Question

如何创建垂直排列的价值 - 频率组合表

> df = data.frame(fruit=c("apple", "banana", "cherry", "cherry", "apple", "banana", "apple", "date"));
> table(df$fruit)

 apple banana cherry   date 
     3      2      2      1

到目前为止一切顺利。但是我想要像这样的东西（例如基于频率的基本操作和值的子集化）：

Fruit   Freq
"apple"   3
"banana"  2
"cherry"  2
"date"    1

在SQL中，那将是SELECT fruit, COUNT(*) AS Freq FROM df GROUP BY fruit，并且会产生一个类似于这个问题起点的表：https://stats.stackexchange.com/questions/15574/how-to-convert-a-frequency-table-into-a-vector-of-values

在 R 中有一种简单的方法吗？（或者，这是否表明心态太'SQL'且不够'R'？）

Answer 1

data.frame(table(df))
#       df Freq
# 1  apple    3
# 2 banana    2
# 3 cherry    2
# 4   date    1

或者

setNames(data.frame(table(df)), c("Fruit", "Freq"))
#    Fruit Freq
# 1  apple    3
# 2 banana    2
# 3 cherry    2
# 4   date    1

Answer 2

只使用基数R，您只需将其转换为矩阵，该矩阵将以列为导向，其值为row.names。

as.matrix(table(df))

Answer 3

使用reshape2包中的melt功能

DF <- table(df$fruit)

library(reshape2)
melt(DF)
colnames(result) <- c('Fruit', 'Freq')
result


  Fruit Freq
1  apple    3
2 banana    2
3 cherry    2
4   date    1

Answer 4

如果你想坚持使用SQL思维模式，总会有“sqldf”包！

df = data.frame(fruit=c("apple", "banana", "cherry", "cherry", 
                        "apple", "banana", "apple", "date"))
library(sqldf)
sqldf("SELECT fruit, COUNT(*) AS Freq FROM df GROUP BY fruit")
#    fruit Freq
# 1  apple    3
# 2 banana    2
# 3 cherry    2
# 4   date    1

有！你已经知道了答案;）

Answer 5

只是添加一个不同的答案：

df = data.frame(fruit=c("apple", "banana", "cherry", "cherry", "apple", "banana", "apple", "date"));
df
   fruit
1  apple
2 banana
3 cherry
4 cherry
5  apple
6 banana
7  apple
8   date
freq <- unlist(lapply(unique(df$fruit),function(x) length(which(df$fruit ==x))))

freq
[1] 3 2 2 1
df.new <- data.frame(fruits = unique(df$fruit),freq)
df.new
  fruits freq
1  apple    3
2 banana    2
3 cherry    2
4   date    1

如何在R中创建垂直排列的值 - 频率组合表

5 个答案: