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

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)

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))

从dplyr 1.0.0开始,可以使用across()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

以及使用select helper来选择变量:

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)
library(plyr)
ddply(tbl, .(Category), summarise, sum = sum(Frequency))

你可以用函数群。sum来自包Rfast。

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有许多组函数和组。和就是其中之一。

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