rcs提供的答案很有效，也很简单。然而，如果你正在处理更大的数据集，需要性能提升，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 

让我们用data.frame和上面的比较一下:

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 

如果你想保留列，这是语法:

data[,list(Frequency=sum(Frequency)),by=Category]
#    Category Frequency
# 1:    First        30
# 2:   Second         5
# 3:    Third        34

数据集越大，这种差异就越明显，如下图所示:

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，可以按如下方式组合lapply和.SD

data[, lapply(.SD, sum), by = Category]
#    Category Frequency
# 1:    First        30
# 2:   Second         5
# 3:    Third        34

你也可以使用dplyr包来实现这个目的:

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，对于多个摘要列(也适用于一个列):

x %>% 
  group_by(Category) %>% 
  summarise(across(everything(), sum))

下面是一些关于如何使用dplyr函数(使用内置数据集mtcars)按组总结数据的例子:

# 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%>%操作符，请参阅dplyr介绍。

你也可以使用by()函数:

x2 <- by(x$Frequency, x$Category, sum)
do.call(rbind,as.list(x2))

其他那些包(plyr，重塑)的好处是返回data.frame，但是by()值得熟悉一下，因为它是一个基函数。

当你需要在不同的列上应用不同的聚合函数(并且你必须/想要坚持以R为基底)时，我发现它非常有用(并且有效):

e.g.

假设输入如下:

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

我们要按1类和2类进行分组，并计算Freq的样本和均值。 下面是使用ave的一个可能的解决方案:

# 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

另一种解决方案是在矩阵或数据帧中按组返回和，并且简短快速:

rowsum(x$Frequency, x$Category)

如果x是一个包含你的数据的数据框架，那么下面的代码将完成你想要的:

require(reshape)
recast(x, Category ~ ., fun.aggregate=sum)