I'm not used to using R. I already asked a question on stack overflow and got a great answer.
I'm sorry to post a similar question, but I tried many times and got the output that I didn't expect.
This time, I want to do slightly different from my previous question.
Merge two data with respect to date and week using R
I have two data. One has a year_month_week column and the other has a date column.
df1<-data.frame(id=c(1,1,1,2,2,2,2),
year_month_week=c(2022051,2022052,2022053,2022041,2022042,2022043,2022044),
points=c(65,58,47,21,25,27,43))
df2<-data.frame(id=c(1,1,1,2,2,2),
date=c(20220503,20220506,20220512,20220401,20220408,20220409),
temperature=c(36.1,36.3,36.6,34.3,34.9,35.3))
For df1, 2022051 means 1st week of May,2022. Likewise, 2022052 means 2nd week of May,2022. For df2,20220503 means May 3rd, 2022. What I want to do now is merge df1 and df2 with respect to year_month_week. In this case, 20220503 and 20220506 are 1st week of May,2022.If more than one date are in year_month_week, I will just include the first of them. Now, here's the different part. Even if there is no date inside year_month_week,just leave it NA. So my expected output has a same number of rows as df1 which includes the column year_month_week.So my expected output is as follows:
df<-data.frame(id=c(1,1,1,2,2,2,2),
year_month_week=c(2022051,2022052,2022053,2022041,2022042,2022043,2022044),
points=c(65,58,47,21,25,27,43),
temperature=c(36.1,36.6,NA,34.3,34.9,NA,NA))
First we can convert the dates in df2 into year-month-date format, then join the two tables:
library(dplyr);library(lubridate)
df2$dt = ymd(df2$date)
df2$wk = day(df2$dt) %/% 7 + 1
df2$year_month_week = as.numeric(paste0(format(df2$dt, "%Y%m"), df2$wk))
df1 %>%
left_join(df2 %>% group_by(year_month_week) %>% slice(1) %>%
select(year_month_week, temperature))
Result
Joining, by = "year_month_week"
id year_month_week points temperature
1 1 2022051 65 36.1
2 1 2022052 58 36.6
3 1 2022053 47 NA
4 2 2022041 21 34.3
5 2 2022042 25 34.9
6 2 2022043 27 NA
7 2 2022044 43 NA
You can build off of a previous answer here by taking the function to count the week of the month, then generate a join key in df2. See here
df1 <- data.frame(
id=c(1,1,1,2,2,2,2),
year_month_week=c(2022051,2022052,2022053,2022041,2022042,2022043,2022044),
points=c(65,58,47,21,25,27,43))
df2 <- data.frame(
id=c(1,1,1,2,2,2),
date=c(20220503,20220506,20220512,20220401,20220408,20220409),
temperature=c(36.1,36.3,36.6,34.3,34.9,35.3))
# Take the function from the previous StackOverflow question
monthweeks.Date <- function(x) {
ceiling(as.numeric(format(x, "%d")) / 7)
}
# Create a year_month_week variable to join on
df2 <-
df2 %>%
mutate(
date = lubridate::parse_date_time(
x = date,
orders = "%Y%m%d"),
year_month_week = paste0(
lubridate::year(date),
0,
lubridate::month(date),
monthweeks.Date(date)),
year_month_week = as.double(year_month_week))
# Remove duplicate year_month_weeks
df2 <-
df2 %>%
arrange(year_month_week) %>%
distinct(year_month_week, .keep_all = T)
# Join dataframes
df1 <-
left_join(
df1,
df2,
by = "year_month_week")
Produces this result
id.x year_month_week points id.y date temperature
1 1 2022051 65 1 2022-05-03 36.1
2 1 2022052 58 1 2022-05-12 36.6
3 1 2022053 47 NA <NA> NA
4 2 2022041 21 2 2022-04-01 34.3
5 2 2022042 25 2 2022-04-08 34.9
6 2 2022043 27 NA <NA> NA
7 2 2022044 43 NA <NA> NA
>
Edit: forgot to mention that you need tidyverse loaded
library(tidyverse)
Related
I practice data merging using R nowadays. Here are simple two data df1 and df2.
df1<-data.frame(id=c(1,1,1,2,2,2,2),
year_month=c(202205,202206,202207,202204,202205,202206,202207),
points=c(65,58,47,21,25,27,43))
df2<-data.frame(id=c(1,1,1,2,2,2),
year_month_week=c(2022052,2022053,2022061,2022043,2022051,2022052),
temperature=c(36.1,36.3,36.6,34.3,34.9,35.3))
For df1, 202205 in year_month column means May 2022.
