使用aggregate：
aggregate(x$Frequency, by=list(Category=x$Category), FUN=sum)
Category x
1 First 30
2 Second 5
3 Third 34
在上面的示例中，可以在list. 可以通过以下方式合并相同数据类型的多个聚合指标cbind：
aggregate(cbind(x$Frequency, x$Metric2, x$Metric3) ...
（嵌入@thelatemail 评论），aggregate也有公式界面
aggregate(Frequency ~ Category, x, sum)
或者，如果您想聚合多列，您可以使用.符号（也适用于一列）
aggregate(. ~ Category, x, sum)
或tapply：
tapply(x$Frequency, x$Category, FUN=sum)
First Second Third
30 5 34
使用这些数据：
x <- data.frame(Category=factor(c("First", "First", "First", "Second",
"Third", "Third", "Second")),
Frequency=c(10,15,5,2,14,20,3))
您还可以为此目的使用dplyr包：
library(dplyr)
x %>%
group_by(Category) %>%
summarise(Frequency = sum(Frequency))
#Source: local data frame [3 x 2]
#
# Category Frequency
#1 First 30
#2 Second 5
#3 Third 34
或者，对于多个汇总列（也适用于一列）：
x %>%
group_by(Category) %>%
summarise(across(everything(), sum))
以下是一些关于如何使用内置数据集的 dplyr 函数按组汇总数据的更多示例mtcars：
# several summary columns with arbitrary names
mtcars %>%
group_by(cyl, gear) %>% # multiple group columns
summarise(max_hp = max(hp), mean_mpg = mean(mpg)) # multiple summary columns
# summarise all columns except grouping columns using "sum"
mtcars %>%
group_by(cyl) %>%
summarise(across(everything(), sum))
# summarise all columns except grouping columns using "sum" and "mean"
mtcars %>%
group_by(cyl) %>%
summarise(across(everything(), list(mean = mean, sum = sum)))
# multiple grouping columns
mtcars %>%
group_by(cyl, gear) %>%
summarise(across(everything(), list(mean = mean, sum = sum)))
# summarise specific variables, not all
mtcars %>%
group_by(cyl, gear) %>%
summarise(across(c(qsec, mpg, wt), list(mean = mean, sum = sum)))
# summarise specific variables (numeric columns except grouping columns)
mtcars %>%
group_by(gear) %>%
summarise(across(where(is.numeric), list(mean = mean, sum = sum)))
有关更多信息，包括%>%运算符，请参阅dplyr 简介。
rcs 提供的答案很简单。但是，如果您正在处理更大的数据集并需要提高性能，则有一个更快的替代方案：
library(data.table)
data = data.table(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
data[, sum(Frequency), by = Category]
# Category V1
# 1: First 30
# 2: Second 5
# 3: Third 34
system.time(data[, sum(Frequency), by = Category] )
# user system elapsed
# 0.008 0.001 0.009
让我们将其与使用 data.frame 和上面的内容进行比较：
data = data.frame(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
system.time(aggregate(data$Frequency, by=list(Category=data$Category), FUN=sum))
# user system elapsed
# 0.008 0.000 0.015
如果您想保留该列，请使用以下语法：
data[,list(Frequency=sum(Frequency)),by=Category]
# Category Frequency
# 1: First 30
# 2: Second 5
# 3: Third 34
使用更大的数据集，这种差异将变得更加明显，如下面的代码所示：
data = data.table(Category=rep(c("First", "Second", "Third"), 100000),
Frequency=rnorm(100000))
system.time( data[,sum(Frequency),by=Category] )
# user system elapsed
# 0.055 0.004 0.059
data = data.frame(Category=rep(c("First", "Second", "Third"), 100000),
Frequency=rnorm(100000))
system.time( aggregate(data$Frequency, by=list(Category=data$Category), FUN=sum) )
# user system elapsed
# 0.287 0.010 0.296
对于多个聚合，可以组合lapply和.SD如下
data[, lapply(.SD, sum), by = Category]
# Category Frequency
# 1: First 30
# 2: Second 5
# 3: Third 34
您还可以使用by()函数：
x2 <- by(x$Frequency, x$Category, sum)
do.call(rbind,as.list(x2))
那些其他包（plyr、reshape）具有返回 data.frame 的好处，但值得熟悉 by()，因为它是一个基本函数。
几年后，只是为了添加另一个简单的基本 R 解决方案，由于某种原因这里不存在 -xtabs
xtabs(Frequency ~ Category, df)
# Category
# First Second Third
# 30 5 34
或者如果你想要一个data.frame背部
