使用以下数据帧，我需要获得以下两个变量的每月总和：“CallsHandled”和“Engaged” 通过以下分组变量：“月”，“ID”，“位置”，“语言”，“MemRegion”
structure(list(Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Week = c(1L, 2L,
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L, 7L,
8L, 9L, 10L), ID = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("A1234",
"F1234"), class = "factor"), Location = structure(c(2L, 1L, 1L,
1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L,
1L), .Label = c("Corona", "Denver"), class = "factor"), LANGUAGE = structure(c(1L,
2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L,
2L, 2L, 1L), .Label = c("English", "Spanish"), class = "factor"),
MemRegion = structure(c(1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L,
1L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("NCAL",
"SCAL"), class = "factor"), CallsHandled = c(1L, 1L, 8L,
1L, 1L, 2L, 1L, 1L, 1L, 1L, 10L, 1L, 3L, 1L, 8L, 1L, 6L,
1L, 1L, 2L), Engaged = c(120L, 30L, 1243L, 75L, 45L, 55L,
200L, 120L, 30L, 230L, 2065L, 45L, 55L, 200L, 1483L, 30L,
1243L, 75L, 45L, 55L), QueueA = c(0L, 0L, 0L, 1L, 1L, 0L,
0L, 0L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L),
QueueB = c(1L, 0L, 1L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L,
0L, 0L, 1L, 1L, 0L, 1L, 0L, 1L), QueueC = c(0L, 1L, 0L, 0L,
0L, 0L, 1L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L,
0L)), .Names = c("Month", "Week", "ID", "Location", "LANGUAGE",
"MemRegion", "CallsHandled", "Engaged", "QueueA", "QueueB", "QueueC"
), class = "data.frame", row.names = c(NA, -20L))
此外，为了将“队列A：C”作为分组变量包含在内，我是否必须将它们组合成一个列？如果是这样，怎么样？
答案 0 :(得分：1)
因此，这个问题有2个部分，首先是如何对事物进行分组和总结，其次如何将队列A：C组合成一列。
对于第一个问题，您可以使用库dplyr，这样可以更轻松，更直观。
df <- structure(list(Month = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L),
Week = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L),
ID = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("A1234", "F1234"), class = "factor"),
Location = structure(c(2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 1L, 1L), .Label = c("Corona", "Denver"), class = "factor"),
LANGUAGE = structure(c(1L, 2L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 1L), .Label = c("English", "Spanish"), class = "factor"),
MemRegion = structure(c(1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("NCAL", "SCAL"), class = "factor"),
CallsHandled = c(1L, 1L, 8L, 1L, 1L, 2L, 1L, 1L, 1L, 1L, 10L, 1L, 3L, 1L, 8L, 1L, 6L, 1L, 1L, 2L),
Engaged = c(120L, 30L, 1243L, 75L, 45L, 55L, 200L, 120L, 30L, 230L, 2065L, 45L, 55L, 200L, 1483L, 30L, 1243L, 75L, 45L, 55L),
QueueA = c(0L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L),
QueueB = c(1L, 0L, 1L, 0L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 0L, 1L, 0L, 1L),
QueueC = c(0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 1L, 0L)), .Names = c("Month", "Week", "ID", "Location", "LANGUAGE", "MemRegion", "CallsHandled", "Engaged", "QueueA", "QueueB", "QueueC"),
class = "data.frame", row.names = c(NA, -20L))
library(dplyr)
df %>% group_by(Month, ID, Location, LANGUAGE) %>%