For df2, 2022052 in year_month_week column means 2nd week of May, 2022.
I want to merge df1 and df2 with respect to year_month_week. So, all the elements of df2 are left, but some values of df2 can be copied.
For example, 202205 in year_month includes 2022052 and 2022053. There is no column points in df2. In this case, 65 is copied. My expected output looks like this:
df<-data.frame(id=c(1,1,1,2,2,2),
year_month_week=c(2022052,2022053,2022061,2022043,2022051,2022052),
temperature=c(36.1,36.3,36.6,34.3,34.9,35.3),
points=c(65,65,58,21,25,25))
Create a temporary year_month column in df2 by taking the first six characters of year_month_week, then do a left join on df1 by year_month and id before removing the temporary column.
Using tidyverse, we could do this as follows:
library(tidyverse)
df2 %>%
mutate(year_month = as.numeric(substr(year_month_week, 1, 6))) %>%
left_join(df1, by = c('year_month', 'id')) %>%
select(-year_month)
#> id year_month_week temperature points
#> 1 1 2022052 36.1 65
#> 2 1 2022053 36.3 65
#> 3 1 2022061 36.6 58
#> 4 2 2022043 34.3 21
#> 5 2 2022051 34.9 25
#> 6 2 2022052 35.3 25
Or in base R using merge:
df2$year_month <- substr(df2$year_month_week, 1, 6)
merge(df2, df1, by = c('year_month', 'id'))[-1]
#> id year_month_week temperature points
#> 1 2 2022043 34.3 21
#> 2 1 2022052 36.1 65
#> 3 1 2022053 36.3 65
#> 4 2 2022051 34.9 25
#> 5 2 2022052 35.3 25
#> 6 1 2022061 36.6 58
Following is what my dataframe/data.table looks like. The rank column is my desired calculated field.
library(data.table)
df <- fread('
Name Score Date Rank
John 42 1/1/2018 3
Rob 85 12/31/2017 2
Rob 89 12/26/2017 1
Rob 57 12/24/2017 1
Rob 53 08/31/2017 1
Rob 72 05/31/2017 2
Kate 87 12/25/2017 1
Kate 73 05/15/2017 1
')
df[,Date:= as.Date(Date, format="%m/%d/%Y")]
I am trying to calculate the rank of each student at every given point in time in the data within a 30 day windows. For that, I need to fetch the most recent scores of all students at a given point in time and then pass the rank function.
In the 1st row, as of 1/1/2018, John has two more competitors in a past 30 day window: Rob with the most recent score of 85 in 12/31/2017 AND Kate with the most recent score of 87 in 12/25/2017 and both of these dates fall within the 1/1/2018 - 30 Day Window. John gets a rank of 3 with the lowest score of 42. If only one students falls within date(at a given row) - 30 day window, then the rank is 1.
In the 3rd row the date is 12/26/2017. So Rob's score as of 12/26/2017 is 89. There is only one case of another student that falls in the time window of 12/26/2017 - 30 days and that is the most recent score(87) of kate on 12/25/2017. So within the time window of (12/26/2017) - 30 , Rob's score of 89 is higher than Kate's score of 87 and therefore Rob gets rank 1.
I was thinking about using the framework from here Efficient way to perform running total in the last 365 day window but struggling to think of a way to fetch all recent score of all students at a given point in time before using rank.
