r-为多变量分类数据创建比例矩阵

Question

假设我从下面的R代码中模拟了这些数据：

library(RNGforGPD)
set.seed(1)
sample.size = 10; no.gpois = 3
lambda.vec = c(-0.2, 0.2, -0.3); theta.vec = c(1, 3, 4)
M = c(0.352, 0.265, 0.342); N = diag(3); N[lower.tri(N)] = M
TV = N + t(N); diag(TV) = 1
cstar = CmatStarGpois(TV, theta.vec, lambda.vec, verbose = TRUE)
data = GenMVGpois(sample.size, no.gpois, cstar, theta.vec, lambda.vec, details = FALSE)
> prop.table(table(data[,1]))
0   1   2 
0.3 0.4 0.3 
> prop.table(table(data[,2]))
2   3   6   8  10 
0.2 0.4 0.1 0.2 0.1 
> prop.table(table(data[,3]))
2   3   4   5   6 
0.2 0.3 0.1 0.3 0.1 
> table(data)
data
0  1  2  3  4  5  6  8 10 
3  4  7  7  1  3  2  2  1 

我想为三个分类变量中的每一个创建一个比例矩阵。如果某个特定列缺少该类别，则会将其标识为0。

>

Cat	X1	X2	X3
0	0.3	0.0
1	0.4	0.0	0.0
2	0.3	0.2	0.2
3	0.0	0.4	0.3
4	0.0	0.0		0.1
5	0.0	0.0	0.3
6	0.0	0.1	0.1
8	0.0	0.2	0.0
10	0.0	0.1	0.0

尝试将逻辑放在代码序列中的适当位置。

props <- data.frame(Cat = sort(unique(c(data))) )  # Just the Cat column
#Now fill in the entries
# the entries will be obtained with table function
apply(data, 2, table)  # run `table(.)` over the columns individually
[[1]]
0 1 2   # these are actually character valued names
3 4 3   # while these are the count values
[[2]]
2  3  6  8 10 
2  4  1  2  1 
[[3]]
2 3 4 5 6 
2 3 1 3 1 

现在对该列表进行迭代，以填充与Cat列匹配的值：

props2 <- cbind(props,  # using dfrm first argument returns dataframe object 
lapply( apply(data, 2, table) ,   # irregular results are a list
function(col) { # first make a named vector of zeros
x <- setNames(rep(0,length(props$Cat)), props$Cat)
# could have skipped that step by using `tabulate`
# then fill with values using names as indices
x[names(col)] <- col # values to matching names
x}) )
props2
#-------------
Cat V1 V2 V3
0    0  3  0  0
1    1  4  0  0
2    2  3  2  2
3    3  0  4  3
4    4  0  0  1
5    5  0  0  3
6    6  0  1  1
8    8  0  2  0
10  10  0  1  0
#---
# now just "proportionalize" those counts
props2[2:4] <- prop.table(data.matrix(props2[2:4]), margin=2)
props2
#-------------
Cat  V1  V2  V3
0    0 0.3 0.0 0.0
1    1 0.4 0.0 0.0
2    2 0.3 0.2 0.2
3    3 0.0 0.4 0.3
4    4 0.0 0.0 0.1
5    5 0.0 0.0 0.3
6    6 0.0 0.1 0.1
8    8 0.0 0.2 0.0
10  10 0.0 0.1 0.0

colnames(data) <- c("X1", "X2", "X3")
as_tibble(data) %>%
pivot_longer(cols = "X1":"X3", values_to = "Cat") %>%
group_by(name, Cat) %>%
count() %>%
ungroup(Cat) %>%
summarize(name, Cat, proportion = n / sum(n)) %>%
pivot_wider(names_from = name, values_from = proportion) %>%
arrange(Cat) %>%
replace(is.na(.), 0)
# A tibble: 9 × 4
Cat    X1    X2    X3
<dbl> <dbl> <dbl> <dbl>
1     0   0.3   0     0  
2     1   0.4   0     0  
3     2   0.3   0.2   0.2
4     3   0     0.4   0.3
5     4   0     0     0.1
6     5   0     0     0.3
7     6   0     0.1   0.1
8     8   0     0.2   0  
9    10   0     0.1   0 

如果你想把它作为一个矩阵，你可以使用as.matrix()