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
可以使用rowsum函数来计算频率。
data("mtcars")
df <- mtcars
df$cyl <- as.factor(df$cyl)
头部看起来如下:
wt mpg cyl
<dbl> <dbl> <fct>
Mazda RX4 2.620 21.0 6
Mazda RX4 Wag 2.875 21.0 6
Datsun 710 2.320 22.8 4
然后,
rowsum(df$mpg, df$cyl) #values , group
4 293.3
6 138.2
8 211.4
其他回答
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$Category, FUN=sum)
First Second Third
30 5 34
使用这些数据:
x <- data.frame(Category=factor(c("First", "First", "First", "Second",
"Third", "Third", "Second")),
Frequency=c(10,15,5,2,14,20,3))
另一种解决方案是在矩阵或数据帧中按组返回和，并且简短快速:
rowsum(x$Frequency, x$Category)
当你需要在不同的列上应用不同的聚合函数(并且你必须/想要坚持以R为基底)时，我发现它非常有用(并且有效):
e.g.
假设输入如下:
DF <-
data.frame(Categ1=factor(c('A','A','B','B','A','B','A')),
Categ2=factor(c('X','Y','X','X','X','Y','Y')),
Samples=c(1,2,4,3,5,6,7),
Freq=c(10,30,45,55,80,65,50))
> DF
Categ1 Categ2 Samples Freq
1 A X 1 10
2 A Y 2 30
3 B X 4 45
4 B X 3 55
5 A X 5 80
6 B Y 6 65
7 A Y 7 50
我们要按1类和2类进行分组，并计算Freq的样本和均值。 下面是使用ave的一个可能的解决方案:
# create a copy of DF (only the grouping columns)
DF2 <- DF[,c('Categ1','Categ2')]
# add sum of Samples by Categ1,Categ2 to DF2
# (ave repeats the sum of the group for each row in the same group)
DF2$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$Freq,DF2,FUN=mean)
# remove the duplicates (keep only one row for each group)
DF2 <- DF2[!duplicated(DF2),]
结果:
> DF2
Categ1 Categ2 GroupTotSamples GroupAvgFreq
1 A X 6 45
2 A Y 9 40
3 B X 7 50
6 B Y 6 65
rcs提供的答案很有效，也很简单。然而，如果你正在处理更大的数据集，需要性能提升，有一个更快的替代方案:
library(data.table)
data = data.table(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
data[, sum(Frequency), by = Category]
# Category V1
# 1: First 30
# 2: Second 5
# 3: Third 34
system.time(data[, sum(Frequency), by = Category] )
# user system elapsed
# 0.008 0.001 0.009
让我们用data.frame和上面的比较一下:
data = data.frame(Category=c("First","First","First","Second","Third", "Third", "Second"),
Frequency=c(10,15,5,2,14,20,3))
system.time(aggregate(data$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
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)
