为因子的每个水平附加一行总和
Appending a row of sums for each level of a factor
我想像这样为每个 Reg 添加一行总和
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
已编辑
我想要 dplyr 和 data.table 两种解决方案。
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
$E
Reg Res Pop
1 E Total 1000303
9 E Urban 499917
10 E Rural 500386
如果需要，您可以使用 do.call(rbind, ...) 将整个内容转换为 data.frame
您可以在摘要中添加额外的 Res 列，然后 bind_rows 使用原始数据框：
df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop), Res = 'Total') %>%
bind_rows(df) %>%
arrange(Reg)
# A tibble: 15 x 3
# Reg Pop Res
# <chr> <int> <chr>
# 1 A 1000915 Total
# 2 A 500414 Urban
# 3 A 500501 Rural
# 4 B 999938 Total
# 5 B 499922 Urban
# 6 B 500016 Rural
# 7 C 1000912 Total
# 8 C 501638 Urban
# 9 C 499274 Rural
#10 D 999629 Total
#11 D 499804 Urban
#12 D 499825 Rural
#13 E 1000303 Total
#14 E 499917 Urban
#15 E 500386 Rural
一个对应的data.table解法：
dt <- setDT(df)
rbindlist(list(dt[, .(Pop = sum(Pop), Res = 'Total'), Reg], dt), use.names = TRUE)
堆叠和重新排列将起作用：
library(dplyr)
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop, stringsAsFactors = FALSE)
sums <- df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop)) %>%
mutate(Res = "Total")
df_sums <- bind_rows(df, sums) %>%
arrange(Reg, Res)
您的数据：
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop)
require(dplyr)
df1 <-
df %>%
group_by(Reg) %>%
summarise(Total = sum(Pop))
我的解决方案（注：我也将之前的管道发送到df1）：
df <- rbind(df, data.frame(Reg=df1$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
如果要在不更改数据类型的情况下在组之间使用换行符打印它们：
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
您还请求了 data.table 解决方案。这与上面的相同，除了像这样创建 df1：
dt <- as.data.table(df)
df1 <- dt[,sum(Pop),by=dt$Reg]
我们可以使用dplyr和purrr。这类似于 d.b 的方法，但 map_dfr 的输出将是一个数据框。因此不需要从列表到数据框的进一步转换。请注意，我使用了 data_frame 函数来构建 df，因为对于这个分析因素来说不需要。 df2 是最终输出。
library(dplyr)
library(purrr)
df <- data_frame(Reg, Res, Pop)
df2 <- df %>%
split(.$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
data.table 包的开发版本 1.10.5（参见 here for installation instructions）具有三个新函数，用于计算不同级别分组的聚合，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，这表明 OP 期望一个 data.frame 或 data.table，而不是 首选的列表。但是，我们将在下面展示 data.table 也可以以对眼睛友好的方式打印。
rollup()
使用新的 rollup() 函数并按 Reg
排序
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)]
我们确实得到了
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
各自的总数用NA表示（包括总计）。如果我们想更好地重现预期结果，可以删除总计，将 NA 替换为 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 行出现在 下方 的详细信息行中，这与 OP 的预期结果不完全一致。
groupingsets()
使用groupingsets()功能，可以对聚合进行非常详细的控制：
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，根本没有创建总计。
印刷精美的“经典”data.table解决方案
到目前为止， and .
发布了两个“经典”data.table 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)]
结果可以使用
以“对眼睛友好”的方式打印，组与组之间有空行
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
我想像这样为每个 Reg 添加一行总和
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
已编辑
我想要 dplyr 和 data.table 两种解决方案。
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
$E
Reg Res Pop
1 E Total 1000303
9 E Urban 499917
10 E Rural 500386
如果需要，您可以使用 do.call(rbind, ...) 将整个内容转换为 data.frame
您可以在摘要中添加额外的 Res 列，然后 bind_rows 使用原始数据框：
df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop), Res = 'Total') %>%
bind_rows(df) %>%
arrange(Reg)
# A tibble: 15 x 3
# Reg Pop Res
# <chr> <int> <chr>
# 1 A 1000915 Total
# 2 A 500414 Urban
# 3 A 500501 Rural
# 4 B 999938 Total
# 5 B 499922 Urban
# 6 B 500016 Rural
# 7 C 1000912 Total
# 8 C 501638 Urban
# 9 C 499274 Rural
#10 D 999629 Total
#11 D 499804 Urban
#12 D 499825 Rural
#13 E 1000303 Total
#14 E 499917 Urban
#15 E 500386 Rural
一个对应的data.table解法：
dt <- setDT(df)
rbindlist(list(dt[, .(Pop = sum(Pop), Res = 'Total'), Reg], dt), use.names = TRUE)
堆叠和重新排列将起作用：
library(dplyr)
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop, stringsAsFactors = FALSE)
sums <- df %>%
group_by(Reg) %>%
summarise(Pop = sum(Pop)) %>%
mutate(Res = "Total")
df_sums <- bind_rows(df, sums) %>%
arrange(Reg, Res)
您的数据：
Reg <- rep(LETTERS[1:5], each = 2)
Res <- rep(c("Urban", "Rural"), times = 5)
set.seed(12345)
Pop <- rpois(n = 10, lambda = 500000)
df <- data.frame(Reg, Res, Pop)
require(dplyr)
df1 <-
df %>%
group_by(Reg) %>%
summarise(Total = sum(Pop))
我的解决方案（注：我也将之前的管道发送到df1）：
df <- rbind(df, data.frame(Reg=df1$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
如果要在不更改数据类型的情况下在组之间使用换行符打印它们：
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
您还请求了 data.table 解决方案。这与上面的相同，除了像这样创建 df1：
dt <- as.data.table(df)
df1 <- dt[,sum(Pop),by=dt$Reg]
我们可以使用dplyr和purrr。这类似于 d.b 的方法，但 map_dfr 的输出将是一个数据框。因此不需要从列表到数据框的进一步转换。请注意，我使用了 data_frame 函数来构建 df，因为对于这个分析因素来说不需要。 df2 是最终输出。
library(dplyr)
library(purrr)
df <- data_frame(Reg, Res, Pop)
df2 <- df %>%
split(.$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
data.table 包的开发版本 1.10.5（参见 here for installation instructions）具有三个新函数，用于计算不同级别分组的聚合，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，这表明 OP 期望一个 data.frame 或 data.table，而不是
rollup()
使用新的 rollup() 函数并按 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)]
我们确实得到了
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
各自的总数用NA表示（包括总计）。如果我们想更好地重现预期结果，可以删除总计，将 NA 替换为 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 行出现在 下方 的详细信息行中，这与 OP 的预期结果不完全一致。
groupingsets()
使用groupingsets()功能，可以对聚合进行非常详细的控制：
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，根本没有创建总计。
印刷精美的“经典”data.table解决方案
到目前为止，
data.table 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)]
结果可以使用
以“对眼睛友好”的方式打印，组与组之间有空行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