Rolling cumulative product with NAs R - r

I have the following database (X) that contains monthly stock returns over time. I show the first 12 rows. The stock returns can contain random NAs.
Obs. Asset Date Ret
1 DJ 1997-10-06 NA
2 DJ 1997-10-07 NA
3 DJ 1997-10-08 -1.13
4 DJ 1997-10-09 -0.136
5 DJ 1997-10-10 NA
6 DJ 1997-10-14 NA
7 DJ 1997-10-15 NA
8 DJ 1997-10-16 -0.225
9 DJ 1997-10-17 -0.555
10 DJ 1997-10-20 NA
11 DJ 1997-10-21 0.102
12 DJ 1997-10-22 NA
I want to calculate the cumulative return over a 5 day window. So I get a cumulative return from observation 5 on, ignoring NAs. The cumulative return will only be a NA when the returns within the window are also NA.
I tried:
Y <- Y %>%
mutate(product = (as.numeric(rollapply(1 + ret/100, 5, prod,
partial = TRUE, na.rm = TRUE, align = "right"))-1)*100)
Which gives an undesired result:
> 1 1997-10-06 DJ NA 0.000000000
> 2 1997-10-07 DJ NA 0.000000000
> 3 1997-10-08 DJ -1.1277917526 -1.127791753
> 4 1997-10-09 DJ -0.1364864885 -1.262738958
> 5 1997-10-10 DJ NA -1.262738958
> 6 1997-10-14 DJ NA -1.262738958
> 7 1997-10-15 DJ NA -1.262738958
> 8 1997-10-16 DJ -0.2250333841 -0.361212732
> 9 1997-10-17 DJ -0.5545946845 -0.778380045
> 10 1997-10-20 DJ NA -0.778380045
> 11 1997-10-21 DJ 0.1022404757 -0.676935389
> 12 1997-10-22 DJ NA -0.676935389
I want to get NAs before the 5th observation, so row 1-4 are NA. Row 5 computes the cumulative return over row 1-5, Row 6 computes the cumulative return over 2-6 etc.
Reprex:
X <- data.frame(Date=c("1997-10-06" ,"1997-10-07", "1997-10-08" ,"1997-10-09", "1997-10-10",
"1997-10-14", "1997-10-15" ,"1997-10-16", "1997-10-17","1997-10-20", "1997-10-21" ,"1997-10-22"),
ret=c(NA,NA,-1.1277918,-0.1364865, NA , NA , NA ,-0.2250334 ,-0.5545947, NA, 0.1022405, NA))

You could replace the NA with 0 and then use zoo::rollsum default behavior:
library(zoo)
library(tidyr)
library(dplyr)
df %>%
dplyr::mutate(ret = zoo::rollsum(tidyr::replace_na(ret, 0), k = 5, na.pad = T, align = "right))
Date ret
1 1997-10-06 NA
2 1997-10-07 NA
3 1997-10-08 NA
4 1997-10-09 NA
5 1997-10-10 -1.2642783
6 1997-10-14 -1.2642783
7 1997-10-15 -1.2642783
8 1997-10-16 -0.3615199
9 1997-10-17 -0.7796281
10 1997-10-20 -0.7796281
11 1997-10-21 -0.6773876
12 1997-10-22 -0.6773876

Here's some code that takes your data, changes all the NA in ret to 0 and then calculates a rolling 5-period sum using the RcppRoll and tidyverse packages.
# Load libraries
library('RcppRoll')
library('tidyverse')
# Load data
df <- data.frame(Date=c("1997-10-06" ,"1997-10-07", "1997-10-08" ,"1997-10-09", "1997-10-10",
"1997-10-14", "1997-10-15" ,"1997-10-16", "1997-10-17","1997-10-20", "1997-10-21" ,"1997-10-22"),
ret=c(NA,NA,-1.1277918,-0.1364865, NA , NA , NA ,-0.2250334 ,-0.5545947, NA, 0.1022405, NA), stringsAsFactors = F)
# Change NA's to Zeros
df[is.na(df[,2]),2] <- 0
# Calculate Rolling Sum
df_new <- df %>% mutate(rollsum = roll_sum(ret, n=5, align = 'right', fill=NA))

