R 汇总折叠数据。表

Question

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))

并希望实现一个新的数据表，例如：

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))

当 CAT、FOX 和 DOG 等于 1 时，dataWANT 按 SCHOOL、GRADE 和 SCHOOL X GRADE 取平均值。

我知道如何一次做到这一点，但这不利于使用大数据。

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])

---

请只使用这个：

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，

当 MEAN 由 GRADE 估计时，TYPE 等于 2，

当 MEAN 由 SCHOOL 和 GRADE 估计时，TYPE 等于 3

Answer 1

你的意思是得到这样的东西吗？

library(data.table)

melt(data, measure.vars = c('CAT', 'FOX', 'DOG'))[, 
        .(MEAN = mean(value, na.rm = TRUE)), .(School, Grade, variable)]

---

要按不同的列分组，我们可以这样做：

cols <- c('CAT', 'FOX', 'DOG')
data1 <- melt(data, measure.vars = cols)
list_cols <- list('School', 'Grade', c('School', 'Grade'))

lapply(list_cols, function(x)  
         data1[, .(MEAN = mean(value, na.rm = TRUE)), c(x, 'variable')])

Answer 2

我们可以在创建list后使用rbindlist ，方法是在melt数据集后获取MEAN （如另一篇文章中所述）

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

Answer 3

您可以先使用lapply(.SD,...)子集并计算您的均值，然后将其融合到您的 output 中：

melt(data[School != 0 | Grade != 0, lapply(.SD, mean), by = .(School, Grade)], id.vars = c("School", "Grade"))

添加此之后还会添加 TYPE 变量

...][, TYPE := School + (2*Grade)]

将它们放在一起并整理一下，它与您想要的 output 匹配

dataWANT <- melt(data[School != 0 | Grade != 0, lapply(.SD, mean), by = .(School, Grade)], id.vars = c("School", "Grade"))[, TYPE := School + (2*Grade)][order(variable, TYPE), .("VARIABLE" = variable, TYPE, "MEAN" = value)]