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In R, how can you count the number of observations fulfilling a condition over a time range?
Specifically, I want to count the number of different id by country over the last 8 months, but only if id occurs at least twice during these 8 months. Hence, for the count, it does not matter whether an id occurs 2x or 100x (doing this in 2 steps is maybe easier). NA exists both in id and country. Since this could otherwise be taken care off, accounting for this is not necessary but still helpful.
My current best try is, but does not account for the restriction (ID must appear at least twice in the previous 8 months) and also I find its counting odd when looking at the dates="2017-12-12", where desired_unrestricted should be equal to 4 according to my counting but the code gives 2.
dt[, date := as.Date(date)][
, totalids := sapply(date,
function(x) length(unique(id[between(date, x - lubridate::month(8), x)]))),
by = country]
Data
library(data.table)
library(lubridate)
ID <- c("1","1","1","1","1","1","2","2","2","3","3",NA,"4")
Date <- c("2017-01-01","2017-01-01", "2017-01-05", "2017-05-01", "2017-05-01","2018-05-02","2017-01-01", "2017-01-05", "2017-05-01", "2017-05-01","2017-05-01","2017-12-12","2017-12-12" )
Value <- c(2,4,3,5,2,5,8,17,17,3,7,5,3)
Country <- c("UK","UK","US","US",NA,"US","UK","UK","US","US","US","US","US")
Desired <- c(1,1,0,2,NA,0,1,2,2,2,2,1,1)
Desired_unrestricted <- c(2,2,1,3,NA,1,2,2,3,3,3,4,4)
dt <- data.frame(id=ID, date=Date, value=Value, country=Country, desired_output=Desired, desired_unrestricted=Desired_unrestricted)
setDT(dt)
Thanks in advance.
This data.table-only answer is motivated by a comment,
dt[, date := as.Date(date)] # if not already `Date`-class
dt[, date8 := do.call(c, lapply(dt$date, function(z) seq(z, length=2, by="-8 months")[2]))
][, results := dt[dt, on = .(country, date > date8, date <= date),
length(Filter(function(z) z > 1, table(id))), by = .EACHI]$V1
][, date8 := NULL ]
# id date value country desired_output desired_unrestricted results
# <char> <Date> <num> <char> <num> <num> <int>
# 1: 1 2017-01-01 2 UK 1 2 1
# 2: 1 2017-01-01 4 UK 1 2 1
# 3: 1 2017-01-05 3 US 0 1 0
# 4: 1 2017-05-01 5 US 1 3 2
# 5: 1 2017-05-01 2 <NA> NA NA 0
# 6: 1 2018-05-02 5 US 0 1 0
# 7: 2 2017-01-01 8 UK 1 2 1
# 8: 2 2017-01-05 17 UK 2 2 2
# 9: 2 2017-05-01 17 US 1 3 2
# 10: 3 2017-05-01 3 US 2 3 2
# 11: 3 2017-05-01 7 US 2 3 2
# 12: <NA> 2017-12-12 5 US 2 4 1
# 13: 4 2017-12-12 3 US 2 4 1
That's a lot to absorb.
Quick walk-through:
"8 months ago":
seq(z, length=2, by="-8 months")[2]
seq.Date (inferred by calling seq with a Date-class first argument) starts at z (current date for each row) and produces a sequence of length 2 with 8 months between them. seq always starts at the first argument, so length=1 won't work (it'll only return z); length=2 guarantees that the second value in the returned vector will be the "8 months before date" that we need.
Date subtraction:
[, date8 := do.call(c, lapply(dt$date, function(z) seq(...)[2])) ]
A simple base-R method for subtracting 8 months is seq(date, length=2, by="-8 months")[2]. seq.Date requires its first argument to be length-1, so we need to sapply or lapply it; unfortunately, sapply drops the class, so we lapply it and then programmatically combine them with do.call(c, ...) (since c(..) creates a list-column, and unlist will de-class it). (Perhaps this part can be improved.)
We need that in dt first since we do a non-equi (range-based) join based on this value.
Counting id with 2 or more visits:
length(Filter(function(z) z > 1, table(id)))
We produce a table(id), which gives us the count of each id within the join-period. Filter(fun, ...) allows us to reduce those that have a count below 2, and we're left with a named-vector of ids that had 2 or more visits. Retrieving the length is what we need.