Related

Reference the previous non-zero row, find the difference and divide by nrows

I must be asking the question terribly because I can't find what I looking for!
I have a large excel file that looks like this for every day of the month:
Date
Well1
1/1/16
10
1/2/16
NA
1/3/16
NA
1/4/16
NA
1/5/16
20
1/6/16
NA
1/7/16
25
1/8/16
NA
1/9/16
NA
1/10/16
35
etc
NA
I want to make a new column that has the difference between the non-zero rows and divide that by the number of rows between each non zero row. Aiming for something like this:
Date
Well1
Adjusted
1/1/16
10
=(20-10)/4 = 2.5
1/2/16
NA
1.25
1/3/16
NA
1.25
1/4/16
NA
1.25
1/5/16
20
=(25-20)/2= 2.5
1/6/16
NA
2.5
1/7/16
25
=(35-25)/3 = 3.3
1/8/16
NA
3.3
1/9/16
NA
3.3
1/10/16
35
etc
etc
NA
etc
I'm thinking I should use lead or lag, but the thing is that the steps are different between each nonzero row (so I'm not sure how to use n in the lead/lag function). I've used group_by so that each month stands alone, as well as attempted case_when and ifelse Mostly need ideas on translating excel format into a workable R format.
With some diff-ing and repeating of values, you should be able to get there.
dat$Date <- as.Date(dat$Date, format="%m/%d/%y")
nas <- is.na(dat$Well1)
dat$adj <- with(dat[!nas,],
diff(Well1) / as.numeric(diff(Date), units="days")
)[cumsum(!nas)]
# Date Well1 adj
#1 2016-01-01 10 2.5
#2 2016-01-02 NA 2.5
#3 2016-01-03 NA 2.5
#4 2016-01-04 NA 2.5
#5 2016-01-05 20 2.5
#6 2016-01-06 NA 2.5
#7 2016-01-07 25 5.0
#8 2016-01-08 NA 5.0
#9 2016-01-09 NA 5.0
#10 2016-01-10 40 NA
dat being used is:
dat <- read.table(text="Date Well1
1/1/16 10
1/2/16 NA
1/3/16 NA
1/4/16 NA
1/5/16 20
1/6/16 NA
1/7/16 25
1/8/16 NA
1/9/16 NA
1/10/16 40", header=TRUE, stringsAsFactors=FALSE)
Base R in the same vein as #thelatemail but with transformations all in one expression:
nas <- is.na(dat$Well1)
res <- within(dat, {
Date <- as.Date(Date, "%m/%d/%y")
Adjusted <- (diff(Well1[!nas]) /
as.numeric(diff(Date[!nas]), units = "days"))[cumsum(!nas)]
}
)
Data:
dat <- read.table(text="Date Well1
1/1/16 10
1/2/16 NA
1/3/16 NA
1/4/16 NA
1/5/16 20
1/6/16 NA
1/7/16 25
1/8/16 NA
1/9/16 NA
1/10/16 40", header=TRUE, stringsAsFactors=FALSE)
Maybe this should work
library(dplyr)
df1 %>%
#// remove the rows with NA
na.omit %>%
# // create a new column with the lead values of Well1
transmute(Date, Well2 = lead(Well1)) %>%
# // join with original data
right_join(df1 %>%
mutate(rn = row_number())) %>%
# // order by the original order
arrange(rn) %>%
# // create a grouping column based on the NA values
group_by(grp = cumsum(!is.na(Well1))) %>%
# // subtract the first element of Well2 with Well1 and divide
# // by number of rows - n() in the group
mutate(Adjusted = (first(Well2) - first(Well1))/n()) %>%
ungroup %>%
select(-grp, - Well2)

