() 中使用 names_from 和 values_from 参数。
示例：pivot_wider(names_from = 'Name', values_from = "Value").
英文:
A tidyverse solution:
df <- data.frame(name = c("A", "B"), value = c(1,2))df |>bind_cols(df |> pivot_wider()) |>transmute(across(A:B, \(x) x - value))A B1 0 12 -1 0
*Note name and value are lowercase in this example. If they are capitalized or if there are other column names, use the names_from and values_from args in pivot_wider().
Example: pivot_wider(names_from = 'Name', values_from = "Value").
答案4
得分: 1
使用 col() 进行索引的另一种方法：
mat <- array(df$Value, rep(nrow(df), 2), dimnames = list(df$Name, df$Name))df$Value[col(mat)] - mat# A B# A 0 1# B -1 0
请注意，我已经删除了代码部分并提供了翻译的内容。
英文:
An alternative approach with col() for indexing:
mat <- array(df$Value, rep(nrow(df), 2), dimnames = list(df$Name, df$Name))df$Value[col(mat)] - mat# A B# A 0 1# B -1 0
答案5
得分: 1
这只是吗？
col(df) - df$value[,1] [,2][1,] 0 1[2,] -1 0with(df, structure(col(df) - value, .Dimnames = list(name, name)))A BA 0 1B -1 0
英文:
Isnt this just?
col(df) - df$value[,1] [,2][1,] 0 1[2,] -1 0with(df, structure(col(df) - value, .Dimnames = list(name, name)))A BA 0 1B -1 0
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