This seems to work:
ranks = df[.(d_dn = Date - 30L, d_up = Date), on=.(Date >= d_dn, Date <= d_up), allow.cart=TRUE][,
.(LatestScore = last(Score)), by=.(Date = Date.1, Name)]
setorder(ranks, Date, -LatestScore)
ranks[, r := rowid(Date)]
df[ranks, on=.(Name, Date), r := i.r]
Name Score Date Rank r
1: John 42 2018-01-01 3 3
2: Rob 85 2017-12-31 2 2
3: Rob 89 2017-12-26 1 1
4: Rob 57 2017-12-24 1 1
5: Rob 53 2017-08-31 1 1
6: Rob 72 2017-05-31 2 2
7: Kate 87 2017-12-25 1 1
8: Kate 73 2017-05-15 1 1
... using last since the Cartesian join seems to sort and we want the latest measurement.
How the update join works
The i. prefix means it's a column from i in the x[i, ...] join, and the assignment := is always in x. So it's looking up each row of i in x and where matches are found, copying values from i to x.
Another way that is sometimes useful is to look up x rows in i, something like df[, r := ranks[df, on=.(Name,Date), x.r]] in which case x.r is still from the ranks table (now in the x position relative to the join).
There's also...
ranks = df[CJ(Name = Name, Date = Date, unique=TRUE), on=.(Name, Date), roll=30, nomatch=0]
setnames(ranks, "Score", "LatestScore")
# and then use the same last three lines above
I'm not sure about efficiency of one vs another, but I guess it depends on number of Names, frequency of measurement and how often measurement days coincide.
A solution that uses data.table though not sure if it is the most efficient usage:
df[.(iName=Name, iScore=Score, iDate=Date, StartDate=Date-30, EndDate=Date),
.(Rank=frank(-c(iScore[1L], .SD[Name != iName, max(Score), by=.(Name)]$V1),
ties.method="first")[1L]),
by=.EACHI,
on=.(Date >= StartDate, Date <= EndDate)]
Explanation:
1) The outer square brackets do a non-equi join within a date range (i.e. 30days ago and latest date for each row). Try studying the below output against the input data:
df[.(iName=Name, iScore=Score, iDate=Date, StartDate=Date-30, EndDate=Date),
c(.(RowGroup=.GRP),
.SD[, .(Name, Score, Date, OrigDate, iName, iScore, iDate, StartDate, EndDate)]),
by=.EACHI,
on=.(Date >= StartDate, Date <= EndDate)]
2) .EACHI is to perform j calculations for each row of i.
3) Inside j, iScore[1L] is the score for the current row, .SD[Name != iName] means taking scores not corresponding to the student in the current row. Then, we use the max(Score) for each student of those students within the 30days window.
4) Concatenate all these scores and calculate the rank for the score of the current row while taking care of ties by taking the first tie.
Note:
see ?data.table to understand what i, j, by, on and .EACHI refers to.
EDIT after comments by OP:
I would add a OrigDate column and find those that matches the latest date.
df[, OrigDate := Date]
df[.(iName=Name, iScore=Score, iDate=Date, StartDate=Date-30, EndDate=Date),
.(Name=iName, Score=iScore, Date=iDate,
Rank=frank(-c(iScore[1L],
.SD[Name != iName, Score[OrigDate==max(OrigDate)], by=.(Name)]$V1),
ties.method="first")[1L]),
by=.EACHI,
on=.(Date >= StartDate, Date <= EndDate)]
I came up with following partial solution, encountered however problem - is it possible that there will be two people occuring with the same date?
if not, have a look at following piece of code:
library(tidyverse) # easy manipulation
library(lubridate) # time handling
# This function can be added to
get_top <- function(df, date_sel) {
temp <- df %>%
filter(Date > date_sel - months(1)) %>% # look one month in the past from given date
group_by(Name) %>% # and for each occuring name
summarise(max_score = max(Score)) %>% # find the maximal score
arrange(desc(max_score)) %>% # sort them
mutate(Rank = 1:n()) # and rank them
temp
}
Now, you have to find the name in the table, for given date and return its rank.
library(data.table)
library(magrittr)
setorder(df, -Date)
fun <- function(i){
df[i:nrow(df), head(.SD, 1), by = Name] %$%
rank(-Score[Date > df$Date[i] - 30])[1]
}
df[, rank := sapply(1:.N, fun)]
This can be done by joining to df those rows of df that are within 30 days behind it or the same date and have higher or equal scores. Then for each original row and joined row Name get the joined row Name that is the most recent. The count of the remaining joined rows for each of the original df rows is the rank.
