Making a for loop over unique id's using lapply or aggregate

Question

For each unique id, I'd like to calculate the maximum difference between each value of SkinTemp using lapply (or aggregate) of a series of curves (Time,SkinTemp) designated by a unique id in a data.frame.

So far I have the following but it returns a single value which is not right:

df<-data.frame(Time=seq(100),
               SkinTemp=rnorm(100,37,0.5),
               id=rep(1:10,each=10))



  maxDiff<-function(id,df) {
    a<-max(diff(df$SkinTemp))
    a
  }
maxA<-lapply(id,maxDiff,df)

Any thoughts on why it doesn't retrieve a unique max value of SkinTemp for each id?

Edit:

Using aggregate no problem (I think it's ddplyr package)

aggregate(data=df,SkinTemp~id,function(x)max(diff(x)))

So what am I doing wrong with lapply?

The normal aggregate approach would be `aggregate(SkinTemp ~ id, df, max)` — talat, Feb 07 '18 at 19:09
So your `lapply` shouldn't even work because `id` is not defined except as a column in df. The *apply function you want is `tapply` except it doesn't work on data.frames (and thus your current function doesn't work with it) so you need the wrapper function `by`, but that function is awkward to use for the post part. The simplest version using `tapply` is `tapply(df$SkinTemp, df$id, function(x) max(diff(x)))` — Vlo, Feb 07 '18 at 19:10

score 0 · Accepted Answer · answered Feb 07 '18 at 19:07

0

The problem is related to your maxDiff function. The maxDiff function is applied over the entire dataset because an id is not specified. If you want to do a single id you need to specify it in the function before you call lapply.

maxDiff<-function(id,df) {
 a<-max(diff(df$SkinTemp[df["id"]==id]))
 a }

lapply(unique(id),maxDiff,df)

Or alternatively without a maxDiff you could specify the entire function in lapply/sapply like so

sapply(unique(df$id), function(x) max(diff(df$SkinTemp[df["id"]==x])))

answered Feb 07 '18 at 19:07

jasbner

2,253
12
24

Thank you for your answer, I'd like to accept this one as it's clearest for me to understand. Is there a way of catching or removing NA from either Time or SkinTemp within the function? – HCAI Feb 07 '18 at 20:00
you can use `is.na` to check for NA from the Time or Skin Temp and remove with something like `df[!is.na(df$Time),]` – jasbner Feb 07 '18 at 20:05
inside the function itself? I would prefer not to alter the initial data. – HCAI Feb 07 '18 at 20:08
Yes you can do this inside the function, check out https://stackoverflow.com/questions/4862178/remove-rows-with-nas-missing-values-in-data-frame – jasbner Feb 07 '18 at 20:20
1

Probably not the best example but here is a shot: `sapply(unique(df$id), function(x) max(diff(as.numeric(df[complete.cases(df),]$SkinTemp[df[complete.cases(df),]["id"]==x])))` – jasbner Feb 07 '18 at 20:24

score 0 · Answer 2 · answered Feb 07 '18 at 19:10

If I read this correctly, you are trying to find the max of each id, for that you can use a for loop.

get_Max_skin_temp_per_id<-function(){
df<-data.frame(Time=seq(100),
           SkinTemp=rnorm(100,37,0.5),
           id=rep(1:10,each=10))
maxSkin<-vector()
for (ids in 1:10) {
  a<-as.numeric(max(df$SkinTemp[ids]))
  maxSkin<-c(maxSkin, rep(a, 10))
}
df$maxSkin<-maxSkin
return(df)
}
get_Max_skin_temp_per_id()

this will rep each id's max in a column after the id column. hope it helps. to just get a list of the items. change to:

get_Max_skin_temp_per_id<-function(){
df<-data.frame(Time=seq(100),
           SkinTemp=rnorm(100,37,0.5),
           id=rep(1:10,each=10))
maxSkin<-vector()
for (ids in 1:10) {
  a<-as.numeric(max(df$SkinTemp[ids]))
  maxSkin<-c(maxSkin, a)
}

return(a)
}
get_Max_skin_temp_per_id()

Thank you for your answer, I appreciate your time. I've decided to go with one further down as it is easier for me as a novice to read at first glance. — HCAI, Feb 07 '18 at 20:01

Making a for loop over unique id's using lapply or aggregate

2 Answers2