Converting this ugly for-loop to something more R-friendly - r

Question

Been using SO as a resource constantly for my work. Thanks for holding together such a great community.
I'm trying to do something kinda complex, and the only way I can think to do it right now is with a pair of nested for-loops (I know that's frowned upon in R)... I have records of three million-odd course enrollments: student UserID's paired with CourseID's. In each row, there's a bunch of data including start/end dates and scores and so forth. What I need to do is, for each enrollment, calculate the average score for that user across the courses she's taken before the course in the enrollment.
The code I'm using for the for-loop follows:
data$Mean.Prior.Score <- 0
for (i in as.numeric(rownames(data)) {
sum <- 0
count <- 0
for (j in as.numeric(rownames(data[data$UserID == data$UserID[i],]))) {
if (data$Course.End.Date[j] < data$Course.Start.Date[i]) {
sum <- sum + data$Score[j]
count <- count + 1
}
}
if (count != 0)
data$Mean.Prior.Score[i] <- sum / count
}
I'm pretty sure this would work, but it runs incredibly slowly... my data frame has over three million rows, but after a good 10 minutes of chugging, the outer loop has only run through 850 of the records. That seems way slower than the time complexity would suggest, especially given that each user has only 5 or 6 courses to her name on average.
Oh, and I should mention that I converted the date strings with as.POSIXct() before running the loop, so the date comparison step shouldn't be too terribly slow...
There's got to be a better way to do this... any suggestions?
Edit: As per mnel's request... finally got dput to play nicely. Had to add control = NULL. Here 'tis:
structure(list(Username = structure(1:20, .Label = c("100225",
"100226", "100228", "1013170", "102876", "105796", "106753",
"106755", "108568", "109038", "110150", "110200", "110350", "111873",
"111935", "113579", "113670", "117562", "117869", "118329"), class = "factor"),
User.ID = c(2313737L, 2314278L, 2314920L, 9708829L, 2325896L,
2315617L, 2314644L, 2314977L, 2330148L, 2315081L, 2314145L,
2316213L, 2317734L, 2314363L, 2361187L, 2315374L, 2314250L,
2361507L, 2325592L, 2360182L), Course.ID = c(2106468L, 2106578L,
2106493L, 5426406L, 2115455L, 2107320L, 2110286L, 2110101L,
2118574L, 2106876L, 2110108L, 2110058L, 2109958L, 2108222L,
2127976L, 2106638L, 2107020L, 2127451L, 2117022L, 2126506L
), Course = structure(c(1L, 7L, 10L, 15L, 11L, 19L, 4L, 6L,
3L, 12L, 2L, 9L, 17L, 8L, 20L, 18L, 13L, 16L, 5L, 14L), .Label = c("ACCT212_A",
"BIOS200_N", "BIS220_T", "BUSN115_A", "BUSN115_T", "CARD205_A",
"CIS211_A", "CIS275_X", "CIS438_S", "ENGL112_A", "ENGL112_B",
"ENGL227_K", "GM400_A", "GM410_A", "HUMN232_M", "HUMN432_W",
"HUMN445_A", "MATH100_X", "MM575_A", "PSYC110_Y"), class = "factor"),
Course.Start.Date = structure(c(1098662400, 1098662400, 1098662400,
1309737600, 1099267200, 1098662400, 1099267200, 1099267200,
1098662400, 1098662400, 1099267200, 1099267200, 1099267200,
1098662400, 1104105600, 1098662400, 1098662400, 1104105600,
1098662400, 1104105600), class = c("POSIXct", "POSIXt"), tzone = "GMT"),
Term.ID = c(12056L, 12056L, 12056L, 66282L, 12057L, 12056L,
12057L, 12057L, 12056L, 12056L, 12057L, 12057L, 12057L, 12056L,
13469L, 12056L, 12056L, 13469L, 12056L, 13469L), Term.Name = structure(c(2L,
2L, 2L, 4L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 1L, 2L, 3L, 2L,
2L, 3L, 2L, 3L), .Label = c("Fall 2004", "Fall 2004 Session A",
"Fall 2004 Session B", "Summer Session A 2011"), class = "factor"),
Term.Start.Date = structure(c(1L, 1L, 1L, 4L, 2L, 1L, 2L,
2L, 1L, 1L, 2L, 2L, 2L, 1L, 3L, 1L, 1L, 3L, 1L, 3L), .Label = c("2004-10-21",
"2004-10-28", "2004-12-27", "2011-06-26"), class = "factor"),
Score = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0.125,
0, 0, 0, 0, 0), First.Course.Date = structure(c(1L, 1L, 1L,
4L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 3L, 1L, 1L, 3L,
1L, 3L), .Label = c("2004-10-25", "2004-11-01", "2004-12-27",
"2011-07-04"), class = "factor"), First.Term.Date = structure(c(1L,
1L, 1L, 4L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 1L, 3L, 1L,
1L, 3L, 1L, 3L), .Label = c("2004-10-21", "2004-10-28", "2004-12-27",
"2011-06-26"), class = "factor"), First.Timer = c(TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE,
TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE, TRUE), Course.Code = structure(c(1L,
6L, 9L, 13L, 9L, 17L, 4L, 5L, 3L, 10L, 2L, 8L, 15L, 7L, 18L,
16L, 11L, 14L, 4L, 12L), .Label = c("ACCT212", "BIOS200",
"BIS220", "BUSN115", "CARD205", "CIS211", "CIS275", "CIS438",
"ENGL112", "ENGL227", "GM400", "GM410", "HUMN232", "HUMN432",
"HUMN445", "MATH100", "MM575", "PSYC110"), class = "factor"),
Course.End.Date = structure(c(1L, 1L, 1L, 4L, 2L, 1L, 2L,
2L, 1L, 1L, 2L, 2L, 2L, 1L, 3L, 1L, 1L, 3L, 1L, 3L), .Label = c("2004-12-19",
"2005-02-27", "2005-03-26", "2011-08-28"), class = "factor")), .Names = c("Username",
"User.ID", "Course.ID", "Course", "Course.Start.Date", "Term.ID",
"Term.Name", "Term.Start.Date", "Score", "First.Course.Date",
"First.Term.Date", "First.Timer", "Course.Code", "Course.End.Date"
), row.names = c(NA, 20L), class = "data.frame")