library(sqldf)
sqldf("with X as
(select a.rowid r, a.*, max(b.Date) Date
from df a join df b
on b.Date between a.Date - 30 and a.Date and b.Score >= a.Score
group by a.rowid, b.Name)
select Name, Date, Score, count(*) Rank
from X
group by r
order by r")
giving:
Name Date Score Rank
1 John 2018-01-01 42 3
2 Rob 2017-12-31 85 2
3 Rob 2017-12-26 89 1
4 Rob 2017-12-24 57 1
5 Rob 2017-08-31 53 1
6 Rob 2017-05-31 72 2
7 Kate 2017-12-25 87 1
8 Kate 2017-05-15 73 1
A tidyverse solution (dplyr + tidyr):
df %>%
complete(Name,Date) %>%
group_by(Name) %>%
mutate(last_score_date = `is.na<-`(Date,is.na(Score))) %>%
fill(Score,last_score_date) %>%
filter(!is.na(Score) & Date-last_score_date <30) %>%
group_by(Date) %>%
mutate(Rank = rank(-Score)) %>%
right_join(df)
# # A tibble: 8 x 5
# # Groups: Date [?]
# Name Date Score last_score_date Rank
# <chr> <date> <int> <date> <dbl>
# 1 John 2018-01-01 42 2018-01-01 3
# 2 Rob 2017-12-31 85 2017-12-31 2
# 3 Rob 2017-12-26 89 2017-12-26 1
# 4 Rob 2017-12-24 57 2017-12-24 1
# 5 Rob 2017-08-31 53 2017-08-31 1
# 6 Rob 2017-05-31 72 2017-05-31 2
# 7 Kate 2017-12-25 87 2017-12-25 1
# 8 Kate 2017-05-15 73 2017-05-15 1
We add all missing combinations of Date and Name
then we create a column for the last_score_date, equal to Date when score isn't NA.
by filling NAs down Score has become the latest score
we filter out NAs and keep only scores that have < 30 days of age
That's our table of valid scores by dates
From there it's easy to add ranks
and a final right_join on the original table gives us the expected output
data
library(data.table)
df <- fread('
Name Score Date
John 42 01/01/2018
Rob 85 12/31/2017
Rob 89 12/26/2017
Rob 57 12/24/2017
Rob 53 08/31/2017
Rob 72 05/31/2017
Kate 87 12/25/2017
Kate 73 05/15/2017
')
df[,Date:= as.Date(Date, format="%m/%d/%Y")]
I have a dataset of a hypothetical exam.
id <- c(1,1,3,4,5,6,7,7,8,9,9)
test_date <- c("2012-06-27","2012-07-10","2013-07-04","2012-03-24","2012-07-22", "2013-09-16","2012-06-21","2013-10-18", "2013-04-21", "2012-02-16", "2012-03-15")
result_date <- c("2012-07-29","2012-09-02","2013-08-01","2012-04-25","2012-09-01","2013-10-20","2012-07-01","2013-10-31", "2013-05-17", "2012-03-17", "2012-04-20")
data1 <- as_data_frame(id)
data1$test_date <- test_date
data1$result_date <- result_date
colnames(data1)[1] <- "id"
"id" indicates the ID of the students who have taken a particular exam. "test_date" is the date the students took the test and "result_date" is the date when the students' results are posted. I'm interested in finding out which students retook the exam BEFORE the result of that exam session was released, e.g. students who knew that they have underperformed and retook the exam without bothering to find out their scores. For example, student with "id" 1 took the exam for the second time on "2012-07-10" which was before the result date for his first exam - "2012-07-29".
I tried to:
data1%>%
group_by(id) %>%
arrange(id, test_date) %>%
filter(n() >= 2) %>% #To only get info on students who have taken the exam more than once and then merge it back in with the original data set using a join function
So essentially, I want to create a new column called "re_test" where it would equal 1 if a student retook the exam BEFORE receiving the result of a previous exam and 0 otherwise (those who retook after seeing their marks or those who did not retake).