Efficient way for insertion of multiple rows at given indices & with repetitions

I have a data frame (DATA) with > 2 million rows (observations at different time points) and another data frame (INSERTION) which gives info about missing observations. The latter object contains 2 columns: 1st column with row indices after which empty (NA) rows should be inserted into DATA, and 2nd column with the number of empty rows that should be inserted at that position.
Below is a minimum working example:
DATA <- data.frame(datetime=strptime(as.character(c(201301011700, 201301011701, 201301011703, 201301011704, 201301011705, 201301011708, 201301011710, 201301011711, 201301011715, 201301011716, 201301011718, 201301011719, 201301011721, 201301011722, 201301011723, 201301011724, 201301011725, 201301011726, 201301011727, 201301011729, 201301011730, 201301011731, 201301011732, 201301011733, 201301011734, 201301011735, 201301011736, 201301011737, 201301011738, 201301011739)), format="%Y%m%d%H%M"), var1=rnorm(30), var2=rnorm(30), var3=rnorm(30))
INSERTION <- data.frame(index=c(2, 5, 6, 8, 10, 12, 19), repetition=c(1, 2, 1, 3, 1, 1, 1))
Now I'm looking for an efficient (and thus fast) way to insert the n empty rows at given row indices of the original file. How can I additionally complement the correct datetimes for these empty rows (add 1 minute for every new row; however, every weekend and bank holidays there are some regular gaps which are not contained in INSERTION!)?
Any help is appreciated!
Looking at the pattern in INSERTION and matching it with DATA most probably you are trying to fill the missing minutes in datetime of DATA. You can create a dataframe with every minute sequence from min to max value of datetime from DATA and then merge
merge(data.frame(datetime = seq(min(DATA$datetime), max(DATA$datetime),
by = "1 min")),DATA, all.x = TRUE)
# datetime var1 var2 var3
#1 2013-01-01 17:00:00 -1.063326 0.11925 -0.788622
#2 2013-01-01 17:01:00 1.263185 0.24369 -0.502199
#3 2013-01-01 17:02:00 NA NA NA
#4 2013-01-01 17:03:00 -0.349650 1.23248 1.496061
#5 2013-01-01 17:04:00 -0.865513 -0.51606 -1.137304
#6 2013-01-01 17:05:00 -0.236280 -0.99251 -0.179052
#7 2013-01-01 17:06:00 NA NA NA
#8 2013-01-01 17:07:00 NA NA NA
#9 2013-01-01 17:08:00 -0.197176 1.67570 1.902362
#10 2013-01-01 17:09:00 NA NA NA
#...
#...
Or using similar logic with tidyr::complete
tidyr::complete(DATA, datetime = seq(min(datetime), max(datetime), by = "1 min"))
If performance is a factor on a large data frame, this approach avoids joins:
# Generate new data.frame containing missing datetimes
tmp <- data.frame(datetime = DATA$datetime[with(INSERTION, rep(index, repetition))] + sequence(INSERTION$repetition)*60)
# Create variables filled with NA to match main data.frame
tmp[setdiff(names(DATA), names(tmp))] <- NA
# Bind and sort
new_df <- rbind(DATA, tmp)
new_df <- new_df[order(new_df$datetime),]
head(new_df, 15)
datetime var1 var2 var3
1 2013-01-01 17:00:00 0.98789253 0.68364933 0.70526985
2 2013-01-01 17:01:00 -0.68307496 0.02947599 0.90731512
31 2013-01-01 17:02:00 NA NA NA
3 2013-01-01 17:03:00 -0.60189915 -1.00153188 0.06165694
4 2013-01-01 17:04:00 -0.87329313 -1.81532302 -2.04930719
5 2013-01-01 17:05:00 -0.58713154 -0.42313098 0.37402224
32 2013-01-01 17:06:00 NA NA NA
33 2013-01-01 17:07:00 NA NA NA
6 2013-01-01 17:08:00 2.41350911 -0.13691754 1.57618578
34 2013-01-01 17:09:00 NA NA NA
7 2013-01-01 17:10:00 -0.38961552 0.83838954 1.18283382
8 2013-01-01 17:11:00 0.02290672 -2.10825367 0.87441448
35 2013-01-01 17:12:00 NA NA NA
36 2013-01-01 17:13:00 NA NA NA
37 2013-01-01 17:14:00 NA NA NA