Answer 1

I found that data.table worked well.
# Create some data.
library(data.table)
set.seed(1)
n=3e6
numCourses=5 # Average courses per student
data=data.table(UserID=as.character(round(runif(n,1,round(n/numCourses)))),course=1:n,Score=runif(n),CourseStartDate=as.Date('2000-01-01')+round(runif(n,1,365)))
data$CourseEndDate=data$CourseStartDate+round(runif(n,1,100))
setkey(data,UserID)
# test=function(CourseEndDate,Score,CourseStartDate) sapply(CourseStartDate, function(y) mean(Score[y>CourseEndDate]))
# I vastly reduced the number of comparisons with a better "test" function.
test2=function(CourseEndDate,Score,CourseStartDate) {
o.end = order(CourseEndDate)
run.avg = cumsum(Score[o.end])/seq_along(CourseEndDate)
idx=findInterval(CourseStartDate,CourseEndDate[o.end])
idx=ifelse(idx==0,NA,idx)
run.avg[idx]
}
system.time(data$MeanPriorScore<-data[,test2(CourseEndDate,Score,CourseStartDate),by=UserID]$V1)
# For three million courses, at an average of 5 courses per student:
# user system elapsed
# 122.06 0.22 122.45
Running a test to see if it looks the same as your code:
set.seed(1)
n=1e2
data=data.table(UserID=as.character(round(runif(n,1,1000))),course=1:n,Score=runif(n),CourseStartDate=as.Date('2000-01-01')+round(runif(n,1,365)))
data$CourseEndDate=data$CourseStartDate+round(runif(n,1,100))
setkey(data,UserID)
data$MeanPriorScore<-data[,test2(CourseEndDate,Score,CourseStartDate),by=UserID]$V1
data["246"]
# UserID course Score CourseStartDate CourseEndDate MeanPriorScore
#1: 246 54 0.4531314 2000-08-09 2000-09-20 0.9437248
#2: 246 89 0.9437248 2000-02-19 2000-03-02 NA
# A comparison with your for loop (slightly modified)
data$MeanPriorScore.old<-NA # Set to NaN instead of zero for easy comparison.
# I think you forgot a bracket here. Also, There is no need to work with the rownames.
for (i in seq(nrow(data))) {
sum <- 0
count <- 0
# I reduced the complexity of figuring out the vector to loop through.
# It will result in the exact same thing if there are no rownames.
for (j in which(data$UserID == data$UserID[i])) {
if (data$CourseEndDate[j] <= data$CourseStartDate[i]) {
sum <- sum + data$Score[j]
count <- count + 1
}
}
# I had to add "[i]" here. I think that is what you meant.
if (count != 0) data$MeanPriorScore.old[i] <- sum / count
}
identical(data$MeanPriorScore,data$MeanPriorScore.old)
# [1] TRUE

Answer 2

This seems to be what you want
library(data.table)
# create a data.table object
DT <- data.table(data)
# key by userID
setkeyv(DT, 'userID')
# for each userID, where the Course.End.Date < Course.Start.Date
# return the mean score
# This is too simplistic
# DT[Course.End.Date < Course.Start.Date,
# list(Mean.Prior.Score = mean(Score)) ,
# by = list(userID)]
As per #jorans comment, this will be more complex than the code above.