Self non-equi join:
dt[dt, on = .(country, date > date8, date <= date), ... ]
Relatively straight-forward. This is an open/closed ranging, it can be changed to both-closed if you prefer.
Self non-equi join but count ids by-row: by=.EACHI.
Retrieve the results of that and assign into the original dt:
[, results := dt[...]$V1 ]
Since the non-equi join included a value (length(Filter(...))) without a name, it's named V1, and all we want is that. (To be honest, I don't know exactly why assigning it more directly doesn't work ... but the counts are all wrong. Perhaps it's backwards by-row tallying.)
Cleanup:
[, date8 := NULL ]
(Nothing fancy here, just proper data-stewardship :-)
There are some discrepancies in my counts versus your desired_output, I wonder if those are just typos in the OP; I think the math is right ...
Here is another option:
setkey(dt, country, date, id)
dt[, date := as.IDate(date)][,
eightmthsago := as.IDate(sapply(as.IDate(date), function(x) seq(x, by="-8 months", length.out=2L)[2L]))]
dt[, c("out", "out_unres") :=
dt[dt, on=.(country, date>=eightmthsago, date<=date),
by=.EACHI, {
v <- id[!is.na(id)]
.(uniqueN(v[duplicated(v)]), uniqueN(v))
}][,1L:3L := NULL]
]
dt
output (like r2evans, I am also getting different output from desired as there seems to be a miscount in the desired output):
id date value country desired_output desired_unrestricted eightmthsago out out_unres
1: 1 2017-05-01 2 <NA> NA NA 2016-09-01 0 1
2: 1 2017-01-01 2 UK 1 2 2016-05-01 1 2
3: 1 2017-01-01 4 UK 1 2 2016-05-01 1 2
4: 2 2017-01-01 8 UK 1 2 2016-05-01 1 2
5: 2 2017-01-05 17 UK 2 2 2016-05-05 2 2
6: 1 2017-01-05 3 US 0 1 2016-05-05 0 1
7: 1 2017-05-01 5 US 1 3 2016-09-01 2 3
8: 2 2017-05-01 17 US 1 3 2016-09-01 2 3
9: 3 2017-05-01 3 US 2 3 2016-09-01 2 3
10: 3 2017-05-01 7 US 2 3 2016-09-01 2 3
11: <NA> 2017-12-12 5 US 2 4 2017-04-12 1 4
12: 4 2017-12-12 3 US 2 4 2017-04-12 1 4
13: 1 2018-05-02 5 US 0 1 2017-09-02 0 2
Although this question is tagged with data.table, here is a dplyr::rowwise solution to the problem. Is this what you had in mind? The output looks valid to me: The number of ìds in the last 8 months which have a count of at least greater than 2.
library(dplyr)
library(lubridate)
dt <- dt %>% mutate(date = as.Date(date))
dt %>%
group_by(country) %>%
group_modify(~ .x %>%
rowwise() %>%
mutate(totalids = .x %>%
filter(date <= .env$date, date >= .env$date %m-% months(8)) %>%
pull(id) %>%
table() %>%
`[`(. >1) %>%
length
))
#> # A tibble: 13 x 7
#> # Groups: country [3]
#> country id date value desired_output desired_unrestricted totalids
#> <chr> <chr> <date> <dbl> <dbl> <dbl> <int>
#> 1 UK 1 2017-01-01 2 1 2 1
#> 2 UK 1 2017-01-01 4 1 2 1
#> 3 UK 2 2017-01-01 8 1 2 1
#> 4 UK 2 2017-01-05 17 2 2 2
#> 5 US 1 2017-01-05 3 0 1 0
#> 6 US 1 2017-05-01 5 1 3 2
#> 7 US 1 2018-05-02 5 0 1 0
#> 8 US 2 2017-05-01 17 1 3 2
#> 9 US 3 2017-05-01 3 2 3 2
#> 10 US 3 2017-05-01 7 2 3 2
#> 11 US <NA> 2017-12-12 5 2 4 1
#> 12 US 4 2017-12-12 3 2 4 1
#> 13 <NA> 1 2017-05-01 2 NA NA 0
Created on 2021-09-02 by the reprex package (v2.0.1)
Following data.table
df <- data.table(id=c(1,2,2,2,3,3,4,4,4),
start_date=c("2019-05-08","2019-08-01","2019-07-12","2017-05-24","2016-05-08","2017-08-01","2019-06-12","2017-02-24","2017-08-24"),
end_date=c("2019-09-08","2019-12-01","2019-07-30","2017-11-24","2017-07-25","2018-08-01","2019-12-12","2017-08-24","2018-08-24"),
