I currently have a dataset with 50,000+ rows of data for which I need to find rolling sums. I have completed this using rollaply which has worked perfectly. I need to apply these rolling sums across a range of widths (600, 1200, 1800...6000) which I have done by cut and pasting each line of script and changing the width. While it works, I'd like to tidy my script but applying a loop, or similar, if possible so that once the rollapply function has completed it's first 'pass' at 600 width, it then completes the same with 1200 and so on. Example:
Var1 Var2 Var3
1 11 19
43 12 1
4 13 47
21 14 29
41 15 42
16 16 5
17 17 16
10 18 15
20 19 41
44 20 27
width_2 <- rollapply(x$Var1, FUN = sum, width = 2)
width_3 <- rollapply(x$Var1, FUN = sum, width = 3)
width_4 <- rollapply(x$Var1, FUN = sum, width = 4)
Is there a way to run widths 2, 3, then 4 in a simpler way rather than cut and paste, particularly when I have up to 10 widths, and then need to run this across other cols. Any help would be appreciated.
We can use lapply in base R
lst1 <- lapply(2:4, function(i) rollapply(x$Var1, FUN = sum, width = i))
names(lst1) <- paste0('width_', 2:4)
list2env(lst1, .GlobalEnv)
NOTE: It is not recommended to create multiple objects in the global environment. Instead, the list would be better
Or with a for loop
for(v in 2:4) {
assign(paste0('width_', v), rollapply(x$Var1, FUN = sum, width = v))
}
Create a function to do this for multiple dataset
f1 <- function(col1, i) {
rollapply(col1, FUN = sum, width = i)
}
lapply(x[c('Var1', 'Var2')], function(x) lapply(2:4, function(i)
f1(x, i)))
Instead of creating separate vectors in global environment probably you can add these as new columns in the already existing dataframe.
Note that rollaplly(..., FUN = sum) is same as rollsum.
library(dplyr)
library(zoo)
bind_cols(x, purrr::map_dfc(2:4,
~x %>% transmute(!!paste0('Var1_roll_', .x) := rollsumr(Var1, .x, fill = NA))))
# Var1 Var2 Var3 Var1_roll_2 Var1_roll_3 Var1_roll_4
#1 1 11 19 NA NA NA
#2 43 12 1 44 NA NA
#3 4 13 47 47 48 NA
#4 21 14 29 25 68 69
#5 41 15 42 62 66 109
#6 16 16 5 57 78 82
#7 17 17 16 33 74 95
#8 10 18 15 27 43 84
#9 20 19 41 30 47 63
#10 44 20 27 64 74 91
You can use seq to generate the variable window size.
seq(600, 6000, 600)
#[1] 600 1200 1800 2400 3000 3600 4200 4800 5400 6000
Related
I have a data frame like this.
> abc
ID 1.x 2.x 1.y 2.y
1 4 10 20 30 40
2 16 5 10 5 10
3 42 16 17 18 19
4 91 20 20 20 20
5 103 103 42 56 84
How do I create two additional columns '1' and '2' by multiplying 1.x * 1.y and 2.x * 2.y in a generalized way?
I am trying to get a generalized solution where number of columns can be too many. So I want to multiply all x with all y. While x and y are fixed, n has to be figured out from data frame.
For simplicity lets assume n is also fixed however it is a large number.
One thing i can try is :-
abc[,c(6,7)]=abc[,c(2,3)]*abc[,c(4,5)]
It will work only if col positions are contiguous. This is good enough for me. If anyone can have more generalized solution, it will benefit us all.
If there are only couple of variables to multiply, we can do this with transform by multiplying the columns of interest
transform(abc, new1 = `1.x`*`1.y`, new2 = `2.x`*`2.y`, check.names = FALSE)
# ID 1.x 2.x 1.y 2.y new1 new2
#1 4 10 20 30 40 300 800
#2 16 5 10 5 10 25 100
#3 42 16 17 18 19 288 323
#4 91 20 20 20 20 400 400
#5 103 103 42 56 84 5768 3528
If we have lots of columns, then one approach is to split the dataset into a list of data.frames by taking the substring of names and then loop through the list and multiply the rows with do.call
abc[paste0("new", 1:2)] <- lapply(split.default(abc[-1],
sub("\\.[a-z]+$", "", names(abc)[-1])), function(x) do.call(`*`, x))
Or another option is (based on the pairwise column multiplication)
apply(aperm(array(unlist(abc[-1]), c(5, 2, 2)),
c(3, 1, 2)), 3, matrixStats::colProds)
Mutate will preserve the original variables. Mutate_all will allow you to multiply all columns in your dataframe.
abc %>%
mutate(new_vary1 = `1.x`* `2.x`,
new_vary2 = `1.y`* `2.y`) %>%
mutate_all(funs(.*`1.x`))
I have a data.table of integers with values between 1 and 60.