Compute the variance of a moving window in a dataframe

Hey I want to compute the variance of column. My dataframe is sorted by the as.Date() format. Here you can see a snippet of it:
Date USA ARG BRA CHL COL MEX PER
2012-04-01 1 0.2271531 0.4970299 0.001956865 0.0005341452 0.07341428 NA
2012-05-01 1 0.2218906 0.4675895 0.001911405 0.0005273186 0.07026524 NA
2012-06-01 1 0.2054076 0.4531661 0.001891352 0.0005292575 0.06897811 NA
2012-07-01 1 0.2033470 0.4596730 0.001950686 0.0005312600 0.07269619 NA
2012-08-01 1 0.1993882 0.4596039 0.001980537 0.0005271514 0.07268987 NA
2012-09-01 1 0.1967152 0.4593390 0.002011212 0.0005305549 0.07418838 NA
2012-10-01 1 0.1972730 0.4597584 0.002002203 0.0005284380 0.07428555 NA
2012-11-01 1 0.1937618 0.4519187 0.001979805 0.0005238670 0.07329656 NA
2012-12-01 1 0.1854037 0.4500448 0.001993309 0.0005323795 0.07453949 NA
2013-01-01 1 0.1866007 0.4607501 0.002013112 0.0005412329 0.07551040 NA
2013-02-01 1 0.1855950 0.4712956 0.002011067 0.0005359562 0.07554661 NA
The dataframe ranges from january 2004 up to dezember 2018. But I do not want to compute the compute the variance of the whole columnes.
I want to compute the variance of one year (or 12 values) which is moving month by month.
I do not really know how to start. I can imagine using the zoo package and the rollapply. But here the problem is (I think) that R computes uses the values around it and not past it?
I also found this question: R: create a data frame out of a rolling window, so my idea was to get rid of the date column. It is easy to build the matrix, but now I do not understand how to apply the variance function to my data...
Is there a smart way to compute it all in one and also using the information of the date? If not, I also appreciate any other solution from you!
We can use rollappyr to perform the rolling computations. Since there are only 11 rows in the data in the question we can't take 12 month averages but using 3 month averages instead we can illustrate it. Remove fill = NA if you want to omit the NA rows or replace it with partial = TRUE if you want variances using fewer than 12 near the beginning. If you want a data frame result use fortify.zoo(zv) .
library(zoo)
z <- read.zoo(DF)
zv <- rollapplyr(z, 3, var, fill = NA)
zv
giving this zoo object:
USA ARG BRA CHL COL MEX PER
2012-04-01 NA NA NA NA NA NA NA
2012-05-01 NA NA NA NA NA NA NA
2012-06-01 0 1.287083e-04 4.998008e-04 1.126781e-09 1.237524e-11 5.208793e-06 NA
2012-07-01 0 1.033001e-04 5.217420e-05 9.109406e-10 3.883996e-12 3.565057e-06 NA
2012-08-01 0 9.358558e-06 1.396497e-05 2.060928e-09 4.221043e-12 4.600220e-06 NA
2012-09-01 0 1.113297e-05 3.108380e-08 9.159058e-10 4.826929e-12 7.453672e-07 NA
2012-10-01 0 1.988357e-06 4.498977e-08 2.485889e-10 2.953403e-12 8.001948e-07 NA
2012-11-01 0 3.560373e-06 1.944961e-05 2.615387e-10 1.168389e-11 2.971477e-07 NA
2012-12-01 0 3.717777e-05 2.655440e-05 1.271886e-10 1.814869e-11 4.312436e-07 NA
2013-01-01 0 2.042867e-05 3.268476e-05 2.806455e-10 7.540331e-11 1.231438e-06 NA
2013-02-01 0 4.134729e-07 1.129013e-04 1.186146e-10 1.983651e-11 3.263780e-07 NA
We can plot the log of the variances like this:
library(ggplot2)
autoplot(log(zv), facet = NULL) + geom_point() + ylab("log(var(.))")
Note
We assume that the starting point is the data frame generated reproducibly below:
Lines <- "Date USA ARG BRA CHL COL MEX PER
2012-04-01 1 0.2271531 0.4970299 0.001956865 0.0005341452 0.07341428 NA
2012-05-01 1 0.2218906 0.4675895 0.001911405 0.0005273186 0.07026524 NA
2012-06-01 1 0.2054076 0.4531661 0.001891352 0.0005292575 0.06897811 NA
2012-07-01 1 0.2033470 0.4596730 0.001950686 0.0005312600 0.07269619 NA
2012-08-01 1 0.1993882 0.4596039 0.001980537 0.0005271514 0.07268987 NA
2012-09-01 1 0.1967152 0.4593390 0.002011212 0.0005305549 0.07418838 NA
2012-10-01 1 0.1972730 0.4597584 0.002002203 0.0005284380 0.07428555 NA
2012-11-01 1 0.1937618 0.4519187 0.001979805 0.0005238670 0.07329656 NA
2012-12-01 1 0.1854037 0.4500448 0.001993309 0.0005323795 0.07453949 NA
2013-01-01 1 0.1866007 0.4607501 0.002013112 0.0005412329 0.07551040 NA
2013-02-01 1 0.1855950 0.4712956 0.002011067 0.0005359562 0.07554661 NA"
DF <- read.table(text = Lines, header = TRUE)

How to do a data.table rolling join?

