I have a data.table dt:
library(data.table)
dt = data.table(a=LETTERS[c(1,1:3)],b=4:7)
a b
1: A 4
2: A 5
3: B 6
4: C 7
The result of dt[, .N, by=a] is
a N
1: A 2
2: B 1
3: C 1
I know the by=a or by="a" means grouped by a column and the N column is the sum of duplicated times of a. However, I don't use nrow() but I get the result. The .N is not just the column name? I can't find the document by ??".N" in R. I tried to use .K, but it doesn't work. What does .N means?
Think of .N as a variable for the number of instances. For example:
dt <- data.table(a = LETTERS[c(1,1:3)], b = 4:7)
dt[.N] # returns the last row
# a b
# 1: C 7
Your example returns a new variable with the number of rows per case:
dt[, new_var := .N, by = a]
dt
# a b new_var
# 1: A 4 2 # 2 'A's
# 2: A 5 2
# 3: B 6 1 # 1 'B'
# 4: C 7 1 # 1 'C'
For a list of all special symbols of data.table, see also https://www.rdocumentation.org/packages/data.table/versions/1.10.0/topics/special-symbols
Related
Suppose I have the following data.table:
dt <- data.table(a = 1:2, b = 1:2, c = c(1, 1))
# dt
# a b c
# 1: 1 1 1
# 2: 2 2 1
What would be the fastest way to create a fourth column d indicating that the preexisting values in each row are all identical, so that the resulting data.table will look like the following?
# dt
# a b c d
# 1: 1 1 1 identical
# 2: 2 2 1 not_identical
I want to avoid using duplicated function and want to stick to using identical or a similar function even if it means iterating through items within each row.
uniqueN can be applied grouped by row and create a logical expression (== 1)
library(data.table)
dt[, d := c("not_identical", "identical")[(uniqueN(unlist(.SD)) == 1) +
1], 1:nrow(dt)]
-output
dt
# a b c d
#1: 1 1 1 identical
#2: 2 2 1 not_identical
Or another efficient approach might be to do comparison with the first column, and create an expression with rowSums
dt[, d := c("identical", "not_identical")[1 + rowSums(.SD[[1]] != .SD) > 0 ] ]
Here is another data.table option using var
dt[, d := ifelse(var(unlist(.SD)) == 0, "identical", "non_identical"), seq(nrow(dt))]
which gives
> dt
a b c d
1: 1 1 1 identical
2: 2 2 1 non_identical
When grouping by an expression involving a column (e.g. DT[...,.SD[c(1,.N)],by=expression(col)]), I want to keep the value of col in .SD.
For example, in the following I am grouping by the remainder of a divided by 3, and keeping the first and last observation in each group. However, a is no longer present in .SD
f <- function(x) x %% 3
Q <- data.table(a = 1:20, x = rnorm(20), y = rnorm(20))
Q[, .SD[c(1., .N)], by = f(a)]
f x y
1: 1 0.2597929 1.0256259
2: 1 2.1106619 -1.4375193
3: 2 1.2862501 0.7918292
4: 2 0.6600591 -0.5827745
5: 0 1.3758503 1.3122561
6: 0 2.6501140 1.9394756
The desired output is as if I had done the following
Q[, f := f(a)]
tmp <- Q[, .SD[c(1, .N)], by=f]
Q[, f := NULL]
tmp[, f := NULL]
tmp
a x y
1: 1 0.2597929 1.0256259
2: 19 2.1106619 -1.4375193
3: 2 1.2862501 0.7918292
4: 20 0.6600591 -0.5827745
5: 3 1.3758503 1.3122561
6: 18 2.6501140 1.9394756
Is there a way to do this directly, without creating a new variable and creating a new intermediate data.table?
Instead of .SD, use .I to get the row index, extract that column ($V1) and subset the original dataset
library(data.table)
Q[Q[, .I[c(1., .N)], by = f(a)]$V1]
# a x y
#1: 1 0.7265238 0.5631753
#2: 19 1.7110611 -0.3141118
#3: 2 0.1643566 -0.4704501
#4: 20 0.5182394 -0.1309016
#5: 3 -0.6039137 0.1349981
#6: 18 0.3094155 -1.1892190
NOTE: The values in columns 'x', 'y' would be different as there was no set.seed
I would like to know, if I have data that I can group by a variable, how can I get the last observation of the previous group?
I have the following data:
dt <- data.table(a=c(1,1,1,2,2,2,2,2,3,3,3,3,3,3,4,4,5,5,5,5,5), b=sample.int(21))
I would like to create a new data.table that has the group ID and the difference between the last observation of the group from the last observation of the previous group. So that from the above I'd get:
a c
1: 1 NA
2: 2 9
3: 3 1
4: 4 -8
5: 5 5
Thanks!