My question is about flooring or ceiling any number to the following values: 12 18 24 30 36 ... 60.
For example, let's say my data.table contains the number 13. I want R to "transform" this number into 12 and 18 as 13 lies in between those numbers. Moreover, if I have 18 I want R to keep it at 18.
If my data.table contains the value 50, I want R to convert that number into 48 and 54 and so on.
My goal is to get two different data.tables. One where the floored values are saved and one where the ceiled values are saved.
Any idea how one could do this in R?
EDIT: Numbers smaller than 12 should always be transformed to 12.
Example output:
If have the following data.table data.table(c(1,28,29,41,53,53,17,41,41,53))
I want the following two output data.tables: floored values data.table(c(12,24,24,36,48,48,12,36,36,48))
I want the following two output data.tables: ceiled values data.table(c(12,30,30,42,54,54,18,42,42,54))
Here is a fairly direct way (edited to round up to 12 if any values are below):
df <- data.frame(nums = 10:20)
df$floors <- with(df,pmax(12,6*floor(nums/6)))
df$ceils <- with(df,pmax(12,6*ceiling(nums/6)))
Leading to:
> df
nums floors ceils
1 10 12 12
2 11 12 12
3 12 12 12
4 13 12 18
5 14 12 18
6 15 12 18
7 16 12 18
8 17 12 18
9 18 18 18
10 19 18 24
11 20 18 24
Here's a way we could do this, using sapply and the which.min functions. From your question, it's not immediately clear how values < 12 should be handled.
x <- 1:60
num_list <- seq(12, 60, 6)
floorr <- sapply(x, function(x){
diff_vec <- x - num_list
diff_vec <- ifelse(diff_vec < 0, Inf, diff_vec)
num_list[which.min(diff_vec)]
})
ceill <- sapply(x, function(x){
diff_vec <- num_list - x
diff_vec <- ifelse(diff_vec < 0, Inf, diff_vec)
num_list[which.min(diff_vec)]
})
tail(cbind(x, floorr, ceill))
x floorr ceill
[55,] 55 54 60
[56,] 56 54 60
[57,] 57 54 60
[58,] 58 54 60
[59,] 59 54 60
[60,] 60 60 60
I am trying to assign a column name to a variable using mutate.
df <-data.frame(x = sample(1:100, 50), y = rnorm(50))
new <- function(name){
df%>%mutate(name = ifelse(x <50, "small", "big"))
}
When I run
new(name = "newVar")
it doesn't work. I know mutate_() could help but I'm struggling in using it together with ifelse.
Any help would be appreciated.
Using dplyr 0.7.1 and its advances in NSE, you have to UQ the argument to mutate and then use := when assigning. There is lots of info on programming with dplyr and NSE here: https://cran.r-project.org/web/packages/dplyr/vignettes/programming.html
I've changed the name of the function argument to myvar to avoid confusion. You could also use case_when from dplyr instead of ifelse if you have more categories to recode.