I have tried to mutate in order to find cases where dates are either positive or negative by subtracting the 2nd test_date from the 1st result_date:
mutate(data1, re_test = result_date - lead(test_date, default = first(test_date)))
However, this leads to mixing up students with different id's. I tried to split but mutate won't work on a list of dataframes so now I'm stuck:
split(data1, data1$id)
Just to add on, this is a part of the desired result:
data2 <- as_data_frame(id <- c(1,1,3,4))
data2$test_date_result <- c("2012-06-27","2012-07-10", "2013-07-04","2012-03-24")
data2$result_date_result <- c("2012-07-29","2012-09-02","2013-08-01","2012-04-25")
data2$re_test <- c(1, 0, 0, 0)
Apologies for the verbosity and hope I was clear enough.
Thanks a lot in advance!
library(reshape2)
library(dplyr)
# first melt so that we can sequence by date
data1m <- data1 %>%
melt(id.vars = "id", measure.vars = c("test_date", "result_date"), value.name = "event_date")
# any two tests in a row is a flag - use dplyr::lag to comapre the previous
data1mc <- data1m %>%
arrange(id, event_date) %>%
group_by(id) %>%
mutate (multi_test = (variable == "test_date" & lag(variable == "test_date"))) %>%
filter(multi_test)
# id variable event_date multi_test
# 1 1 test_date 2012-07-10 TRUE
# 2 9 test_date 2012-03-15 TRUE
## join back to the original
data1 %>%
left_join (data1mc %>% select(id, event_date, multi_test),
by=c("id" = "id", "test_date" = "event_date"))
I have a piecewise answer that may work for you. I first create a data.frame called student that contains the re-test information, and then join it with the data1 object. If students re-took the test multiple times, it will compare the last test to the first, which is a flaw, but I'm unsure if students have the ability to re-test multiple times?
student <- data1 %>%
group_by(id) %>%
summarise(retest=(test_date[length(test_date)] < result_date[1]) == TRUE)
Some re-test values were NA. These were individuals that only took the test once. I set these to FALSE here, but you can retain the NA, as they do contain information.
student$retest[is.na(student$retest)] <- FALSE
Join the two data.frames to a single object called data2.
data2 <- left_join(data1, student, by='id')
I am sure there are more elegant ways to approach this. I did this by taking advantage of the structure of your data (sorted by id) and the lag function that can refer to the previous records while dealing with a current record.
### Ensure Data are sorted by ID ###
data1 <- arrange(data1,id)
### Create Flag for those that repeated ###
data1$repeater <- ifelse(lag(data1$id) == data1$id,1,0)
### I chose to do this on all data, you could filter on repeater flag first ###
data1$timegap <- as.Date(data1$result_date) - as.Date(data1$test_date)
data1$lagdate <- as.Date(data1$test_date) - lag(as.Date(data1$result_date))
### Display results where your repeater flag is 1 and there is negative time lag ###
data1[data1$repeater==1 & !is.na(data1$repeater) & as.numeric(data1$lagdate) < 0,]
# A tibble: 2 × 6
id test_date result_date repeater timegap lagdate
<dbl> <chr> <chr> <dbl> <time> <time>
1 1 2012-07-10 2012-09-02 1 54 days -19 days
2 9 2012-03-15 2012-04-20 1 36 days -2 days
I went with a simple shift comparison. 1 line of code.
