我想为每个Reg添加一行总和，如此
Reg Res Pop
1 Total 1000915
2 A Urban 500414
3 A Rural 500501
4 Total 999938
5 B Urban 499922
6 B Rural 500016
7 Total 1000912
8 C Urban 501638
9 C Rural 499274
10 Total 999629
11 D Urban 499804
12 D Rural 499825
13 Total 1000303
14 E Urban 499917
15 E Rural 500386
MWE如下：
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop)
df
Reg Res Pop
1 A Urban 500414
2 A Rural 500501
3 B Urban 499922
4 B Rural 500016
5 C Urban 501638
6 C Rural 499274
7 D Urban 499804
8 D Rural 499825
9 E Urban 499917
10 E Rural 500386
df %>%
group_by(Reg) %>%
summarise(Total = sum(Pop))
# A tibble: 5 x 2
Reg Total
<fctr> <int>
1 A 1000915
2 B 999938
3 C 1000912
4 D 999629
5 E 1000303
被修改
我希望同时拥有dplyr和data.table解决方案。
答案 0 :(得分：3)
lapply(split(df, df$Reg),
function(a) rbind(data.frame(Reg = a$Reg[1],
Res = "Total",
Pop = sum(a$Pop)),
a))
$A
Reg Res Pop
1 A Total 1000915
2 A Urban 500414
3 A Rural 500501
$B
Reg Res Pop
1 B Total 999938
3 B Urban 499922
4 B Rural 500016
$C
Reg Res Pop
1 C Total 1000912
5 C Urban 501638
6 C Rural 499274
$D
Reg Res Pop
1 D Total 999629
7 D Urban 499804
8 D Rural 499825
$E
Reg Res Pop
1 E Total 1000303
9 E Urban 499917
10 E Rural 500386
如果您需要
，可以使用do.call(rbind, ...)将整个内容转换为data.frame
答案 1 :(得分：3)
您可以在摘要中添加额外的 Res 列，然后在原始数据框中添加bind_rows：
df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop), Res = 'Total') %>%
bind_rows(df) %>%
arrange(Reg)
# A tibble: 15 x 3
# Reg Pop Res
# <chr> <int> <chr>
# 1 A 1000915 Total
# 2 A 500414 Urban
# 3 A 500501 Rural
# 4 B 999938 Total
# 5 B 499922 Urban
# 6 B 500016 Rural
# 7 C 1000912 Total
# 8 C 501638 Urban
# 9 C 499274 Rural
#10 D 999629 Total
#11 D 499804 Urban
#12 D 499825 Rural
#13 E 1000303 Total
#14 E 499917 Urban
#15 E 500386 Rural
相应的data.table解决方案：
dt <- setDT(df)
rbindlist(list(dt[, .(Pop = sum(Pop), Res = 'Total'), Reg], dt), use.names = TRUE)
答案 2 :(得分：1)
堆叠和重新排列将起作用：
library(dplyr)
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop, stringsAsFactors = FALSE)
sums <- df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop)) %>%
mutate(Res = "Total")
df_sums <- bind_rows(df, sums) %>%
arrange(Reg, Res)
答案 3 :(得分：1)
您的数据：
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop)
require(dplyr)
df1 <-
df %>%
group_by(Reg) %>%
summarise(Total = sum(Pop))
我的解决方案（注意：我还将早期的管道发送到df1）：
df <- rbind(df, data.frame(Reg=df1$Reg, Res="Total", Pop=df1$Total))
df <- df[order(as.character(df$Reg), decreasing = T),]
df <- df[seq(dim(df)[1],1),]
结果：
print(df, row.names = F)
Reg Res Pop A Total 1000915 A Rural 500501 A Urban 500414 B Total 999938 B Rural 500016 B Urban 499922 C Total 1000912 C Rural 499274 C Urban 501638 D Total 999629 D Rural 499825 D Urban 499804 E Total 1000303 E Rural 500386 E Urban 499917
如果要在组之间使用换行符打印它们，而不更改数据类型：
for(g in unique(df$Reg)){
print(df[df$Reg==g,], row.names = F)
cat("\n")
}
Reg Res Pop A Total 1000915 A Rural 500501 A Urban 500414 Reg Res Pop B Total 999938 B Rural 500016 B Urban 499922 Reg Res Pop C Total 1000912 C Rural 499274 C Urban 501638 Reg Res Pop D Total 999629 D Rural 499825 D Urban 499804 Reg Res Pop E Total 1000303 E Rural 500386 E Urban 499917
