Vectorising function on subset of dataframe based on other columns

Question

I have a dataframe from a psychology experiment with the time since the beginning of the experiment for each subject, and what I want is to set from that the time since the beginning of each trial for each subject. To do so I'm basically just substracting the minimum time value for each trial/subject to all the values for that same trial/subject.

I'm currently doing it with two for loops, I was just wondering if there's a way to vectorise it. What I have at the minute:

for (s in 1:max(df$Subject)){
  subject <- df[df$Subject==s,]
  for (t in 1:max(subject$TrialId)){
    trial <- subject[subject$TrialId==t,]
    start_offset <- min(trial$timestamp)
    df$timestamp[df$Subject==s & df$TrialId==t] <- df$timestamp[df$Subject==s &
                                                                df$TrialId==t]
                                                     - start_offset
  }
}

And what I would like is something like

df$timestamp <- df$timestamp - min_per_trial_per_subject(df$timestamp)

Can you add a sample of your dataset? https://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example — Ale, Nov 08 '17 at 16:05
The full dataset is technically on Github, would that be sufficient? — Arthur Spoon, Nov 08 '17 at 16:06

score 3 · Accepted Answer · answered Nov 08 '17 at 16:08

3

With dplyr

library(dplyr)
df %>% group_by(Subject, TrialId) %>%
  mutate(modified_timestamp = timestamp - min(timestamp))

Should work. If it doesn't, please share a reproducible example so we can test.

answered Nov 08 '17 at 16:08

Gregor Thomas

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Works just fine! Although you still need to put `df <-` at the beginning of your line (not familiar with the `%>%` notation so didn't know that). Thanks a lot! – Arthur Spoon Nov 08 '17 at 16:17
3

I don't like to make assumptions about assignment - you might prefer `df_modified <- ...` – Gregor Thomas Nov 08 '17 at 16:20

Vectorising function on subset of dataframe based on other columns

1 Answers1