variable1=c("a","c","c","d","a",NA,"a","a","b"))
df
id start_date end_date variable1
1: 1 2019-05-08 2019-09-08 a
2: 2 2019-08-01 2019-12-01 c
3: 2 2019-07-12 2019-07-30 c
4: 2 2017-05-24 2017-11-24 d
5: 3 2016-05-08 2017-07-25 a
6: 3 2017-08-01 2018-08-01 <NA>
7: 4 2019-06-12 2019-12-12 a
8: 4 2017-02-24 2017-08-24 a
9: 4 2017-08-24 2018-08-24 b
Within the same ID, I want to compare the start_date and end_date. If the end_date of one row is within 30 days of the start_date of another row, I want to combine the rows. So that it looks like this:
id start_date end_date variable1
1: 1 2019-05-08 2019-09-08 a
2: 2 2019-07-12 2019-12-01 c
3: 2 2017-05-24 2017-11-24 d
4: 3 2016-05-08 2018-08-01 a
5: 4 2019-06-12 2019-12-12 a
6: 4 2017-02-24 2017-08-24 a
7: 4 2017-08-24 2018-08-24 b
If the other variables of the rows are the same, rows should be combined with the earliest start_date and latest end_date as id number 2. If the variable1 is NA it should be replaced with values from the matching row as id number 3. If the variable1 has different values, rows should remain separate as id number 4.
The data.table contains more variables and objects than displayed here. Preferable a function in data.table.
Not clear what happens if an id has 3 overlapping rows with variable1 = c('a', NA, 'b'), what should the variable1 be for the NA for this case? a or b?
If we just choose the first variable1 when there are multiple matches, here is an option to first fill the NA and then borrow the idea from David Aurenburg's solution here
setorder(df, id, start_date, end_date)
df[, end_d := end_date + 30L]
df[is.na(variable1), variable1 :=
df[!is.na(variable1)][.SD, on=.(id, start_date<=start_date, end_d>=start_date), mult="first", x.variable1]]
df[, g:= c(0L, cumsum(shift(start_date, -1L) > cummax(as.integer(end_d)))[-.N]), id][,
.(start_date=min(start_date), end_date=max(end_date)), .(id, variable1, g)]
output:
id variable1 g start_date end_date
1: 1 a 0 2019-05-08 2019-09-08
2: 2 d 0 2017-05-24 2017-11-24
3: 2 c 1 2019-07-12 2019-12-01
4: 3 a 0 2016-05-08 2018-08-01
5: 4 a 0 2017-02-24 2017-08-24
6: 4 b 0 2017-08-24 2018-08-24
7: 4 a 1 2019-06-12 2019-12-12
data:
library(data.table)
df <- data.table(id=c(1,2,2,2,3,3,4,4,4),
start_date=as.IDate(c("2019-05-08","2019-08-01","2019-07-12","2017-05-24","2016-05-08","2017-08-01","2019-06-12","2017-02-24","2017-08-24")),
end_date=as.IDate(c("2019-09-08","2019-12-01","2019-07-30","2017-11-24","2017-07-25","2018-08-01","2019-12-12","2017-08-24","2018-08-24")),
variable1=c("a","c","c","d","a",NA,"a","a","b"))
I'm trying to add missing lines for "day" and extrapolate the data for "value". In my data each subject ("id") has 2 periods (period 1 and period 2) and values for consecutive days.
An example of my data looks like this:
df <- data.frame(
id = c(1,1,1,1, 1,1,1,1, 2,2,2,2, 2,2,2,2, 3,3,3,3, 3,3,3,3),
period = c(1,1,1,1, 2,2,2,2, 1,1,1,1, 2,2,2,2, 1,1,1,1, 2,2,2,2),
day= c(1,2,4,5, 1,3,4,5, 2,3,4,5, 1,2,3,5, 2,3,4,5, 1,2,3,4),
value =c(10,12,15,16, 11,14,15,17, 13,14,15,16, 15,16,18,20, 16,17,19,29, 14,16,18,20))
For each id and period I am missing data for days 3,2,1,4,1,5, respectively. I want to expand the data to let's say 10 days and extrapolate the data on value column (e.g. with linear regression).