mutate(TotalCallsHandled = sum(CallsHandled),
TotalEngaged = sum(Engaged))
首先，我们使用group_by专门对您列出的变量进行分组，然后进行变异以对所有内容进行求和，这将直观地实现您的想法。
要将所有内容组合成一列，可能有很多方法可以做到这一点，但最直接的方法可能是为每列创建某种唯一标识符，并将所有列合并为一列。
df$Queue <- as.factor(df$QueueA + df$QueueB*2 + df$QueueC*3)
levels(df$Queue) <- c("A", "B", "C")
由于一切都应该是0,1标志，我们可以将标志重新创建为1 -> A，2 -> B，3 -> C，然后将因子重新分配为A，B，C再次。然后我们可以再次使用group_by函数来获得上述预期结果。
df %>% group_by(Month, ID, Location, LANGUAGE, Queue) %>%
mutate(TotalCallsHandled = sum(CallsHandled),
TotalEngaged = sum(Engaged)) %>%
select(-QueueA, -QueueB, -QueueC)
输出：
Source: local data frame [20 x 11]
Groups: Month, ID, Location, LANGUAGE, Queue
Month Week ID Location LANGUAGE MemRegion CallsHandled Engaged Queue
1 1 1 F1234 Denver English NCAL 1 120 B
2 1 2 F1234 Corona Spanish SCAL 1 30 C
3 1 3 F1234 Corona English NCAL 8 1243 B
4 1 4 F1234 Corona Spanish NCAL 1 75 A
5 1 5 F1234 Corona Spanish SCAL 1 45 A
6 1 6 F1234 Denver English SCAL 2 55 B
7 1 7 F1234 Corona English NCAL 1 200 C
8 1 8 F1234 Corona English NCAL 1 120 B
9 1 9 F1234 Denver English NCAL 1 30 A
10 1 10 F1234 Corona Spanish NCAL 1 230 C
11 1 1 A1234 Corona English NCAL 10 2065 C
12 1 2 A1234 Corona English SCAL 1 45 A
13 1 3 A1234 Corona Spanish NCAL 3 55 A
14 1 4 A1234 Corona English NCAL 1 200 A
15 1 5 A1234 Corona English SCAL 8 1483 B
16 1 6 A1234 Denver English SCAL 1 30 B
17 1 7 A1234 Corona Spanish SCAL 6 1243 C
18 1 8 A1234 Corona Spanish SCAL 1 75 B
19 1 9 A1234 Corona Spanish SCAL 1 45 C
20 1 10 A1234 Corona English SCAL 2 55 B
Variables not shown: TotalCallsHandled (int), TotalEngaged (int)
答案 1 :(得分：0)
要将Queue变量转换为单个因子变量，您可以这样做：
queues <- which(dat[ , c("QueueA", "QueueB", "QueueC")]==1, arr.ind=TRUE)
queues<-queues[
order(queues[,"row"]), "col"]
queues<-factor(queues, labels=c("QueueA", "QueueB", "QueueC"))
dat <- data.frame(dat, queues)
虽然，@ chappers的方法更好。
然后，您可以使用aggregate：
aggregate(dat[,c("CallsHandled", "Engaged")],
by=list(dat$Month, dat$ID, dat$Location, dat$LANGUAGE, dat$MemRegion, dat$queues),
sum)
# Group.1 Group.2 Group.3 Group.4 Group.5 Group.6 CallsHandled Engaged
#1 1 A1234 Corona English NCAL QueueA 1 200
#2 1 F1234 Denver English NCAL QueueA 1 30
#3 1 A1234 Corona Spanish NCAL QueueA 3 55
#4 1 F1234 Corona Spanish NCAL QueueA 1 75
#5 1 A1234 Corona English SCAL QueueA 1 45
#6 1 F1234 Corona Spanish SCAL QueueA 1 45
#7 1 F1234 Corona English NCAL QueueB 9 1363
#8 1 F1234 Denver English NCAL QueueB 1 120
#9 1 A1234 Corona English SCAL QueueB 10 1538
#10 1 A1234 Denver English SCAL QueueB 1 30
#11 1 F1234 Denver English SCAL QueueB 2 55
#12 1 A1234 Corona Spanish SCAL QueueB 1 75
#13 1 A1234 Corona English NCAL QueueC 10 2065
#14 1 F1234 Corona English NCAL QueueC 1 200
#15 1 F1234 Corona Spanish NCAL QueueC 1 230
#16 1 A1234 Corona Spanish SCAL QueueC 7 1288
#17 1 F1234 Corona Spanish SCAL QueueC 1 30
答案 2 :(得分：0)
@chappers解决方案正确聚合，但由于某些我无法弄清楚的原因，我留下了一堆重复的行。这适用于因子并减少实际数据帧中的行数（无重复）：
aggregate(cbind(CallsHandled,Engaged~Month + ID + Location + LANGUAGE + MemRegion, data=df, sum, na.rm=TRUE)