I have a dataset that has 3 different conditions. Data within condition 1 will need to be divided by 15, data within conditions 2 and 3 will need to be divided by 10. I tried to do for() in order to create separate datasets for each condition and then merge the two groups (group 1 is composed of condition 1, group 2 is composed of conditions 2 and 3). This is what I have so far for condition 1. Is there an easier way to do this that does not require creating subgroups?
Group1 <- NULL
for (val in ParticipantID) {
ParticipantID_subset_Group1 <- subset(PronounData, ParticipantID == val & Condition == "1")
I_Words_PPM <- (ParticipantID_subset_Group1$I_Words/"15")
YOU_Words_PPM <- (ParticipantID_subset_Group1$YOU_Words/"15")
WE_Words_PPM <- (ParticipantID_subset_Group1$WE_Words/"15")
df <- data.frame(val, Group, I_Words_PPM, YOU_Words_PPM, WE_Words_PPM)
Group1 <- rbind(Group1, df)
}
dim(Group1)
colnames(Group1) <- c("ParticipantID", "Condition", "I_Words_PPM", "YOU_Words_PPM", "WE_Words_PPM")
View(Group1)
Couldn't fully test this solution without example data, but this should do what you want:
# make some fake data
PronounData <- data.frame(
ParticipantID = 1:9,
Condition = rep(1:3, 3),
I_Words = sample(0:20, 9, replace = TRUE),
YOU_Words = sample(0:40, 9, replace = TRUE),
WE_Words = sample(0:10, 9, replace = TRUE)
)
# if Condition 1, divide by 15
PronounData[PronounData$Condition == 1, c("I_Words_PPM", "YOU_Words_PPM", "WE_Words_PPM")] <-
PronounData[PronounData$Condition == 1, c("I_Words", "YOU_Words", "WE_Words")] / 15
# if Condition 2 or 3, divide by 10
PronounData[PronounData$Condition %in% 2:3, c("I_Words_PPM", "YOU_Words_PPM", "WE_Words_PPM")] <-
PronounData[PronounData$Condition %in% 2:3, c("I_Words", "YOU_Words", "WE_Words")] / 10
# result
PronounData
# ParticipantID Condition I_Words YOU_Words WE_Words I_Words_PPM YOU_Words_PPM WE_Words_PPM
# 1 1 1 17 40 6 1.1333 2.6667 0.4000
# 2 2 2 14 1 6 1.4000 0.1000 0.6000
# 3 3 3 2 34 8 0.2000 3.4000 0.8000
# 4 4 1 0 33 1 0.0000 2.2000 0.0667
# 5 5 2 4 15 0 0.4000 1.5000 0.0000
# 6 6 3 1 7 6 0.1000 0.7000 0.6000
# 7 7 1 6 10 1 0.4000 0.6667 0.0667
# 8 8 2 1 33 9 0.1000 3.3000 0.9000
# 9 9 3 9 40 0 0.9000 4.0000 0.0000
NB, R is built on vectorized operations, so looping through each row is rarely the best solution. Instead, you generally want to find a way of modifying whole vectors/columns at once, or at least subsets of them. This will usually be faster and simpler.