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I have a 2 column data frame with "date" and "ID" headings. Some IDs are listed more than once. I want to create a new column "Attempt" that denotes the number of attempts that each ID has taken, ordered by the date of occurrence.
Here is my sample data:
ID <- c(1,2,5,8,4,9,1,11,15,32,54,1,4,2,14)
Date <- c("2021-04-12", "2021-04-12", "2021-04-13", "2021-04-14", "2021-04-19",
"2021-04-19", "2021-04-20", "2021-04-21", "2021-04-22", "2021-04-28",
"2021-04-28", "2021-04-29", "2021-04-29", "2021-05-06", "2021-05-07")
Data <- data.frame(ID, Date)
Data$Date <- as.Date(Data$Date, format="%Y-%m-%d")
I tried various iterations of duplicated(). I can remove all duplicates or make every instance of a duplicated value "2" or "3" for example, but I want each occurrence to be ordered based on the date of the attempt taken.
Here is my expected result column to be added onto the original data frame:
Attempt <- c(1,1,1,1,1,1,2,1,1,1,1,3,2,2,1)
Data %>%
group_by(ID)
mutate(Attempt1 = row_number())
ID Date Attempt
1 1 2021-04-12 1
2 2 2021-04-12 1
3 5 2021-04-13 1
4 8 2021-04-14 1
5 4 2021-04-19 1
6 9 2021-04-19 1
7 1 2021-04-20 2
8 11 2021-04-21 1
9 15 2021-04-22 1
10 32 2021-04-28 1
11 54 2021-04-28 1
12 1 2021-04-29 3
13 4 2021-04-29 2
14 2 2021-05-06 2
15 14 2021-05-07 1
If you have the latest version of dplyr use
Data %>%
mutate(Attempt = row_number(), .by = ID)
Using data.table
library(data.table)
setDT(Data)[, Attempt := rowid(ID)]
-output
> Data
ID Date Attempt
1: 1 2021-04-12 1
2: 2 2021-04-12 1
3: 5 2021-04-13 1
4: 8 2021-04-14 1
5: 4 2021-04-19 1
6: 9 2021-04-19 1
7: 1 2021-04-20 2
8: 11 2021-04-21 1
9: 15 2021-04-22 1
10: 32 2021-04-28 1
11: 54 2021-04-28 1
12: 1 2021-04-29 3
13: 4 2021-04-29 2
14: 2 2021-05-06 2
15: 14 2021-05-07 1
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)
Question: In data.table is there any way to fill an incomplete date sequence with zeros? For instance, in the toy example some dates does not show up and I want to have a complete date sequence with y = 0 for these cases. Is there something like a forward fill?
Remark: Note I do not want to use merges where you create first the full date sequence and then merge it back to the initial data.table object (I think this is inefficient and rather slow).
library(data.table)
dt <- data.table(
x = c("2020-03-28", "2020-03-29", "2020-03-31", "2020-04-05"),
y = c(1, 5, 3, 70)
)
## Output:
x y
1: 2020-03-28 1
2: 2020-03-29 5
3: 2020-03-31 3
4: 2020-04-05 70
## Desired Output:
x y
1: 2020-03-28 1
2: 2020-03-29 5
3: 2020-03-30 0
4: 2020-03-31 3
5: 2020-04-01 0
6: 2020-04-02 0
7: 2020-04-03 0
8: 2020-04-04 0
9: 2020-04-05 70
How about this?
# convert to data.table's integer date type
dt[ , x := as.IDate(x)]
# find the range of dates
date_bounds = range(dt$x)
# construct a sequence of all dates
# NB: this will be integers as attributes are stripped
all_dates = date_bounds[1L]:date_bounds[2L]
# construct a table with the missing dates,
# with y filled to 0
missing = data.table(
# as.IDate uses the right origin for integer input
x = as.IDate(setdiff(all_dates, dt$x)),
y = 0
)
dt = rbind(dt, missing)
# x y
# <IDat> <num>
# 1: 2020-03-28 1
# 2: 2020-03-29 5
# 3: 2020-03-31 3
# 4: 2020-04-05 70
# 5: 2020-03-30 0
# 6: 2020-04-01 0
# 7: 2020-04-02 0
# 8: 2020-04-03 0
# 9: 2020-04-04 0
Afterwards you can setorder(dt, x) if you want the dates to be in order
Use CJ and tidyr::full_seq to create a join data table.