data1 <- data.frame(id = c(1,1,3,4,5,6,7,7,8,9,9), test_date = c("2012-06-27","2012-07-10","2013-07-04","2012-03-24","2012-07-22", "2013-09-16","2012-06-21","2013-10-18", "2013-04-21", "2012-02-16", "2012-03-15"), result_date = c("2012-07-29","2012-09-02","2013-08-01","2012-04-25","2012-09-01","2013-10-20","2012-07-01","2013-10-31", "2013-05-17", "2012-03-17", "2012-04-20"))
data1$re_test <- unlist(lapply(split(data1,data1$id), function(x)
ifelse(as.Date(x$test_date) > c(NA, as.Date(x$result_date[-nrow(x)])), 0, 1)))
data1
id test_date result_date re_test
1 1 2012-06-27 2012-07-29 NA
2 1 2012-07-10 2012-09-02 1
3 3 2013-07-04 2013-08-01 NA
4 4 2012-03-24 2012-04-25 NA
5 5 2012-07-22 2012-09-01 NA
6 6 2013-09-16 2013-10-20 NA
7 7 2012-06-21 2012-07-01 NA
8 7 2013-10-18 2013-10-31 0
9 8 2013-04-21 2013-05-17 NA
10 9 2012-02-16 2012-03-17 NA
11 9 2012-03-15 2012-04-20 1
I think there is benefit in leaving NAs but if you really want all others as zero, simply:
data1$re_test <- ifelse(is.na(data1$re_test), 0, data1$re_test)
data1
id test_date result_date re_test
1 1 2012-06-27 2012-07-29 0
2 1 2012-07-10 2012-09-02 1
3 3 2013-07-04 2013-08-01 0
4 4 2012-03-24 2012-04-25 0
5 5 2012-07-22 2012-09-01 0
6 6 2013-09-16 2013-10-20 0
7 7 2012-06-21 2012-07-01 0
8 7 2013-10-18 2013-10-31 0
9 8 2013-04-21 2013-05-17 0
10 9 2012-02-16 2012-03-17 0
11 9 2012-03-15 2012-04-20 1
Let me know if you have any questions, cheers.
I have a data frame (track) with the position (longitude - Latitude) and date (number of the day in the year) of tracking point for different animals and an other data frame (var) which gives a the mean temperature for every day of the year in different locations.
I would like to add a new column TEMP to my data frame (Track) where the value would be from (var) and correspond to the date and GPS location of each tracking points in (track).
Here are a really simple subset of my data and what I would like to obtain.
track = data.frame(
animals=c(1,1,1,2,2),
Longitude=c(117,116,117,117,116),
Latitude=c(18,20,20,18,20),
Day=c(1,3,4,1,5))
Var = data.frame(
Longitude=c(117,117,116,116),
Latitude=c(18,20,18,20),
Day1=c(22,23,24,21),
Day2=c(21,28,27,29),
Day3=c(12,13,14,11),
Day4=c(17,19,20,23),
Day5=c(32,33,34,31)
)
TrackPlusVar = data.frame(
animals=c(1,1,1,2,2),
Longitude=c(117,116,117,117,116),
Latitude=c(18,20,20,18,20),
Day=c(1,3,4,1,5),
Temp= c(22,11,19,22,31)
)
I've no idea how to assign the value from the same date and GPS location as it is a column name. Any idea would be very useful !
This is a dplyr and tidyr approach.
library(dplyr)
library(tidyr)
# reshape table Var
Var %>%
gather(Day,Temp,-Longitude, -Latitude) %>%
mutate(Day = as.numeric(gsub("Day","",Day))) -> Var2
# join tables
track %>% left_join(Var2, by=c("Longitude", "Latitude", "Day"))
# animals Longitude Latitude Day Temp
# 1 1 117 18 1 22
# 2 1 116 20 3 11
# 3 1 117 20 4 19
# 4 2 117 18 1 22
# 5 2 116 20 5 31
If the process that creates your tables makes sure that all your cases belong to both tables, then you can use inner_join instead of left_join to make the process faster.
If you're still not happy with the speed you can use a data.table join process to check if it is faster, like:
library(data.table)
Var2 = setDT(Var2, key = c("Longitude", "Latitude", "Day"))
track = setDT(track, key = c("Longitude", "Latitude", "Day"))
Var2[track][order(animals,Day)]
# Longitude Latitude Day Temp animals
# 1: 117 18 1 22 1
# 2: 116 20 3 11 1
# 3: 117 20 4 19 1
# 4: 117 18 1 22 2
# 5: 116 20 5 31 2
Okay, so I have two different data frames (df1 and df2) which, to simplify it, have an ID, a date, and the score on a test. In each data frame the person (ID) have taken the test on multiple dates. When looking between the two data frames, some of the people are listed in df1 but not in df2, and vice versa, but some are listed in both and they can overlap differently.