We group by 'a', get the last element of 'b', then take the lag of 'c' by shifting
dt[, .(c = last(b)), a][, c:= shift(c)][]
Here is a way:
dt[, c := b * (1:.N == .N), by = a] ## get last row within the group
dt <- dt[b == c] ## filter data.table to get rows of interest
dt[, c := shift(c, type = "lag") - c][] ## getting difference using shift with lag argument
# a b c
#1: 1 11 NA
#2: 2 10 NA
#3: 3 18 9
#4: 4 19 -7
#5: 5 12 -8
data
set.seed(1)
dt <- data.table(a=c(1,1,1,2,2,2,2,2,3,3,3,3,3,3,4,4,5,5,5,5,5), b=sample.int(21))
I had a data.table like this:
library(data.table)
dt <- data.table(a = c(rep("A", 3), rep("B", 3)), b = c(1, 3, 5, 2, 4, 6))
I needed to perform an operation (forecast) on the values for each a, so I decided to put them in a list, like this:
dt <- dt[, x := .(list(b)), by = a][, .SD[1,], by = a, .SDcols = "x"]
Now I wanted to "melt" (that's the thing that comes to mind) dt back into its original form.
I could do it for very few levels of a like this:
dt2 <- rbind(expand.grid(dt[1, a], dt[1, x[[1]]]), expand.grid(dt[2, a], dt[2, x[[1]]]))
but of course, the solution is impractical for more levels of a.
I've tried
dt2 <- dt[, expand.grid(a, x[[1]]), by = a]
which results in
dt2
## a Var1 Var2
## 1: A A 1
## 2: A A 3
## 3: A A 5
## 4: B A 2
## 5: B A 4
## 6: B A 6
it's interesting to notice that Var1 doesn't actually follow the "A - B" pattern expected (but at least a remains).
Is there a better approach to achieve this?
EDITS
Expected output will be the result of
dt2[, .(a, Var2)]
Corrected "melt" for "dcast".
You are looking for a method to nest(convert a column from a atomic vector type to list type) and unnest(the opposite direction) in a data.table way. This is different from reshaping data which either spread a column values to row header(dcast) or gather the row headers to a column values(melt):
In data.table syntax, you can use list and unlist on the target column to summarize or broadcast it along with group variables:
Say if we are starting from:
dt
# a b
# 1: A 1
# 2: A 3
# 3: A 5
# 4: B 2
# 5: B 4
# 6: B 6
To repeat what you have achieved in your first step, i.e. nest column b, you can do:
dt_nest <- dt[, .(b = list(b)), a]
dt_nest
# a b
# 1: A 1,3,5
# 2: B 2,4,6
To go the opposite direction, use unlist with the group variable:
dt_nest[, .(b = unlist(b)), a]
# a b
# 1: A 1
# 2: A 3
# 3: A 5
# 4: B 2
# 5: B 4
# 6: B 6
For a data.table DT grouped by site, sorted by time t, I need to change the last value of a variable in each group. I assume it should be possible to do this by reference using :=, but I haven't found a way that works yet.
Sample data:
require(data.table) # using 1.8.11
DT <- data.table(site=c(rep("A",5), rep("B",4)),t=c(1:5,1:4),a=as.double(c(11:15,21:24)))
setkey(DT, site, t)
DT
# site t a
# 1: A 1 11
# 2: A 2 12
# 3: A 3 13
# 4: A 4 14
# 5: A 5 15
# 6: B 1 21
# 7: B 2 22
# 8: B 3 23
# 9: B 4 24
The desired result is to change the last value of a in each group, for example to 999, so the result looks like:
# site t a
# 1: A 1 11
# 2: A 2 12
# 3: A 3 13
# 4: A 4 14
# 5: A 5 999
# 6: B 1 21
# 7: B 2 22
# 8: B 3 23
# 9: B 4 999
It seems like .I and/or .N should be used, but I haven't found a form that works. The use of := in the same statement as .I[.N] gives an error. The following gives me the row numbers where the assignment is to be made:
DT[, .I[.N], by=site]
# site V1
# 1: A 5
# 2: B 9
but I don't seem to be able to use this with a := assignment. The following give errors:
DT[.N, a:=999, by=site]
# Null data.table (0 rows and 0 cols)
DT[, .I[.N, a:=999], by=site]
# Error in `:=`(a, 999) :
# := and `:=`(...) are defined for use in j, once only and in particular ways.
# See help(":="). Check is.data.table(DT) is TRUE.
DT[.I[.N], a:=999, by=site]
# Null data.table (0 rows and 0 cols)
Is there a way to do this by reference in data.table? Or is this better done another way in R?
Currently you can use:
DT[DT[, .I[.N], by = site][['V1']], a := 999]
# or, avoiding the overhead of a second call to `[.data.table`
set(DT, i = DT[,.I[.N],by='site'][['V1']], j = 'a', value = 999L)
alternative approaches:
use replace...
DT[, a := replace(a, .N, 999), by = site]
or shift the replacement to the RHS, wrapped by {} and return the full vector
DT[, a := {a[.N] <- 999L; a}, by = site]
or use mult='last' and take advantage of by-without-by. This requires the data.table to be keyed by the groups of interest.
DT[unique(site), a := 999, mult = 'last']
There is a feature request #2793 that would allow
DT[, a[.N] := 999]
but this is yet to be implemented