My final df should be something like that:
df2 <- data.frame(
id = c(1,1,1,1,1,1,1, 1,1,1,1,1,1,1, 2,2,2,2,2,2,2, 2,2,2,2,2,2,2, 3,3,3,3,3,3,3, 3,3,3,3,3,3,3),
period = c(1,1,1,1,1,1,1, 2,2,2,2,2,2,2, 1,1,1,1,1,1,1, 2,2,2,2,2,2,2, 1,1,1,1,1,1,1, 2,2,2,2,2,2,2),
day= c(1,2,3,4,5,6,7, 1,2,3,4,5,6,7, 1,2,3,4,5,6,7, 1,2,3,4,5,6,7, 1,2,3,4,5,6,7, 1,2,3,4,5,6,7),
value =c(10,12,13,15,16,17,18, 11,12,14,15,17,18,19, 12,13,14,15,16,18,22, 15,16,18,19,20,22,23, 15,16,17,19,29,39,49, 14,16,18,20,22,24,26))
The most similar example I found doesn't extrapolate by two variables (ID and period in my case), it extrapolates only by year. I tried to adapt the code but no success :(
Another example extrapolates the data by multiple id but doesn't add rows for missing data.
I couldn't combine both codes with my limited experience in R. Any suggestions?
Thanks in advance...
We can use complete
library(dplyr)
library(tidyr)
library(forecast)
df %>%
group_by(id, period) %>%
complete(day =1:7)%>%
mutate(value = as.numeric(na.interp(value)))
#akrun's answer is good, as long as you don't mind using linear interpolation. However, if you do want to use a linear model, you could try this data.table approach.
library(data.table)
model <- lm(value ~ day + period + id,data=df)
dt <- as.data.table(df)[,.SD[,.(day = 1:7,value = value[match(1:7,day)])],by=.(id,period)]
dt[is.na(value), value := predict(model,.SD),]
dt
id period day value
1: 1 1 1 10.00000
2: 1 1 2 12.00000
3: 1 1 3 12.86714
4: 1 1 4 15.00000
5: 1 1 5 16.00000
6: 1 1 6 18.13725
7: 1 1 7 19.89396
8: 1 2 1 11.00000
9: 1 2 2 12.15545
10: 1 2 3 14.00000
11: 1 2 4 15.00000
12: 1 2 5 17.00000
13: 1 2 6 19.18227
14: 1 2 7 20.93898
15: 2 1 1 11.90102
16: 2 1 2 13.00000
17: 2 1 3 14.00000
18: 2 1 4 15.00000
19: 2 1 5 16.00000
20: 2 1 6 20.68455
21: 2 1 7 22.44125
22: 2 2 1 15.00000
23: 2 2 2 16.00000
24: 2 2 3 18.00000
25: 2 2 4 18.21616
26: 2 2 5 20.00000
27: 2 2 6 21.72957
28: 2 2 7 23.48627
29: 3 1 1 14.44831
30: 3 1 2 16.00000
31: 3 1 3 17.00000
32: 3 1 4 19.00000
33: 3 1 5 29.00000
34: 3 1 6 23.23184
35: 3 1 7 24.98855
36: 3 2 1 14.00000
37: 3 2 2 16.00000
38: 3 2 3 18.00000
39: 3 2 4 20.00000
40: 3 2 5 22.52016
41: 3 2 6 24.27686
42: 3 2 7 26.03357
id period day value
I have a list of 100+ time series dataframes my.list with daily observations for each product in its own data frame. Some values are NA without any record of the date. I would like to update each data frame in this list to show the date and NA if it does not have a record on this date.