I have two data tables that I'm trying to merge. One is data on company market values through time and the other is company dividend history through time. I'm trying to find out how much each company has paid each quarter and put that value next to the market value data through time.
library(magrittr)
library(data.table)
library(zoo)
library(lubridate)
set.seed(1337)
# data table of company market values
companies <-
data.table(companyID = 1:10,
Sedol = rep(c("91772E", "7A662B"), each = 5),
Date = (as.Date("2005-04-01") + months(seq(0, 12, 3))) - days(1),
MktCap = c(100 + cumsum(rnorm(5,5)),
50 + cumsum(rnorm(5,1,5)))) %>%
setkey(Sedol, Date)
# data table of dividends
dividends <-
data.table(DivID = 1:7,
Sedol = c(rep('91772E', each = 4), rep('7A662B', each = 3)),
Date = as.Date(c('2004-11-19', '2005-01-13', '2005-01-29',
'2005-10-01', '2005-06-29', '2005-06-30',
'2006-04-17')),
DivAmnt = rnorm(7, .8, .3)) %>%
setkey(Sedol, Date)
I believe this is a situation where you could use a data.table rolling join, something like:
dividends[companies, roll = "nearest"]
to try and get a dataset that looks like
DivID Sedol Date DivAmnt companyID MktCap
1: NA 7A662B <NA> NA 6 61.21061
2: 5 7A662B 2005-06-29 0.7772631 7 66.92951
3: 6 7A662B 2005-06-30 1.1815343 7 66.92951
4: NA 7A662B <NA> NA 8 78.33914
5: NA 7A662B <NA> NA 9 88.92473
6: NA 7A662B <NA> NA 10 87.85067
7: 2 91772E 2005-01-13 0.2964291 1 105.19249
8: 3 91772E 2005-01-29 0.8472649 1 105.19249
9: NA 91772E <NA> NA 2 108.74579
10: 4 91772E 2005-10-01 1.2467408 3 113.42261
11: NA 91772E <NA> NA 4 120.04491
12: NA 91772E <NA> NA 5 124.35588
(note that I've matched the dividends to the company market values by the exact quarter)
But I'm not exactly sure how to execute it. The CRAN pdf is rather vague about what the number is or should be if roll is a value (Can you pass dates? Does a number quantify the days forward to carry? the number of obersvations?) and changing rollends around doesn't seem to get me what I want.
In the end, I ended up mapping the dividend dates to their quarter end and then joining on that. A good solution, but not useful if I end up needing to know how to perform rolling joins. In your answer, could you describe a situation where rolling joins are the only solution as well as help me understand how to perform them?
Instead of a rolling join, you may want to use an overlap join with the foverlaps function of data.table:
# create an interval in the 'companies' datatable
companies[, `:=` (start = compDate - days(90), end = compDate + days(15))]
# create a second date in the 'dividends' datatable
dividends[, Date2 := divDate]
# set the keys for the two datatable
setkey(companies, Sedol, start, end)
setkey(dividends, Sedol, divDate, Date2)
# create a vector of columnnames which can be removed afterwards
deletecols <- c("Date2","start","end")
# perform the overlap join and remove the helper columns
res <- foverlaps(companies, dividends)[, (deletecols) := NULL]
the result:
> res
Sedol DivID divDate DivAmnt companyID compDate MktCap
1: 7A662B NA <NA> NA 6 2005-03-31 61.21061
2: 7A662B 5 2005-06-29 0.7772631 7 2005-06-30 66.92951
3: 7A662B 6 2005-06-30 1.1815343 7 2005-06-30 66.92951
4: 7A662B NA <NA> NA 8 2005-09-30 78.33914
5: 7A662B NA <NA> NA 9 2005-12-31 88.92473
6: 7A662B NA <NA> NA 10 2006-03-31 87.85067
7: 91772E 2 2005-01-13 0.2964291 1 2005-03-31 105.19249
8: 91772E 3 2005-01-29 0.8472649 1 2005-03-31 105.19249
9: 91772E NA <NA> NA 2 2005-06-30 108.74579
10: 91772E 4 2005-10-01 1.2467408 3 2005-09-30 113.42261
11: 91772E NA <NA> NA 4 2005-12-31 120.04491
12: 91772E NA <NA> NA 5 2006-03-31 124.35588
In the meantime the data.table authors have introduced non-equi joins (v1.9.8). You can also use that to solve this problem. Using a non-equi join you just need:
companies[, `:=` (start = compDate - days(90), end = compDate + days(15))]
dividends[companies, on = .(Sedol, divDate >= start, divDate <= end)]
to get the intended result.
Used data (the same as in the question, but without the creation of the keys):
set.seed(1337)
companies <- data.table(companyID = 1:10, Sedol = rep(c("91772E", "7A662B"), each = 5),
compDate = (as.Date("2005-04-01") + months(seq(0, 12, 3))) - days(1),
MktCap = c(100 + cumsum(rnorm(5,5)), 50 + cumsum(rnorm(5,1,5))))
dividends <- data.table(DivID = 1:7, Sedol = c(rep('91772E', each = 4), rep('7A662B', each = 3)),
divDate = as.Date(c('2004-11-19','2005-01-13','2005-01-29','2005-10-01','2005-06-29','2005-06-30','2006-04-17')),
DivAmnt = rnorm(7, .8, .3))

