Question

我有一个如下所示的数据框：

pred1 pred2 pred3 exp
a     b     c     0
a     d     c     0
a     b     c     1

我想要做的是先获取pred1-3的所有独特组合，将它们写入附加表，为每个组合的频率添加一列，然后添加另一列，给出值1的比例每个组合的exp（只能是0或1）。像这样：

pred1 pred2 pred3 freq exp_prop
a     b     c     2    0.5
a     d     c     1    0

plyr的前三个步骤非常简单：

ddply(df, .(pred1, pred2, pred3), summarise, freq=length(exp))

或更短

count(df[,c(pred1, pred2, pred3)])

但我无法弄清楚如何获得exp的比例。

Answer 1

你差不多完成了。只需将exp_prop = mean(exp)添加到ddply命令：

ddply(df, .(pred1, pred2, pred3), summarise,
      freq = length(exp), exp_prop = mean(exp))

  pred1 pred2 pred3 freq exp_prop
1     a     b     c    2      0.5
2     a     d     c    1      0.0

Answer 2

# read in your data
x <- 
read.table(text="pred1 pred2 pred3 exp
a     b     c     0
a     d     c     0
a     b     c     1" , h = T)

library(sqldf)
sqldf( "select pred1, pred2, pred3, count(*) as numtimes, avg( exp ) as prop from x group by pred1, pred2, pred3" )

###### alternative:

# write all the column names according to some pattern
cols <- paste0("pred" , 1:3 , collapse = "," )

# save your data frame to another object
y <-
    sqldf( 
        paste( 
            "select" , 
            cols  , 
            " , count(*) as numtimes, avg( exp ) as prop from x group by" , 
            cols 
        ) 
    )

# print to screen
y

获取具有另一个变量的摘要计数的唯一值组合

2 个答案: