[R] iterating over groups of columns

[Jannis]

you should have found a solution for that in the help page of apply.

just run

min.values = apply(the.data,1,min)

the '1' marks the direction (e.g. whether apply is applied to rows or 
columns), it could be a 2 as well. Check that yourself in the apply 
documentation.

Then run rbind(the.data,min.values) (could be cbind as well, I am not 
sure again ;-) ) and you get what you want.

09wkj schrieb:
> I am mainly a Java/C++ programmer, so my mind is used to iterating over data with for loops. After a long break, I am trying to get back into the "R mindset", but I could not find a solution in the documentation for the applys, aggregate, or by.
>
> I have a data.frame where each row is an entry with 10 groups of measurements. The first measurement spans 1 column, the second spans 2 columns, third 3, and so on (55 total columns). What I want to do is add to my data.frame 10 new columns containing the minimum value of each measurement.
>
> dim(the.data)
> [1] 1679  55
>
>   
>> colnames(the.data)
>>     
>   [1] "k.1.1"   "k.2.1"   "k.2.2"   "k.3.1"   "k.3.2"   "k.3.3"   "k.4.1"  
>   [8] "k.4.2"   "k.4.3"   "k.4.4"   "k.5.1"   "k.5.2"   "k.5.3"   "k.5.4"  
>  [15] "k.5.5"   "k.6.1"   "k.6.2"   "k.6.3"   "k.6.4"   "k.6.5"   "k.6.6"  
>  [22] "k.7.1"   "k.7.2"   "k.7.3"   "k.7.4"   "k.7.5"   "k.7.6"   "k.7.7"  
>  [29] "k.8.1"   "k.8.2"   "k.8.3"   "k.8.4"   "k.8.5"   "k.8.6"   "k.8.7"  
>  [36] "k.8.8"   "k.9.1"   "k.9.2"   "k.9.3"   "k.9.4"   "k.9.5"   "k.9.6"  
>  [43] "k.9.7"   "k.9.8"   "k.9.9"   "k.10.1"  "k.10.2"  "k.10.3"  "k.10.4" 
>  [50] "k.10.5"  "k.10.6"  "k.10.7"  "k.10.8"  "k.10.9"  "k.10.10"
>
> I want to add to the.data new columns: min.k.1, min.k.2, ..., min.k.10
>
> This is the section of code I would like to improve, hopefully getting rid of the eval and the for loop:
>
> for(k in 1:10){
>     s <- subset(the.data, select=paste("k", k, 1:k, sep="."))
>     eval(parse(text = paste("the.data$min.k.", k, "<-as.vector(by(s, 1:nrow(s), min))", sep="")))
> }
>
> Thanks for any help,
> Bill
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