您还要求 data.table 解决方案。这与上面的内容相同，除了像这样创建df1：
dt <- as.data.table(df)
df1 <- dt[,sum(Pop),by=dt$Reg]
答案 4 :(得分：1)
我们可以使用dplyr和purrr。这与d.b的方法类似，但map_dfr的输出将是数据帧。因此不需要从列表到数据帧的进一步转换。请注意，我使用data_frame函数构造df，因为不需要此分析因子。 df2是最终输出。
library(dplyr)
library(purrr)
df <- data_frame(Reg, Res, Pop)
df2 <- df %>%
split(.$Reg) %>%
map_dfr(~bind_rows(.x, data_frame(Reg = .x$Reg[1], Res = "Total", Pop = sum(.x$Pop))))
df2
# A tibble: 15 x 3
Reg Res Pop
<chr> <chr> <int>
1 A Urban 500414
2 A Rural 500501
3 A Total 1000915
4 B Urban 499922
5 B Rural 500016
6 B Total 999938
7 C Urban 501638
8 C Rural 499274
9 C Total 1000912
10 D Urban 499804
11 D Rural 499825
12 D Total 999629
13 E Urban 499917
14 E Rural 500386
15 E Total 1000303
答案 5 :(得分：1)
data.table包的开发版本1.10.5（参见here for installation instructions）有三个新功能，用于计算各种分组级别的聚合，可在此处使用。
请注意，OP的预期结果包含连续的行号1到15，这表示OP期望一个data.frame或data.table而不是Frank首选的列表。但是，我们将在下面显示，data.table也可以用眼睛友好的方式打印。
rollup() 使用新的rollup()功能并按Reg
library(data.table) # development version 1.10.5 as of 2015-09-10
setDT(df)
rollup(df, j = list(Pop = sum(Pop)), by = c("Reg", "Res"))[order(Reg)]
我们得到了
Reg Res Pop 1: A Urban 500414 2: A Rural 500501 3: A NA 1000915 4: B Urban 499922 5: B Rural 500016 6: B NA 999938 7: C Urban 501638 8: C Rural 499274 9: C NA 1000912 10: D Urban 499804 11: D Rural 499825 12: D NA 999629 13: E Urban 499917 14: E Rural 500386 15: E NA 1000303 16: NA NA 5001697
相应的总数由NA表示（包括总计）。如果我们想要更好地重现预期结果，可以删除总计，并NA替换Total：
rollup(df, j = list(Pop = sum(Pop)), by = c("Reg", "Res"))[order(Reg)][
is.na(Res), Res := "Total"][!is.na(Reg)]
Reg Res Pop 1: A Urban 500414 2: A Rural 500501 3: A Total 1000915 4: B Urban 499922 5: B Rural 500016 6: B Total 999938 7: C Urban 501638 8: C Rural 499274 9: C Total 1000912 10: D Urban 499804 11: D Rural 499825 12: D Total 999629 13: E Urban 499917 14: E Rural 500386 15: E Total 1000303
请注意，Total行在下面显示详细信息行，这些行不完全符合OP的预期结果。
groupingsets() 使用groupingsets()功能，可以非常详细地控制聚合：
groupingsets(df, j = list(Pop = sum(Pop)), by = c("Reg", "Res"),
sets = list("Reg", c("Reg", "Res")))[order(Reg)][
is.na(Res), Res := "Total"][]
Reg Res Pop 1: A Total 1000915 2: A Urban 500414 3: A Rural 500501 4: B Total 999938 5: B Urban 499922 6: B Rural 500016 7: C Total 1000912 8: C Urban 501638 9: C Rural 499274 10: D Total 999629 11: D Urban 499804 12: D Rural 499825 13: E Total 1000303 14: E Urban 499917 15: E Rural 500386
现在，Total行显示在详细信息行上方，并且根本没有创建总计。
data.table解决方案截至目前，两个＆＃34;经典＆＃34; Psidom和Hack-R发布了data.table个解决方案。
两者都可以更简洁地重写为
rbind(df[, .(Res = "Total", Pop = sum(Pop)), by = Reg], df)[order(Reg)]
结果可以打印在眼睛友好的＆＃34;使用
在组之间使用空行的方式rbind(df[, .(Res = "Total", Pop = sum(Pop)), by = Reg], df)[
order(Reg), {print(data.table(Reg, .SD), row.names = FALSE); cat("\n")}, by = Reg]
Reg Res Pop A Total 1000915 A Urban 500414 A Rural 500501 Reg Res Pop B Total 999938 B Urban 499922 B Rural 500016 Reg Res Pop C Total 1000912 C Urban 501638 C Rural 499274 Reg Res Pop D Total 999629 D Urban 499804 D Rural 499825 Reg Res Pop E Total 1000303 E Urban 499917 E Rural 500386