I want to combine all the data into one frame, but the tricky part is if any of the IDs and scores from df1 and df2 are within 7 days (I can do this with a subtracted dates column), I want to combine that row.
In essence, for every ID there will be one row with both scores written separately if taken within 7 days, and if not it will make two separate rows, one with score from df1 and one from df2 along with all the other scores that might not be listed in both.
EX:
df1
ID Date1(yyyymmdd) Score1
1 20140512 50
1 20140501 30
1 20140703 50
1 20140805 20
3 20140522 70
3 20140530 10
df2
ID Date2(yyyymmdd) Score2
1 20140530 40
1 20140622 20
1 20140702 10
1 20140820 60
2 20140522 30
2 20140530 80
Wanted_df
ID Date1(yyyymmdd) Score1 Date2(yyyymmdd) Score2
1 20140512 50
1 20140501 30
1 20140703 50 20140702 10
1 20140805 20
1 20140530 40
1 20140622 20
1 20140820 60
3 20140522 70
3 20140530 10
2 20140522 30
2 20140530 80
Alright. I feel bad about the bogus outer join answer (which may be possible in a library I don't know about, but there are advantages to using RDBMS sometimes...) so here is a hacky workaround. It assumes that all the joins will be at most one to one, which you've said is OK.
# ensure the date columns are date type
df1$Date1 <- as.Date(as.character(df1$Date1), format="%Y%m%d")
df2$Date2 <- as.Date(as.character(df2$Date2), format="%Y%m%d")
# ensure the dfs are sorted
df1 <- df1[order(df1$ID, df1$Date1),]
df2 <- df2[order(df2$ID, df2$Date2),]
# initialize the output df3, which starts as everything from df1 and NA from df2
df3 <- cbind(df1,Date2=NA, Score2=NA)
library(plyr) #for rbind.fill
for (j in 1:nrow(df2)){
# see if there are any rows of test1 you could join test2 to
join_rows <- which(df3[,"ID"]==df2[j,"ID"] & abs(df3[,"Date1"]-df2[j,"Date2"])<7 )
# if so, join it to the first one (see discussion)
if(length(join_rows)>0){
df3[min(join_rows),"Date2"] <- df2[j,"Date2"]
df3[min(join_rows),"Score2"] <- df2[j,"Score2"]
} # if not, add a new row of just the test2
else df3 <- rbind.fill(df3,df2[j,])
}
df3 <- df3[order(df3$ID,df3$Date1,df3$Date2),]
row.names(df3)<-NULL # i hate these
df3
# ID Date1 Score1 Date2 Score2
# 1 1 2014-05-01 30 <NA> NA
# 2 1 2014-05-12 50 <NA> NA
# 3 1 2014-07-03 50 2014-07-02 10
# 4 1 2014-08-05 20 <NA> NA
# 5 1 <NA> NA 2014-05-30 40
# 6 1 <NA> NA 2014-06-22 20
# 7 1 <NA> NA 2014-08-20 60
# 8 2 <NA> NA 2014-05-22 30
# 9 2 <NA> NA 2014-05-30 80
# 10 3 2014-05-22 70 <NA> NA
# 11 3 2014-05-30 10 <NA> NA
I couldn't get the rows in the same sort order as yours, but they look the same.
Short explanation: For each row in df2, see if there's a row in df1 you can "join" it to. If not, stick it at the bottom of the table. In the initialization and rbinding, you'll see some hacky ways of assigning blank rows or columns as placeholders.
Why this is a bad hacky workaround: for large data sets, the rbinding of df3 to itself will consume more and more memory. The loop is definitely not optimal and its search does not exploit the fact that the tables are sorted. If by some chance the test were taken twice within a week, you would see some unexpected behavior (duplicates from df2, etc).
Use an outer join with an absolute value limit on the date difference. (A outer join B keeps all rows of A and B.) For example:
library(sqldf)
sqldf("select a.*, b.* from df1 a outer join df2 b on a.ID = b.ID and abs(a.Date1 - b.Date2) <=7")
Note that your date variables will have to be true dates. If they are currently characters or integers, you need to do something like df1$Date1 <- as.Date(as.character(df$Date1), format="%Y%M%D) etc.