Dates:
start = as.Date('2016/04/08')
full <- seq(start, by='1 days', length=10)
Sample Time Series Data:
d1 <- data.frame(Date = seq(start, by ='2 days',length=5), Sales = c(5,10,15,20,25))
d2 <- data.frame(Date = seq(start, by= '1 day', length=10),Sales = c(1, 2, 3,4,5,6,7,8,9,10))
my.list <- list(d1, d2)
I want to merge all full date values into each data frame, and if no match exists then sales is NA:
my.list
[[d1]]
Date Sales
2016-04-08 5
2016-04-09 NA
2016-04-10 10
2016-04-11 NA
2016-04-12 15
2016-04-13 NA
2016-04-14 20
2016-04-15 NA
2016-04-16 25
2016-04-17 NA
[[d2]]
Date Sales
2016-04-08 1
2016-04-09 2
2016-04-10 3
2016-04-11 4
2016-04-12 5
2016-04-13 6
2016-04-14 7
2016-04-15 8
2016-04-16 9
2016-04-17 10
If I understand correctly, the OP wants to update each of the dataframes in my.list to contain one row for each date given in the vector of dates full
Base R
In base R, merge() can be used as already mentioned by Hack-R. However, th answer below expands this to work on all dataframes in the list:
# creat dataframe from vector of full dates
full.df <- data.frame(Date = full)
# apply merge on each dataframe in the list
lapply(my.list, merge, y = full.df, all.y = TRUE)
[[1]]
Date Sales
1 2016-04-08 5
2 2016-04-09 NA
3 2016-04-10 10
4 2016-04-11 NA
5 2016-04-12 15
6 2016-04-13 NA
7 2016-04-14 20
8 2016-04-15 NA
9 2016-04-16 25
10 2016-04-17 NA
[[2]]
Date Sales
1 2016-04-08 1
2 2016-04-09 2
3 2016-04-10 3
4 2016-04-11 4
5 2016-04-12 5
6 2016-04-13 6
7 2016-04-14 7
8 2016-04-15 8
9 2016-04-16 9
10 2016-04-17 10
Caveat
The answer assumes that full covers the overall range of Date of all dataframes in the list.
In order to avoid any mishaps, the overall range of Date can be retrieved from the available data in my.list:
overall_date_range <- Reduce(range, lapply(my.list, function(x) range(x$Date)))
full <- seq(overall_date_range[1], overall_date_range[2], by = "1 days")
Using rbindlist()
Alternatively, the list of dataframes which are identical in structure can be stored in a large dataframe. An additional attribute indicates to which product each row belongs to. The homogeneous structure simplifies subsequent operations.
The code below uses the rbindlist() function from the data.table package to create a large data.table. CJ() (cross join) creates all combinations of dates and product id which is then merged / joined to fill in the missing dates:
library(data.table)
all_products <- rbindlist(my.list, idcol = "product.id")[
CJ(product.id = unique(product.id), Date = seq(min(Date), max(Date), by = "1 day")),
on = .(Date, product.id)]
all_products
product.id Date Sales
1: 1 2016-04-08 5
2: 1 2016-04-09 NA
3: 1 2016-04-10 10
4: 1 2016-04-11 NA
5: 1 2016-04-12 15
6: 1 2016-04-13 NA
7: 1 2016-04-14 20
8: 1 2016-04-15 NA
9: 1 2016-04-16 25
10: 1 2016-04-17 NA
11: 2 2016-04-08 1
12: 2 2016-04-09 2
13: 2 2016-04-10 3
14: 2 2016-04-11 4
15: 2 2016-04-12 5
16: 2 2016-04-13 6
17: 2 2016-04-14 7
18: 2 2016-04-15 8
19: 2 2016-04-16 9
20: 2 2016-04-17 10
Subsequent operations can be grouped by product.id, e.g., to determine the number of valid sales data for each product:
all_products[!is.na(Sales), .(valid.sales.data = .N), by = product.id]
product.id valid.sales.data
1: 1 5
2: 2 10
Or, the totals sales per product:
all_products[, .(total.sales = sum(Sales, na.rm = TRUE)), by = product.id]
product.id total.sales
1: 1 75
2: 2 55
If required for some reason the result can be converted back to a list by
split(all_products, by = "product.id")
I am looking for a function in R similar to lag1, lag2 and retain functions in SAS which I can use with data.tables.
I know there are functions like embed and lag in R but they don't return a single value or the previous value . They return a complete set of vectors.
Is there anything in R which I can use with data.table?
More info on the SAS functions :
Retain
Lag
You have to be aware that R works very different from the data step in SAS. The lag function in SAS is used in the data step, and is used within the implicit loop structure of that data step. The same goes for the retain function, which simply keeps the value constant when going through the data looping.
R on the other hand works completely vectorized. This means that you have to rethink what you want to do, and adapt accordingly.
retain is simply useless in R, as R recycles arguments by default. If you want to do this explicitly, you might look at eg rep() to construct a vector with constant values and a certain length.
lag is a matter of using indices, and just shifting position of all values in a vector. In order to keep a vector of the same length, you need to add some NA and remove some extra values.
