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
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
其他回答
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)
library(plyr)
ddply(tbl, .(Category), summarise, sum = sum(Frequency))
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
使用cast代替reccast(注意'Frequency'现在是'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)
得到:
Category (all)
First 30
Second 5
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)