Condition for function and loop

I have a data frame simplified as follow:
head(dendro)
X DateTime ID diameter dendro ring DOY month mday year Rain_mm_Tot Through_Tot temp
1 1 2012-06-21 13:45:00 r1_1 5482 1 1 173 6 22 113 NA NA NA
2 2 2012-06-21 13:45:00 r2_3 NA 3 2 173 6 22 113 NA NA NA
3 3 2012-06-21 13:45:00 r1_2 5534 2 1 173 6 22 113 NA NA NA
4 4 2012-06-21 13:45:00 r2_4 NA 4 2 173 6 22 113 NA NA NA
5 5 2012-06-21 13:45:00 r1_3 5606 3 1 173 6 22 113 NA NA NA
6 6 2012-06-21 13:45:00 r2_5 NA 5 2 173 6 22 113 NA NA NA
The dataframe is first splitted by "ID", so it's a list of IDs
After that I apply a function, that includes a loop, and the result is a new column "Diameter2", with the result I want from the function, that works OK:
dendro_sp <- split(dendro, dendro$ID)
library(changepoint)
dendro_sp <- lapply(dendro_sp, function(x){
x <- subset(x, !is.na(diameter))
cpfit <- cpt.mean(x$diameter, method="BinSeg")
x$diameter2 <- x$diameter
cpts <- cpfit#cpts
means <- param.est(cpfit)$mean
meanZero <- means[1]
for(i in 1:(length(cpts)-1)){
x$diameter2[(cpts[i]+1):cpts[i+1]] <- x$diameter2[(cpts[i]+1):cpts[i+1]] + (meanZero - means[i+1])
}
return(x)
})
dendro2 <- do.call(rbind, dendro_sp)
rownames(dendro2) <- NULL
My problem is that I want it to apply it conditionally, for example to r1_1 and r1_3, and grab the "diameter" value for r3 in the new column "diameter2", instead of applying the function for the rest of IDs:
ifelse(diameter$ID==c("r1_1","r1_3"), apply_the_function_to_r11_and_r13_to_calculate_diameter2, otherwise_write_diameter_value_in_diameter2_column)
Remember that the dataframe "dendro" is splitted by ID, I don't know if that is important to define the condition for several IDs.
Thanks
I am not sure if I understand the problem correctly. I try to answer.
I assume you want to apply a function to the "diameter" field of the "diameter" data.frame, conditioning on the "ID" field and retunr the result in the corresponding diameter2 field. I don't know how the function works, so forgive me if this will not work.
Selected fields
diameter$diameter2[diameter$ID=="r1_1"|diameter$ID=="r1_3"]<- yourfun(diameter$diameter[diameter$ID=="r1_1"|diameter$ID=="r1_3"]
Unselected fields
diameter$diameter2[diameter$ID!="r1_1" & diameter$ID=="r1_3"]<- diameter$diameter[diameter$ID=="r1_1"|diameter$ID=="r1_3"]

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