dt[, x := as.Date(x)] # convert x to the Date type
dt2 <- dt[CJ(x = tidyr::full_seq(x, 1)), on = .(x)] # create the full sequence
dt2[is.na(y), y := 0] # fill NAs with 0s
dt2
# x y
# 1: 2020-03-28 1
# 2: 2020-03-29 5
# 3: 2020-03-30 0
# 4: 2020-03-31 3
# 5: 2020-04-01 0
# 6: 2020-04-02 0
# 7: 2020-04-03 0
# 8: 2020-04-04 0
# 9: 2020-04-05 70
You could use complete from tidyr :
library(dplyr)
library(tidyr)
dt %>%
mutate(x = as.Date(x)) %>%
complete(x = seq(min(x), max(x), by = "day"), fill = list(y = 0))
# x y
# <date> <dbl>
#1 2020-03-28 1
#2 2020-03-29 5
#3 2020-03-30 0
#4 2020-03-31 3
#5 2020-04-01 0
#6 2020-04-02 0
#7 2020-04-03 0
#8 2020-04-04 0
#9 2020-04-05 70
You could also try this:
dt[, x := as.IDate(x)]
dt[.(seq(min(x), max(x), 1)), .(y = fifelse(is.na(y), 0, y)), .EACHI, on = "x"]
# x y
# 1: 2020-03-28 1
# 2: 2020-03-29 5
# 3: 2020-03-30 0
# 4: 2020-03-31 3
# 5: 2020-04-01 0
# 6: 2020-04-02 0
# 7: 2020-04-03 0
# 8: 2020-04-04 0
# 9: 2020-04-05 70
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
By using the data below, I want to create a new unique customer id by considering their contact date.
Rule: After every two days, I want each customer to get a new unique customer id and preserve it on the following record if the following contact date for the same customer is within the following two days if not assign a new id to this same customer.
I couldn't go any further than calculating date differences.
The original dataset I work is bigger; therefore, I prefer a data.table solution if possible.
library(data.table)
treshold <- 2
dt <- structure(list(customer_id = c('10','20','20','20','20','20','30','30','30','30','30','40','50','50'),
contact_date = as.Date(c("2019-01-05","2019-01-01","2019-01-01","2019-01-02",
"2019-01-08","2019-01-09","2019-02-02","2019-02-05",
"2019-02-05","2019-02-09","2019-02-12","2019-02-01",
"2019-02-01","2019-02-05")),
desired_output = c(1,2,2,2,3,3,4,5,5,6,7,8,9,10)),
class = "data.frame",
row.names = 1:14)
setDT(dt)
setorder(dt, customer_id, contact_date)
dt[, date_diff_in_days:=contact_date - shift(contact_date, type = c("lag")), by=customer_id]
dt[, date_diff_in_days:=as.numeric(date_diff_in_days)]
dt
customer_id contact_date desired_output date_diff_in_days
1: 10 2019-01-05 1 NA
2: 20 2019-01-01 2 NA
3: 20 2019-01-01 2 0
4: 20 2019-01-02 2 1
5: 20 2019-01-08 3 6
6: 20 2019-01-09 3 1
7: 30 2019-02-02 4 NA
8: 30 2019-02-05 5 3
9: 30 2019-02-05 5 0
10: 30 2019-02-09 6 4
11: 30 2019-02-12 7 3
12: 40 2019-02-01 8 NA
13: 50 2019-02-01 9 NA
14: 50 2019-02-05 10 4
Rule: After every two days, I want each customer to get a new unique customer id and preserve it on the following record if the following contact date for the same customer is within the following two days if not assign a new id to this same customer.
When creating a new ID, if you set up the by= vectors correctly to capture the rule, the auto-counter .GRP can be used:
thresh <- 2
dt[, g := .GRP, by=.(
customer_id,
cumsum(contact_date - shift(contact_date, fill=first(contact_date)) > thresh)
)]
dt[, any(g != desired_output)]
# [1] FALSE
I think the code above is correct since it works on the example, but you might want to check on your actual data (comparing against results from, eg, Gregor's approach) to be sure.
We use cumsum to increment whenever date_diff_in_days is NA or when the threshold is exceeded.
dt[, result := cumsum(is.na(date_diff_in_days) | date_diff_in_days > treshold)]
# customer_id contact_date desired_output date_diff_in_days result
# 1: 10 2019-01-05 1 NA 1
# 2: 20 2019-01-01 2 NA 2
# 3: 20 2019-01-01 2 0 2
# 4: 20 2019-01-02 2 1 2
# 5: 20 2019-01-08 3 6 3
# 6: 20 2019-01-09 3 1 3
# 7: 30 2019-02-02 4 NA 4
# 8: 30 2019-02-05 5 3 5
# 9: 30 2019-02-05 5 0 5
# 10: 30 2019-02-09 6 4 6
# 11: 30 2019-02-12 7 3 7
# 12: 40 2019-02-01 8 NA 8
# 13: 50 2019-02-01 9 NA 9
# 14: 50 2019-02-05 10 4 10