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How to filter data.frame in R by categorical variable?

Just learn R.

Given data.framein R with two columns, one number and one categorical, how can I extract the part data.frameto use?

str(ex0331)
'data.frame':   36 obs. of  2 variables:
$ Iron      : num  0.71 1.66 2.01 2.16 2.42 ...
$ Supplement: Factor w/ 2 levels "Fe3","Fe4": 1 1 1 1 1 1 1 1 1 1 ...

Basically, I need to be able to work on two factors separately; that is, I need to individually determine the length / average / sd / etc of the iron retention rate by type Supplement( Fe3or Fe4).

What is the easiest way to accomplish this?

I know the team by(). For example, the following gets some of what I need:

by(ex0331, ex0331$Supplement, summary)
ex0331$Supplement: Fe3
     Iron       Supplement
Min.   :0.710   Fe3:18    
1st Qu.:2.420   Fe4: 0    
Median :3.475             
Mean   :3.699             
3rd Qu.:4.472             
Max.   :8.240             
------------------------------------------------------------ 
ex0331$Supplement: Fe4
     Iron        Supplement
Min.   : 2.200   Fe3: 0    
1st Qu.: 3.892   Fe4:18    
Median : 5.750             
Mean   : 5.937             
3rd Qu.: 6.970             
Max.   :12.450

. axis, , log() . , ; . data.frame, , , .

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2

subset:

ex0331 <- data.frame( iron=rnorm(36), supplement=c("Fe3","Fe4"))

subset(ex0331, supplement=="Fe3")
subset(ex0331, supplement=="Fe4")

ex0331[ex0331$supplement=="Fe3",]

split, :

split(ex0331,ex0331$supplement)

, , tapply , :

tapply(ex0331$iron,ex0331$supplement,mean)
        Fe3         Fe4 
-0.15443861 -0.01308835

plyr, . :

library(plyr)
daply(ex0331,.(supplement),function(x)mean(x[1]))
        Fe3         Fe4 
-0.15443861 -0.01308835

:

ex0331 <- data.frame( iron=abs(rnorm(36)), supplement=c("Fe3","Fe4"))

tapply(ex0331$iron,ex0331$supplement,log)

plyr:

library(plyr)
dlply(ex0331,.(supplement),function(x)log(x$iron))

. , , plyr.

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ddply plyr, :

> require(plyr)
> ddply( ex0331, .(Supplement), summarise, 
         mean = mean(Iron), 
         sd = sd(Iron), 
         len = length(Iron))

  Supplement       mean        sd len
1        Fe3 -0.3749169 0.2827360   4
2        Fe4  0.1953116 0.7128129   6

. LogIron, log() Iron, transform:

> transform(ex0331, LogIron = log(Iron))

         Iron Supplement     LogIron
1  0.07185141        Fe3 -2.63315498
2  1.10367297        Fe3  0.09864368
3  0.48592428        Fe3 -0.72170246
4  0.20286918        Fe3 -1.59519393
5  0.80830682        Fe4 -0.21281357

, , " ", :

> ddply( ex0331, .(Supplement), summarise, meanLog = mean(log(Iron)))
  Supplement    meanLog
1        Fe3 -1.0062304
2        Fe4  0.2791507
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