R汇总折叠的数据表 [英] R Summarize Collapsed Data.Table

问题描述

我有这样的数据

    data=data.table("School"=c(1,1,1,1,1,1,0,1,0,0,1,1,1,0,1,0,1,1,1,1,1,0,0,1,0,1,1,1,1,1,1,0,1,0,1,0),
    "Grade"=c(0,1,1,1,0,0,0,1,1,1,0,1,1,0,0,1,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,0,0,0,1,0),
    "CAT"=c(1,0,1,1,0,1,0,1,1,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,1,1,1,1),
    "FOX"=c(1,1,0,1,1,1,1,1,0,0,0,1,1,1,0,0,1,1,1,1,1,1,1,0,1,1,0,0,1,0,0,1,0,0,1,0),
    "DOG"=c(0,0,0,1,0,0,1,0,0,1,0,1,1,1,0,1,1,0,0,1,1,0,0,1,0,1,1,0,1,0,1,1,1,0,1,1))

，并希望建立一个新的数据表，例如：

and wish to achieve a new data table such as this:

dataWANT=data.frame("VARIABLE"=c('CAT', 'CAT', 'CAT', 'FOX', 'FOX', 'FOX', 'DOG', 'DOG', 'DOG'),
"SCHOOL"=c(1, 1, 0, 1, 1, 0, 1, 1, 0),
"GRADE"=c(0, 1, 1, 0, 1, 1, 0, 1, 1),
"MEAN"=c(NA))

dataWANT用等于1的SCHOOL，GRADE和SCHOOL X GRADE取CAT和FOX和DOG的平均值。

dataWANT takes the mean for CAT and FOX and DOG by SCHOOL, GRADE, and SCHOOL X GRADE when they are equal to 1.

我知道一次如何执行此操作，但这不利于处理大数据。

I know how to do this one at a time but that is not good for doing this with a big data.

data[, CAT1:=mean(CAT), by=list(SCHOOL)]
data[, FOX1:=mean(FOX), by=list(GRADE)]
data[, DOG1:=mean(DOG), by=list(SCHOOL, GRADE)]

data$CAT2 = unique(data[SCHOOL==1, CAT1])
data$FOX2 = unique(data[GRADE==1, FOX1])
data$DOG2 = unique(data[SCHOOL==1 & GRADE==1, DOG1])

---

仅使用此命令：

---

Please only use this:

data=data.table("SCHOOL"=c(1,1,1,1,1,1,0,1,0,0,1,1,1,0,1,0,1,1,1,1,1,0,0,1,0,1,1,1,1,1,1,0,1,0,1,0),
                "GRADE"=c(0,1,1,1,0,0,0,1,1,1,0,1,1,0,0,1,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,0,0,0,1,0),
                "CAT"=c(1,0,1,1,0,1,0,1,1,0,1,0,0,1,0,1,0,0,0,0,0,0,1,0,0,1,1,0,0,1,1,0,1,1,1,1),
                "FOX"=c(1,0,0,1,1,1,1,1,0,0,0,1,1,1,0,0,1,1,1,1,1,1,1,0,1,1,0,0,1,0,0,1,0,0,1,0),
                "DOG"=c(0,0,0,1,0,0,1,0,0,1,0,1,1,1,0,1,1,0,0,1,1,0,0,1,0,1,1,0,1,0,1,1,1,0,1,1))


data[, CAT1:=mean(CAT), by=list(SCHOOL)]
data[, CAT2:=mean(CAT), by=list(GRADE)]
data[, CAT3:=mean(CAT), by=list(SCHOOL, GRADE)]

data[, FOX1:=mean(FOX), by=list(SCHOOL)]
data[, FOX2:=mean(FOX), by=list(GRADE)]
data[, FOX3:=mean(FOX), by=list(SCHOOL, GRADE)]

data[, DOG1:=mean(DOG), by=list(SCHOOL)]
data[, DOG2:=mean(DOG), by=list(GRADE)]
data[, DOG3:=mean(DOG), by=list(SCHOOL, GRADE)]

dataWANT=data.frame("VARIABLE"=c('CAT','CAT','CAT','FOX','FOX','FOX','DOG','DOG','DOG'),
                    "TYPE"=c(1,2,3,1,2,3,1,2,3),
                    "MEAN"=c(0.48,0.44,0.428,0.6,0.611,0.6428,0.52,0.61,0.6428))

其中，当SCHOOL估计MEAN时，TYPE等于1，

where TYPE equals to 1 when MEAN in estimated by SCHOOL,

TYPE等于2，

TYPE等于3

推荐答案

我们可以在创建列表后使用 rbindlist ，方法是在融化数据集后获取 MEAN （与另一篇文章一样）

We could use rbindlist after creating a list by taking the MEAN after melting the dataset (as in the other post)

library(data.table)
cols <- c('CAT', 'FOX', 'DOG')
data1 <- melt(data, measure.vars = cols)
list_cols <- list('SCHOOL', 'GRADE', c('SCHOOL', 'GRADE'))
lst1 <- lapply(list_cols, function(x)  
       data1[, .(MEAN = mean(value, na.rm = TRUE)), c(x, 'variable')])
rbindlist(lapply(lst1, function(x)  {
     nm1 <- setdiff(names(x), c('variable', 'MEAN'))
     x[Reduce(`&`, lapply(mget(nm1), as.logical)),
     .(VARIABLE = variable, MEAN)]}), idcol = 'TYPE')[order(VARIABLE)]
#   TYPE VARIABLE      MEAN
#1:    1      CAT 0.4800000
#2:    2      CAT 0.4444444
#3:    3      CAT 0.4285714
#4:    1      FOX 0.6000000
#5:    2      FOX 0.5555556
#6:    3      FOX 0.6428571
#7:    1      DOG 0.5200000
#8:    2      DOG 0.6111111
#9:    3      DOG 0.6428571

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