plyr在转换之后不会返回新的变量

I'm trying to learn how to write function in R/plyr. I am aware that there are easier ways to do what I show below, but that's not the point.

我在学习如何用R/plyr写函数。我知道有更简单的方法来做我下面展示的，但这不是重点。

In the example that follows, PLYR does not return a new variable to my new data frame

在下面的示例中，PLYR不会向我的新数据帧返回一个新变量

library(plyr)
highab <-subset(baseball, ab >= 600)

testfunc1 <-function(x) {
    print(x) #just to show me that the vector does get into the function. Works fine.
    medianAB <- median(x)
    print(medianAB) #just to prove that medianAB was calculated correctly. Works fine   
}


baseball3 <-ddply(highab, .(id), transform, testfunc1(ab))
str(baseball3$medianAB) #No medianAB

What obvious thing am I missing?

我遗漏了什么明显的东西?

R version 2.12.2 (2011-02-25)
Platform: x86_64-pc-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_CA.UTF-8       LC_NUMERIC=C               LC_TIME=en_CA.UTF-8        LC_COLLATE=en_CA.UTF-8    
 [5] LC_MONETARY=C              LC_MESSAGES=en_CA.UTF-8    LC_PAPER=en_CA.UTF-8       LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C             LC_MEASUREMENT=en_CA.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] grid      splines   stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] foreign_0.8-42  ggplot2_0.8.9   proto_0.3-9.1   reshape_0.8.4   plyr_1.4.1      rms_3.3-0       Hmisc_3.8-3    
[8] survival_2.36-5 stringr_0.4    

loaded via a namespace (and not attached):
[1] cluster_1.13.3  lattice_0.19-23 tools_2.12.2

3 个解决方案

#1

Just make two changes

只是让两个变化

Remove the print command inside the function, so that median is returned
删除函数中的print命令，以便返回中位数
Add medianAB = testfunc1(ab) as suggested by Joshua
按照约书亚的建议添加medianAB = testfunc1(ab)

You are done!

你做的!

Here is the simplified code with the output

这是带有输出的简化代码

library(plyr)
highab <-subset(baseball, ab >= 600)
baseball3 <-ddply(highab, .(id), transform, medianAB = median(ab))
summary(baseball3$medianAB)

Min. 1st Qu. Median Mean 3rd Qu.
Max. 600.0 612.0 621.5 623.1 631.5 677.0

第1区:中位数第三区。最大值:600.0 621.5 623.1 631.5 677.0

#2

Sorry. I mis-understood the question.

对不起。我误解了问题。

See ?transform. You need to specify the new variables you want as tag=value pairs. So you need something like

看到了什么?变换。您需要指定您想要的新变量为标记=值对。所以你需要一些类似的东西。

baseball3 <- ddply(highab, .(id), transform, medianAB=testfunc1(ab))

#3

At first I liked the idiom to add derived columns to a data.frame, but I find the usage of transform() unacceptably slow far large sets.

起初，我喜欢将派生列添加到data.frame中，但我发现transform()的使用速度慢得令人无法接受。

Would it be better to use a lambda form in ddply() and a subsequent call to merge merge()? Timing it looks like it's worth it:

在ddply()中使用lambda表单和随后调用merge()会更好吗?时机看起来是值得的:

    > library(plyr)
    > highab <-subset(baseball, ab >= 600)
    > 
    > system.time( 
    +   baseball3.lambda <-merge(highab, 
    +     ddply(highab, .(id), 
    +       function(u) data.frame(medianAB = median(u$ab)))), FALSE)
       user  system elapsed 
      0.336   0.000   0.336 
    > 
    > system.time( 
        baseball3.orig <- ddply(highab, .(id), 
          transform, medianAB = median(ab)), FALSE)
       user  system elapsed 
      0.640   0.000   0.641 
    > 
    > summary(baseball3.lambda$medianAB)
       Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      600.0   612.0   621.5   623.1   631.5   677.0 
    > summary(baseball3.orig$medianAB)
       Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      600.0   612.0   621.5   623.1   631.5   677.0

3 tenths of a second may not seem much but it is halving the execution time. The improvement is even bigger by selecting the whole baseball dataset.

3 / 10秒似乎并不多，但它将执行时间缩短了一半。通过选择整个棒球数据集，改进更大。

#1