A simple example: This SAS code lags a variable x and adds a variable year that has a constant value:
data one;
retain year 2013;
input x ##;
y=lag1(x);
z=lag2(x);
datalines;
1 2 3 4 5 6
;
In R, you could write your own lag function like this:
mylag <- function(x,k) c(rep(NA,k),head(x,-k))
This single line adds k times NA at the beginning of the vector, and drops the last k values from the vector. The result is a lagged vector as given by lag1 etc. in SAS.
this allows something like :
nrs <- 1:6 # equivalent to datalines
one <- data.frame(
x = nrs,
y = mylag(nrs,1),
z = mylag(nrs,2),
year = 2013 # R automatically loops, so no extra command needed
)
The result is :
> one
x y z year
1 1 NA NA 2013
2 2 1 NA 2013
3 3 2 1 2013
4 4 3 2 2013
5 5 4 3 2013
6 6 5 4 2013
Exactly the same would work with a data.table object. The important note here is to rethink your strategy: Instead of thinking loopwise as you do with the DATA step in SAS, you have to start thinking in terms of vectors and indices when using R.
I would say the closet equivalent to retain, lag1, and lag2 would be the Lag function in the quantmod package.
It's very easy to use with data.tables. E.g.:
library(data.table)
library(quantmod)
d <- data.table(v1=c(rep('a', 10), rep('b', 10)), v2=1:20)
setkeyv(d, 'v1')
d[,new_var := Lag(v2, 1), by='v1']
d[,new_var2 := v2-Lag(v2, 3), by='v1']
d[,new_var3 := Next(v2, 2), by='v1']
This yields the following:
print(d)
v1 v2 new_var new_var2 new_var3
1: a 1 NA NA 3
2: a 2 1 NA 4
3: a 3 2 NA 5
4: a 4 3 3 6
5: a 5 4 3 7
6: a 6 5 3 8
7: a 7 6 3 9
8: a 8 7 3 10
9: a 9 8 3 NA
10: a 10 9 3 NA
11: b 11 NA NA 13
12: b 12 11 NA 14
13: b 13 12 NA 15
14: b 14 13 3 16
15: b 15 14 3 17
16: b 16 15 3 18
17: b 17 16 3 19
18: b 18 17 3 20
19: b 19 18 3 NA
20: b 20 19 3 NA
As you can see, Lag lets you look back and Next lets you look forward. Both functions are nice because they pad the result with NAs such that it has the same length as the input.
If you want to get even fancier, and higher-performance, you can look into rolling joins with data.table objects. This is a little bit different thab what you are asking for, but is conceptually related, and so powerful and awesome I have to share.
Start with a data.table:
library(data.table)
library(quantmod)
set.seed(42)
d1 <- data.table(
id=c(rep('a', 10), rep('b', 10)),
time=rep(1:10,2),
value=runif(20))
setkeyv(d1, c('id', 'time'))
print(d1)
id time value
1: a 1 0.9148060
2: a 2 0.9370754
3: a 3 0.2861395
4: a 4 0.8304476
5: a 5 0.6417455
6: a 6 0.5190959
7: a 7 0.7365883
8: a 8 0.1346666
9: a 9 0.6569923
10: a 10 0.7050648
11: b 1 0.4577418
12: b 2 0.7191123
13: b 3 0.9346722
14: b 4 0.2554288
15: b 5 0.4622928
16: b 6 0.9400145
17: b 7 0.9782264
18: b 8 0.1174874
19: b 9 0.4749971
20: b 10 0.5603327
You have another data.table you want to join, but not all time indexes are present in the second table:
d2 <- data.table(
id=sample(c('a', 'b'), 5, replace=TRUE),
time=sample(1:10, 5),
value2=runif(5))
setkeyv(d2, c('id', 'time'))
print(d2)
id time value2
1: a 4 0.811055141
2: a 10 0.003948339
3: b 6 0.737595618
4: b 8 0.388108283
5: b 9 0.685169729
A regular merge yields lots of missing values:
d2[d1,,roll=FALSE]
id time value2 value
1: a 1 NA 0.9148060
2: a 2 NA 0.9370754
3: a 3 NA 0.2861395
4: a 4 0.811055141 0.8304476
5: a 5 NA 0.6417455
6: a 6 NA 0.5190959
7: a 7 NA 0.7365883
8: a 8 NA 0.1346666
9: a 9 NA 0.6569923
10: a 10 0.003948339 0.7050648
11: b 1 NA 0.4577418
12: b 2 NA 0.7191123
13: b 3 NA 0.9346722
14: b 4 NA 0.2554288
15: b 5 NA 0.4622928
16: b 6 0.737595618 0.9400145
17: b 7 NA 0.9782264
18: b 8 0.388108283 0.1174874
19: b 9 0.685169729 0.4749971
20: b 10 NA 0.5603327
However, data.table allows you to roll the secondary index forward, WITHIN THE PRIMARY INDEX!
