r语言 - 组合两个循环结构得到一个矩阵输出

Question

我在r中使用两个密切相关的公式，我想知道是否有可能将B1和B2结合起来以获得我的所需的矩阵输出所示?

z <- "group    y1    y2
1 1         2     3
2 1         3     4
3 1         5     4
4 1         2     5
5 2         4     8
6 2         5     6
7 2         6     7
8 3         7     6
9 3         8     7
10 3        10     8
11 3         9     5
12 3         7     6"
dat <- read.table(text = z, header = T)
(B1 = Reduce("+", group_split(dat, group, .keep = FALSE) %>%
map(~ nrow(.)*(colMeans(.)-colMeans(dat[-1]))^2)))
#     y1       y2 
#61.86667 19.05000
(B2 = Reduce("+",group_split(dat, group, .keep = FALSE) %>%
map(~ nrow(.)*prod(colMeans(.)-colMeans(dat[-1])))))
# 24.4

期望矩阵输出:

matrix(c(61.87,24.40,24.40,19.05),2)
#      [,1]  [,2]
#[1,] 61.87 24.40
#[2,] 24.40 19.05

Answer 1

也许像这样?

mat <- matrix(B2, length(B1), length(B1))
diag(mat) <- B1
mat
#      [,1]  [,2]
#[1,] 61.87 24.40
#[2,] 24.40 19.05

Answer 2

我们也可以在单个链中执行此操作，而无需重新计算。与Reduce中的+相比，使用sum的优点之一是它可以考虑na.rm参数的缺失值，而如果在执行+时存在任何NA，则由于NA

的属性，它返回NA。

library(dplyr)
dat %>% 
# // group by group
group_by(group) %>%
# // create a count column 'n' 
summarise(n = n(), 
# // loop across y1, y2, get the difference between the grouped 
# // column  mean value and the full data column mean
across(c(y1, y2), ~ (mean(.) - mean(dat[[cur_column()]]))),
.groups = 'drop') %>% 
# // create the columns by multiplying the output of y1, y2 with n        
transmute(y1y2 = y1 * y2 * n, 
# //  Get the raised power of y1, y2, and multiply with n
across(c(y1, y2), list(new1 = ~ n * .^2))) %>%
# // then do a columnwise sum, replicate the 'y1y2' clumn
summarise(across(everything(), sum, na.rm = TRUE), y1y2new = y1y2) %>% 
# // rearrange the column order
select(c(2, 1, 4, 3)) %>% 
# // unlist to a vector
unlist %>%
# // create a matrix with 2 rows, 2 columns
matrix(2, 2)
#         [,1]  [,2]
#[1,] 61.86667 24.40
#[2,] 24.40000 19.05