一直在使用SO作为我工作的资源。感谢您召集这样一个伟大的社区。
我正在尝试做一些有点复杂的事情，我现在能想到的唯一方法就是使用一对嵌套的for循环（我知道在R中不赞成）...我有记录三百万多课程注册：学生UserID与CourseID配对。在每一行中，都有一堆数据，包括开始/结束日期和分数等等。我需要做的是，对于每次注册，计算该用户在注册课程之前所参加的课程的平均分数。
我用于for循环的代码如下：
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
}
我很确定这会起作用，但它的运行速度非常慢......我的数据框有超过三百万行，但经过10分钟的好处后，外部循环只运行了850条记录。这似乎比时间复杂性所暗示的要慢，特别是考虑到每个用户平均只有5或6个课程。
哦，我应该提一下，在运行循环之前我用as.POSIXct（）转换了日期字符串，所以日期比较步骤不应该太慢......
必须有更好的方法来做到这一点......有什么建议吗？
编辑：按照mnel的要求......终于让dput玩得很好。不得不添加control = NULL。这是'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")
答案 0 :(得分：2)
我发现data.table效果很好。
# 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]$V1)
# For three million courses, at an average of 5 courses per student:
# user system elapsed
# 122.06 0.22 122.45
运行测试以查看它是否与您的代码相同：
set.seed(1)
n=1e2
data=data.table(UserID=as.character(round(runif(n,1,1000))),course=1:n,Score=runif(n),CourseStartDate=as.Date('2000-01-01')+round(runif(n,1,365)))
data$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
答案 1 :(得分：1)
这似乎是你想要的
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)]
根据@jorans评论，这将比上面的代码更复杂。
答案 2 :(得分：0)
这只是我认为可能所带来的解决方案的概述。为了简单起见，我将仅使用plyr来说明所需的步骤。
让我们仅限于一名学生的情况。如果我们可以为一个学生计算这个，那么通过某种分割应用来扩展它将是微不足道的。
所以我们假设我们有一个特定学生的分数，按课程结束日期排序：
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))
首先，我认为您需要按日期汇总，然后计算累计运行平均值：
dat_sum <- ddply(dat,.(date),summarise,valsum = sum(val),n = length(val))
dat_sum$mn <- with(dat_sum,cumsum(valsum) / cumsum(n))
最后，您将这些值合并回具有重复日期的原始数据：
dat_merge <- merge(dat,dat_sum[,c("date","mn")])
我可能会在 data.table 中使用匿名函数来编写所有这些步骤，但我怀疑其他人可能会更好地做一些简洁快速的事情。 （特别是，我不建议用 plyr 来解决这个问题，因为我怀疑它仍然会非常慢。）
答案 3 :(得分：0)
我认为这样的事情应该有效，尽管每个用户拥有多个课程的测试数据会很有帮助。也可能需要在findInterval的开始日期+1，以使条件为End.Date＆lt; Start.Date而不是＆lt; =。
# in the test data, one is POSIXct and the other a factor
data$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])])
答案 4 :(得分：0)
有三百万行，也许数据库是有用的。这里有一个sqlite示例，我相信它会创建类似于for循环的东西：
# 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$avg.p.score[[i]] <- sum / count
}
})
#variant 2: sqlite
system.time({
library(RSQLite)
con <- dbConnect("SQLite", ":memory:")
dbWriteTable(con, "enroll", enroll, overwrite=TRUE)
sql <- paste("Select e.user, e.course, Avg(p.score)",
"from enroll as e",
"cross join enroll as p",
"where e.user=p.user and p.c_end < e.c_start",
"group by e.user, e.course;")
res <- dbSendQuery(con, sql)
dat <- fetch(res, n=-1)
})
在我的机器上，sqlite快十倍。如果这还不够，就可以索引数据库。
答案 5 :(得分：0)
我无法对此进行测试，因为您的数据似乎不能满足任何组合中的不等式，但我会尝试这样的事情：
library(plyr)
res <- ddply(data, .(User.ID), function(d) {
with(subset(merge(d, d, by=NULL, suffixes=c(".i", ".j")),
Course.End.Date.j < Course.Start.Date.i),
c(Mean.Prior.Score = mean(Score.j)))
})
res$Mean.Prior.Score[is.nan(res$Mean.Prior.Score)] = 0
以下是它的工作原理：
User.ID对数据进行分组，并为每个d行的User.ID行执行函数.i，另一个.j User.ID列和Mean.Prior.Score mean将为零长度向量返回NaN，将其更改为零如果每个User.ID没有太多行，我想这可能会相当快。如果这还不够快，其他答案中提到的data.tables可能会有所帮助。
您的代码在所需的输出上有点模糊：您将data$Mean.Prior.Score视为长度为一的变量，但在循环的每次迭代中都将其赋值给它。我假设这个赋值只适用于一行。您是否需要数据框的每一行，或每个用户只有一个均值？