Answer 3

This is only an outline of what I think a solution might entail. I'm going to use plyr just to illustrate the steps needed, for simplicity.
Let's just restrict ourselves to the case of one student. If we can calculate this for one student, extending it with some sort of split-apply will be trivial.
So let's suppose we have scores for a particular student, sorted by course end date:
d <- sample(seq(as.Date("2011-01-01"),as.Date("2011-01-31"),by = 1),100,replace = TRUE)
dat <- data.frame(date = sort(d),val = rnorm(100))
First, I think you'd need to summarise this by date and then calculate the cumulative running mean:
dat_sum <- ddply(dat,.(date),summarise,valsum = sum(val),n = length(val))
dat_sum$mn <- with(dat_sum,cumsum(valsum) / cumsum(n))
Finally, you'd merge these values back into the original data with the duplicate dates:
dat_merge <- merge(dat,dat_sum[,c("date","mn")])
I could probably write something that does this in data.table using an anonymous function to do all those steps, but I suspect others may be better able to do something that will be concise and fast. (In particular, I don't recommend actually tackling this with plyr, as I suspect it will still be extremely slow.)

Answer 4

I think something like this should work though it'd be helpful to have test data with multiple courses per user. Also might need +1 on the start dates in findInterval to make condition be End.Date < Start.Date instead of <=.
# in the test data, one is POSIXct and the other a factor
data$Course.Start.Date = as.Date(data$Course.Start.Date)
data$Course.End.Date = as.Date(data$Course.End.Date)
data = data[order(data$Course.End.Date), ]
data$Mean.Prior.Score = ave(seq_along(data$User.ID), data$User.ID, FUN=function(i)
c(NA, cumsum(data$Score[i]) / seq_along(i))[1L + findInterval(data$Course.Start.Date[i], data$Course.End.Date[i])])

Answer 5

With three million rows, maybe a database is helpful. Here an sqlite example which I believe creates something similar to your for loop:
# data.frame for testing
user <- sample.int(10000, 100)
course <- sample.int(10000, 100)
c_start <- sample(
seq(as.Date("2004-01-01"), by="3 months", length.ou=12),
100, replace=TRUE
)
c_end <- c_start + as.difftime(11, units="weeks")
c_idx <- sample.int(100, 1000, replace=TRUE)
enroll <- data.frame(
user=sample(user, 1000, replace=TRUE),
course=course[c_idx],
c_start=as.character(c_start[c_idx]),
c_end=as.character(c_end[c_idx]),
score=runif(1000),
stringsAsFactors=FALSE
)
#variant 1: for-loop
system.time({
enroll$avg.p.score <- NA
for (i in 1:nrow(enroll)) {
sum <- 0
count <- 0
for (j in which(enroll$user==enroll$user[[i]]))
if (enroll$c_end[[j]] < enroll$c_start[[i]]) {
sum <- sum + enroll$score[[j]]
count <- count + 1
}
if(count !=0) enroll$avg.p.score[[i]] <- sum / count
}
})
#variant 2: sqlite
system.time({
library(RSQLite)
con <- dbConnect("SQLite", ":memory:")
dbWriteTable(con, "enroll", enroll, overwrite=TRUE)
sql <- paste("Select e.user, e.course, Avg(p.score)",
"from enroll as e",
"cross join enroll as p",
"where e.user=p.user and p.c_end < e.c_start",
"group by e.user, e.course;")
res <- dbSendQuery(con, sql)
dat <- fetch(res, n=-1)
})
On my machine, sqlite is ten times faster. If that is not enough, it would be possible to index the database.

Answer 6

I can't really test this, as your data doesn't appear to satisfy the inequality in any combination, but I'd try something like this:
library(plyr)
res <- ddply(data, .(User.ID), function(d) {
with(subset(merge(d, d, by=NULL, suffixes=c(".i", ".j")),
Course.End.Date.j < Course.Start.Date.i),
c(Mean.Prior.Score = mean(Score.j)))
})
res$Mean.Prior.Score[is.nan(res$Mean.Prior.Score)] = 0
Here is how it works:
ddply: Group data by User.ID and execute function for each subset d of rows for one User.ID
merge: Create two copies of the data for one user, one with columns suffixed .i the other .j
subset: From this outer join, only select those matching the given inequality
mean: Compute the mean for the matched rows
c(…): Give a name to the resulting column
res: Will be a data.frame with columns User.ID and Mean.Prior.Score
is.nan: mean will return NaN for zero-length vectors, change these to zeros
I guess this might be reasonably fast if there are not too many rows for each User.ID. If this isn't fast enough, the data.tables mentioned in other answers might help.
Your code is a bit fuzzy on the desired output: you treat data$Mean.Prior.Score like a length-one variable, but assign to it in every iteration of the loop. I assume that this assignment is meant only for one row. Do you need means for every row of the data frame, or only one mean per user?