如何解决加权比例
如果可能，我想使用 prop.table() 和 xtabs() 函数计算加权比例。
示例：
library(data.table)
set.seed(1)
n <- 20
X <- data.table(color = sample(c("Blue","Green","Red"),n,replace = TRUE),letter = sample(c("a","b"),number = sample(1:3,weight = runif(n))
X[,weight := weight/sum(weight)]
# > X
# color letter number weight
# 1: Blue a 2 0.03300399
# 2: Red a 3 0.08170773
# 3: Blue a 2 0.03374477
# 4: Green a 1 0.03248830
# 5: Blue a 3 0.04636611
# 6: Red b 2 0.08684298
# 7: Red a 1 0.08413131
# 8: Green a 1 0.03796001
# 9: Green b 3 0.07566126
# 10: Red b 2 0.09350268
# 11: Red b 2 0.04230800
# 12: Blue a 3 0.06935330
# 13: Blue b 3 0.03893384
# 14: Blue a 2 0.03166846
# 15: Green a 2 0.07369181
# 16: Green b 2 0.01972925
# 17: Green a 2 0.06921767
# 18: Green b 1 0.01184500
# 19: Red b 2 0.02389486
# 20: Blue b 2 0.01394867
P <- X[,prop.table(xtabs(~color + letter + number),c("color","letter"))]
P <- data.table(P)
setnames(P,"N","percentage")
# > P
# color letter number percentage
# 1: Blue a 1 0.0000000
# 2: Green a 1 0.5000000
# 3: Red a 1 0.5000000
# 4: Blue b 1 0.0000000
# 5: Green b 1 0.3333333
# 6: Red b 1 0.0000000
# 7: Blue a 2 0.6000000
# 8: Green a 2 0.5000000
# 9: Red a 2 0.0000000
# 10: Blue b 2 0.5000000
# 11: Green b 2 0.3333333
# 12: Red b 2 1.0000000
# 13: Blue a 3 0.4000000
# 14: Green a 3 0.0000000
# 15: Red a 3 0.5000000
# 16: Blue b 3 0.5000000
# 17: Green b 3 0.3333333
# 18: Red b 3 0.0000000
P[,sum(percentage),.(color,letter)]
# > P[,letter)]
# color letter V1
# 1: Blue a 1
# 2: Green a 1
# 3: Red a 1
# 4: Blue b 1
# 5: Green b 1
# 6: Red b 1
如您所见，这些是未加权的比例。我想根据我的 weight 变量对它们进行加权。最终，百分比仍应加到 1。我知道使用 prop.table() 和 xtabs() 两个函数可能无法实现此结果，因此如果有其他解决方案也可以。
解决方法
使用data.table，我们通过'color'、'letter'、'number'得到'weight'的sum，然后通过'color'、'letter'进行分组并创建通过将 'n' 与 'n' 的 sum 相除
library(data.table)
X[,.(n = sum(weight)),.(color,letter,number)][,prop := n/sum(n),letter)][order(color)]
-输出
# color letter number n prop
# 1: Blue a 2 0.09841723 0.4596001
# 2: Blue a 3 0.11571941 0.5403999
# 3: Blue b 3 0.03893384 0.7362328
# 4: Blue b 2 0.01394867 0.2637672
# 5: Green a 1 0.07044831 0.3301886
# 6: Green b 3 0.07566126 0.7055616
# 7: Green a 2 0.14290947 0.6698114
# 8: Green b 2 0.01972925 0.1839805
# 9: Green b 1 0.01184500 0.1104578
#10: Red a 3 0.08170773 0.4926930
#11: Red b 2 0.24654852 1.0000000
#12: Red a 1 0.08413131 0.5073070
或使用 dplyr
library(dplyr)
X %>%
count(color,number,wt = weight) %>%
group_by(color,letter) %>%
mutate(prop = n/sum(n))
# A tibble: 12 x 5
# Groups: color,letter [6]
# color letter number n prop
# <chr> <chr> <int> <dbl> <dbl>
# 1 Blue a 2 0.0984 0.460
# 2 Blue a 3 0.116 0.540
# 3 Blue b 2 0.0139 0.264
# 4 Blue b 3 0.0389 0.736
# 5 Green a 1 0.0704 0.330
# 6 Green a 2 0.143 0.670
# 7 Green b 1 0.0118 0.110
# 8 Green b 2 0.0197 0.184
# 9 Green b 3 0.0757 0.706
#10 Red a 1 0.0841 0.507
#11 Red a 3 0.0817 0.493
#12 Red b 2 0.247 1
如果我们想要所有组合，请使用 complete
library(tidyr)
X %>%
count(color,letter) %>%
mutate(prop = n/sum(n)) %>%
ungroup %>%
complete(color,fill = list(n = 0,prop = 0))
# A tibble: 18 x 5
# color letter number n prop
# <chr> <chr> <int> <dbl> <dbl>
# 1 Blue a 1 0 0
# 2 Blue a 2 0.0984 0.460
# 3 Blue a 3 0.116 0.540
# 4 Blue b 1 0 0
# 5 Blue b 2 0.0139 0.264
# 6 Blue b 3 0.0389 0.736
# 7 Green a 1 0.0704 0.330
# 8 Green a 2 0.143 0.670
# 9 Green a 3 0 0
#10 Green b 1 0.0118 0.110
#11 Green b 2 0.0197 0.184
#12 Green b 3 0.0757 0.706
#13 Red a 1 0.0841 0.507
#14 Red a 2 0 0
#15 Red a 3 0.0817 0.493
#16 Red b 1 0 0
#17 Red b 2 0.247 1
#18 Red b 3 0 0
或 CJ 中的 data.table
X[,letter)][order(color)][
CJ(color,unique = TRUE),on = .(color,number)]
或者使用 xtabs/prop.table
setnames(as.data.table(X[,prop.table(xtabs(weight~color +
letter + number),c("color","letter"))])[N > 0][order(color)],"N","prop")[]
# color letter number prop
# 1: Blue a 2 0.4596001
# 2: Blue b 2 0.2637672
# 3: Blue a 3 0.5403999
# 4: Blue b 3 0.7362328
# 5: Green a 1 0.3301886
# 6: Green b 1 0.1104578
# 7: Green a 2 0.6698114
# 8: Green b 2 0.1839805
# 9: Green b 3 0.7055616
#10: Red a 1 0.5073070
#11: Red b 2 1.0000000
#12: Red a 3 0.4926930
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