我的数据框很高：
data = data.frame("id"=c(1,2,3,4,5,6,7,8,9,10),
"group"=c(1,1,2,1,2,2,2,2,1,2),
"type"=c(1,1,2,3,2,2,3,3,3,1),
"score1"=c(sample(1:4,10,r=T)),
"score2"=c(sample(1:4,10,r=T)),
"score3"=c(sample(1:4,10,r=T)),
"score4"=c(sample(1:4,10,r=T)),
"score5"=c(sample(1:4,10,r=T)),
"weight1"=c(173,109,136,189,186,146,173,102,178,174),
"weight2"=c(147,187,125,126,120,165,142,129,144,197),
"weight3"=c(103,192,102,159,128,179,195,193,135,145),
"weight4"=c(114,182,199,101,111,116,198,123,119,181),
"weight5"=c(159,125,104,171,166,154,197,124,180,154))
library(reshape2)
library(plyr)
data1 <- reshape(data, direction = "long",
varying = list(c(paste0("score",1:5)),c(paste0("weight",1:5))),
v.names = c("score","weight"),
idvar = "id", timevar = "count", times = c(1:5))
data1 <- data1[order(data1$id), ]
我要创建的是一个新的数据框，如下所示：
want = data.frame("score"=rep(1:4,6),
"group"=rep(1:2,12),
"type"=rep(1:3,8),
"weightedCOUNT"=NA) # how to calculate this? count(data1, score, wt = weight)
我只是不确定如何计算应将权重应用于分数变量的weightedCOUNT，因此它会在“ weightedCOUNT”列中提供按分数，组和类型汇总的加权计数。
答案 0 :(得分：1)
一个选择是melt（来自data.table-可能需要多个measure patterns，然后按“ group”分组，“ type”得到{ {1}}
count如果我们需要library(data.table)
library(dplyr)
melt(setDT(data), measure = patterns('^score', "^weight"),
value.name = c("score", "weight")) %>%
group_by(group, type) %>%
count(score, wt = weight)
组组合
complete答案 1 :(得分：0)
如果我理解正确，weightedCOUNT是按score，group和type分组的权重之和。
为了完整起见，我想展示一下accepted solution在以纯基本R和纯data.table语法实现时的外观。
OP快要出现了。他已经将data从宽格式改写为长格式，以处理多个值变量。仅缺少最后的聚合步骤：
data1 <- reshape(data, direction = "long",
varying = list(c(paste0("score",1:5)),c(paste0("weight",1:5))),
v.names = c("score","weight"),
idvar = "id", timevar = "count", times = c(1:5))
result <- aggregate(weight ~ score + group + type, data1, FUN = sum)
result
score group type weight 1 1 1 1 479 2 3 1 1 558 3 4 1 1 454 4 1 2 1 378 5 2 2 1 154 6 3 2 1 174 7 4 2 1 145 8 1 2 2 535 9 2 2 2 855 10 3 2 2 248 11 4 2 2 499 12 1 1 3 189 13 2 1 3 351 14 3 1 3 600 15 4 1 3 362 16 1 2 3 596 17 2 2 3 265 18 3 2 3 193 19 4 2 3 522
result可以通过以下方式重新排序
with(result, result[order(score, group, type), ])
score group type weight 1 1 1 1 479 12 1 1 3 189 4 1 2 1 378 8 1 2 2 535 16 1 2 3 596 13 2 1 3 351 5 2 2 1 154 9 2 2 2 855 17 2 2 3 265 2 3 1 1 558 14 3 1 3 600 6 3 2 1 174 10 3 2 2 248 18 3 2 3 193 3 4 1 1 454 15 4 1 3 362 7 4 2 1 145 11 4 2 2 499 19 4 2 3 522
data.table 如akrun所示，可以将melt()软件包中的data.table与dplyr结合使用。另外，我们可以继续使用data.table语法进行聚合：
library(data.table)
cols <- c("score", "weight") # to save typing
melt(setDT(data), measure = patterns(cols), value.name = cols)[
, .(weightedCOUNT = sum(weight)), keyby = .(score, group, type)]
score group type weightedCOUNT 1: 1 1 1 479 2: 1 1 3 189 3: 1 2 1 378 4: 1 2 2 535 5: 1 2 3 596 6: 2 1 3 351 7: 2 2 1 154 8: 2 2 2 855 9: 2 2 3 265 10: 3 1 1 558 11: 3 1 3 600 12: 3 2 1 174 13: 3 2 2 248 14: 3 2 3 193 15: 4 1 1 454 16: 4 1 3 362 17: 4 2 1 145 18: 4 2 2 499 19: 4 2 3 522
keyby参数用于一步一步对输出进行分组和排序。
使用交叉连接函数data.table，也可以用CJ()语法来完成分组变量组合的缺失：
melt(setDT(data), measure = patterns(cols), value.name = cols)[
, .(weightedCOUNT = sum(weight)), keyby = .(score, group, type)][
CJ(score, group, type, unique = TRUE), on = .(score, group, type)][
is.na(weightedCOUNT), weightedCOUNT := 0][]
score group type weightedCOUNT 1: 1 1 1 479 2: 1 1 2 0 3: 1 1 3 189 4: 1 2 1 378 5: 1 2 2 535 6: 1 2 3 596 7: 2 1 1 0 8: 2 1 2 0 9: 2 1 3 351 10: 2 2 1 154 11: 2 2 2 855 12: 2 2 3 265 13: 3 1 1 558 14: 3 1 2 0 15: 3 1 3 600 16: 3 2 1 174 17: 3 2 2 248 18: 3 2 3 193 19: 4 1 1 454 20: 4 1 2 0 21: 4 1 3 362 22: 4 2 1 145 23: 4 2 2 499 24: 4 2 3 522 score group type weightedCOUNT