我有一个有两列的数据帧。第一列包含类别，如“第一”，“第二”，“第三”，第二列有数字，表示我从“类别”中看到特定组的次数。
例如:
Category Frequency
First 10
First 15
First 5
Second 2
Third 14
Third 20
Second 3
我想按类别对数据进行排序，并将所有频率相加:
Category Frequency
First 30
Second 5
Third 34
在R中怎么做呢?
当前回答
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)
其他回答
library(plyr)
ddply(tbl, .(Category), summarise, sum = sum(Frequency))
使用聚合:
aggregate(x$Frequency, by=list(Category=x$Category), FUN=sum)
Category x
1 First 30
2 Second 5
3 Third 34
在上面的例子中，可以在列表中指定多个维度。相同数据类型的多个聚合指标可以通过cbind合并:
aggregate(cbind(x$Frequency, x$Metric2, x$Metric3) ...
(嵌入@thelatemail评论)，聚合也有一个公式界面
aggregate(Frequency ~ Category, x, sum)
或者，如果希望聚合多个列，可以使用。符号(也适用于一列)
aggregate(. ~ Category, x, sum)
或tapply:
tapply(x$Frequency, x$Category, FUN=sum)
First Second Third
30 5 34
使用这些数据:
x <- data.frame(Category=factor(c("First", "First", "First", "Second",
"Third", "Third", "Second")),
Frequency=c(10,15,5,2,14,20,3))
虽然我最近对大多数这些类型的操作都转换为dplyr，但sqldf包对于某些事情仍然非常好(恕我直言，可读性更强)。
下面是一个示例，说明如何使用sqldf回答这个问题
x <- data.frame(Category=factor(c("First", "First", "First", "Second",
"Third", "Third", "Second")),
Frequency=c(10,15,5,2,14,20,3))
sqldf("select
Category
,sum(Frequency) as Frequency
from x
group by
Category")
## Category Frequency
## 1 First 30
## 2 Second 5
## 3 Third 34
rcs提供的答案很有效，也很简单。然而，如果你正在处理更大的数据集，需要性能提升，有一个更快的替代方案:
library(data.table)
data = data.table(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
data[, sum(Frequency), by = Category]
# Category V1
# 1: First 30
# 2: Second 5
# 3: Third 34
system.time(data[, sum(Frequency), by = Category] )
# user system elapsed
# 0.008 0.001 0.009
让我们用data.frame和上面的比较一下:
data = data.frame(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
system.time(aggregate(data$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
几年后，只是为了添加另一个简单的基本R解决方案，由于某种原因这里没有给出——xtabs
xtabs(Frequency ~ Category, df)
# Category
# First Second Third
# 30 5 34
或者你想要回数据帧
as.data.frame(xtabs(Frequency ~ Category, df))
# Category Freq
# 1 First 30
# 2 Second 5
# 3 Third 34