as.data.frame(xtabs(Frequency ~ Category, df))
# Category Freq
# 1 First 30
# 2 Second 5
# 3 Third 34
library(plyr)
ddply(tbl, .(Category), summarise, sum = sum(Frequency))
如果x是包含您的数据的数据框，那么以下内容将执行您想要的操作：
require(reshape)
recast(x, Category ~ ., fun.aggregate=sum)
虽然我最近已经成为dplyr大多数此类操作的转换者，但sqldf对于某些事情，该包仍然非常好（恕我直言，更具可读性）。
这是一个如何回答这个问题的例子sqldf
x <- data.frame(Category=factor(c("First", "First", "First", "Second",
"Third", "Third", "Second")),
Frequency=c(10,15,5,2,14,20,3))
sqldf("select
Category
,sum(Frequency) as Frequency
from x
group by
Category")
## Category Frequency
## 1 First 30
## 2 Second 5
## 3 Third 34
只是添加第三个选项：
require(doBy)
summaryBy(Frequency~Category, data=yourdataframe, FUN=sum)
编辑：这是一个非常古老的答案。现在我建议使用group_byand summarisefrom dplyr，就像@docendo 的答案一样。
另一种在矩阵或数据框中按组返回和的解决方案又短又快：
rowsum(x$Frequency, x$Category)
ave当您需要在不同的列上应用不同的聚合函数（并且您必须/想要坚持使用基础 R）时，我发现非常有帮助（并且有效）：
例如
鉴于此输入：
DF <-
data.frame(Categ1=factor(c('A','A','B','B','A','B','A')),
Categ2=factor(c('X','Y','X','X','X','Y','Y')),
Samples=c(1,2,4,3,5,6,7),
Freq=c(10,30,45,55,80,65,50))
> DF
Categ1 Categ2 Samples Freq
1 A X 1 10
2 A Y 2 30
3 B X 4 45
4 B X 3 55
5 A X 5 80
6 B Y 6 65
7 A Y 7 50
我们想按Categ1和分组Categ2并计算 的总和Samples和 的平均值Freq。
这是一个可能的解决方案ave：
# create a copy of DF (only the grouping columns)
DF2 <- DF[,c('Categ1','Categ2')]
# add sum of Samples by Categ1,Categ2 to DF2
# (ave repeats the sum of the group for each row in the same group)
DF2$GroupTotSamples <- ave(DF$Samples,DF2,FUN=sum)
# add mean of Freq by Categ1,Categ2 to DF2
# (ave repeats the mean of the group for each row in the same group)
DF2$GroupAvgFreq <- ave(DF$Freq,DF2,FUN=mean)
# remove the duplicates (keep only one row for each group)
DF2 <- DF2[!duplicated(DF2),]
结果 ：
> DF2
Categ1 Categ2 GroupTotSamples GroupAvgFreq
1 A X 6 45
2 A Y 9 40
3 B X 7 50
6 B Y 6 65
由于dplyr 1.0.0，across()可以使用该函数：
df %>%
group_by(Category) %>%
summarise(across(Frequency, sum))
Category Frequency
<chr> <int>
1 First 30
2 Second 5
3 Third 34
如果对多个变量感兴趣：
df %>%
group_by(Category) %>%
summarise(across(c(Frequency, Frequency2), sum))
Category Frequency Frequency2
<chr> <int> <int>
1 First 30 55
2 Second 5 29
3 Third 34 190
并使用选择助手选择变量：
df %>%
group_by(Category) %>%
summarise(across(starts_with("Freq"), sum))
Category Frequency Frequency2 Frequency3
<chr> <int> <int> <dbl>
1 First 30 55 110
2 Second 5 29 58
3 Third 34 190 380
样本数据：
df <- read.table(text = "Category Frequency Frequency2 Frequency3
1 First 10 10 20
2 First 15 30 60
3 First 5 15 30
4 Second 2 8 16
5 Third 14 70 140
6 Third 20 120 240
7 Second 3 21 42",
header = TRUE,
stringsAsFactors = FALSE)
您可以使用group.sum包Rfast中的功能。
Category <- Rfast::as_integer(Category,result.sort=FALSE) # convert character to numeric. R's as.numeric produce NAs.
result <- Rfast::group.sum(Frequency,Category)
names(result) <- Rfast::Sort(unique(Category)
# 30 5 34
Rfast有很多组函数，group.sum就是其中之一。
使用cast代替recast（注意'Frequency'现在'value'）
df <- data.frame(Category = c("First","First","First","Second","Third","Third","Second")
, value = c(10,15,5,2,14,20,3))
install.packages("reshape")
result<-cast(df, Category ~ . ,fun.aggregate=sum)
要得到：
Category (all)
First 30
Second 5
Third 34
library(tidyverse)
x <- data.frame(Category= c('First', 'First', 'First', 'Second', 'Third', 'Third', 'Second'),
Frequency = c(10, 15, 5, 2, 14, 20, 3))
count(x, Category, wt = Frequency)