d2[d1,,roll=TRUE]
id time value2 value
1: a 1 NA 0.9148060
2: a 2 NA 0.9370754
3: a 3 NA 0.2861395
4: a 4 0.811055141 0.8304476
5: a 5 0.811055141 0.6417455
6: a 6 0.811055141 0.5190959
7: a 7 0.811055141 0.7365883
8: a 8 0.811055141 0.1346666
9: a 9 0.811055141 0.6569923
10: a 10 0.003948339 0.7050648
11: b 1 NA 0.4577418
12: b 2 NA 0.7191123
13: b 3 NA 0.9346722
14: b 4 NA 0.2554288
15: b 5 NA 0.4622928
16: b 6 0.737595618 0.9400145
17: b 7 0.737595618 0.9782264
18: b 8 0.388108283 0.1174874
19: b 9 0.685169729 0.4749971
20: b 10 0.685169729 0.5603327
This is pretty damn cool: Old observations are rolled forward in time, until they are replaced by new ones. If you want to replace the NA values at the beggining of the series, you can do so by rolling the first observation backwards:
d2[d1,,roll=TRUE, rollends=c(TRUE, TRUE)]
id time value2 value
1: a 1 0.811055141 0.9148060
2: a 2 0.811055141 0.9370754
3: a 3 0.811055141 0.2861395
4: a 4 0.811055141 0.8304476
5: a 5 0.811055141 0.6417455
6: a 6 0.811055141 0.5190959
7: a 7 0.811055141 0.7365883
8: a 8 0.811055141 0.1346666
9: a 9 0.811055141 0.6569923
10: a 10 0.003948339 0.7050648
11: b 1 0.737595618 0.4577418
12: b 2 0.737595618 0.7191123
13: b 3 0.737595618 0.9346722
14: b 4 0.737595618 0.2554288
15: b 5 0.737595618 0.4622928
16: b 6 0.737595618 0.9400145
17: b 7 0.737595618 0.9782264
18: b 8 0.388108283 0.1174874
19: b 9 0.685169729 0.4749971
20: b 10 0.685169729 0.5603327
These rolling joins are absolutely incredible, and I've never seen them implemented in any other open source package (see ?data.table for more info). It will take a little while to turn off your "SAS brain" and turn on your "R brain", but once you get over that initial hump you'll find that the language is much more expressive.
For retain, try this :
retain<-function(x,event,outside=NA)
{
indices <- c(1,which(event==TRUE), nrow(df)+1)
values <- c(outside,x[event==TRUE])
y<- rep(values, diff(indices))
}
With data : I want to retain down the value when w==b
df <- data.frame(w = c("a","b","c","a","b","c"), x = 1:6, y = c(1,1,2,2,2,3), stringsAsFactors = FALSE)
df$z<-retain(df$x-df$y,df$w=="b")
df
And here's the contrary obtain, that does not exist in SAS:
obtain<-function(x,event,outside=NA)
{
indices <- c(0,which(event==TRUE), nrow(df))
values <- c(x[event==TRUE],outside)
y<- rep(values, diff(indices))
}
Here's an example. I want to obtain the value in advance where w==b
df$z2<-obtain(df$x-df$y,df$w=="b")
df
Thanks to Julien for helping.
here's an example: cumulate value with sqldf:
> w_cum <-
sqldf("select t1.id, t1.SomeNumt, SUM(t2.SomeNumt) as cum_sum
from w_cum t1
inner join w_cum t2 on t1.id >= t2.id
group by t1.id, t1.SomeNumt
order by t1.id
")
id SomeNumt cum_sum
1 11 11
2 12 23
3 13 36
4 14 50
5 15 65
6 16 81
7 17 98
8 18 116
9 19 135
10 20 155