df <- data.frame(x = sample(1:100, 50), y = rnorm(50))
new <- function(myvar){
df %>% mutate(UQ(myvar) := ifelse(x < 50, "small", "big"))
}
new(myvar = "newVar")
This returns
x y newVar
1 37 1.82669 small
2 63 -0.04333 big
3 46 0.20748 small
4 93 0.94169 big
5 83 -0.15678 big
6 14 -1.43567 small
7 61 0.35173 big
8 26 -0.71826 small
9 21 1.09237 small
10 90 1.99185 big
11 60 -1.01408 big
12 70 0.87534 big
13 55 0.85325 big
14 38 1.70972 small
15 6 0.74836 small
16 23 -0.08528 small
17 27 2.02613 small
18 76 -0.45648 big
19 97 1.20124 big
20 99 -0.34930 big
21 74 1.77341 big
22 72 -0.32862 big
23 64 -0.07994 big
24 53 -0.40116 big
25 16 -0.70226 small
26 8 0.78965 small
27 34 0.01871 small
28 24 1.95154 small
29 82 -0.70616 big
30 77 -0.40387 big
31 43 -0.88383 small
32 88 -0.21862 big
33 45 0.53409 small
34 29 -2.29234 small
35 54 1.00730 big
36 22 -0.62636 small
37 100 0.75193 big
38 52 -0.41389 big
39 36 0.19817 small
40 89 -0.49224 big
41 81 -1.51998 big
42 18 0.57047 small
43 78 -0.44445 big
44 49 -0.08845 small
45 20 0.14014 small
46 32 0.48094 small
47 1 -0.12224 small
48 66 0.48769 big
49 11 -0.49005 small
50 87 -0.25517 big
Following the dlyr programming vignette, define your function as follows:
new <- function(name)
{
nn <- enquo(name) %>% quo_name()
df %>% mutate( !!nn := ifelse(x <50, "small", "big"))
}
enquo takes its expression argument and quotes it, followed by quo_name converting it into a string. Since nn is now quoted, we need to tell mutate not to quote it a second time. That's what !! is for. Finally, := is a helper operator to make it valid R code. Note that with this definition, you can simply pass newVar instead of "newVar" to your function, maintaining dplyr style.
> new( newVar ) %>% head
x y newVar
1 94 -1.07642088 big
2 85 0.68746266 big
3 80 0.02630903 big
4 74 0.18323506 big
5 86 0.85086915 big
6 38 0.41882858 small
Base R solution
df <-data.frame(x = sample(1:100, 50), y = rnorm(50))
new <- function(name){
df[,name]='s'
df[,name][df$x>50]='b'
return(df)
}
I am using dplyr 0.5 so i just combine base R with mutate
new <- function(Name){
df=mutate(df,ifelse(x <50, "small", "big"))
names(df)[3]=Name
return(df)
}
new("newVar")
I have a data.frame
set.seed(100)
exp <- data.frame(exp = c(rep(LETTERS[1:2], each = 10)), re = c(rep(seq(1, 10, 1), 2)), age1 = seq(10, 29, 1), age2 = seq(30, 49, 1),
h = c(runif(20, 10, 40)), h2 = c(40 + runif(20, 4, 9)))
I'd like to make a lm for each row in a data set (h and h2 ~ age1 and age2)
I do it by loop
exp$modelh <- 0
for (i in 1:length(exp$exp)){
age = c(exp$age1[i], exp$age2[i])
h = c(exp$h[i], exp$h2[i])
model = lm(age ~ h)
exp$modelh[i] = coef(model)[1] + 100 * coef(model)[2]
}
and it works well but takes some time with very large files. Will be grateful for the faster solution f.ex. dplyr
Using dplyr, we can try with rowwise() and do. Inside the do, we concatenate (c) the 'age1', 'age2' to create 'age', likewise, we can create 'h', apply lm, extract the coef to create the column 'modelh'.
library(dplyr)
exp %>%
rowwise() %>%
do({
age <- c(.$age1, .$age2)
h <- c(.$h, .$h2)
model <- lm(age ~ h)
data.frame(., modelh = coef(model)[1] + 100*coef(model)[2])
} )
gives the output
# exp re age1 age2 h h2 modelh
#1 A 1 10 30 19.23298 46.67906 68.85506
#2 A 2 11 31 17.73018 47.55402 66.17050
#3 A 3 12 32 26.56967 46.69174 84.98486
#4 A 4 13 33 11.69149 47.74486 61.98766
#5 A 5 14 34 24.05648 46.10051 82.90167
#6 A 6 15 35 24.51312 44.85710 89.21053
#7 A 7 16 36 34.37208 47.85151 113.37492
#8 A 8 17 37 21.10962 48.40977 74.79483
#9 A 9 18 38 26.39676 46.74548 90.34187
#10 A 10 19 39 15.10786 45.38862 75.07002
#11 B 1 20 40 28.74989 46.44153 100.54666
#12 B 2 21 41 36.46497 48.64253 125.34773
#13 B 3 22 42 18.41062 45.74346 81.70062
#14 B 4 23 43 21.95464 48.77079 81.20773
#15 B 5 24 44 32.87653 47.47637 115.95097
#16 B 6 25 45 30.07065 48.44727 101.10688
#17 B 7 26 46 16.13836 44.90204 84.31080
#18 B 8 27 47 20.72575 47.14695 87.00805
#19 B 9 28 48 20.78425 48.94782 84.25406
#20 B 10 29 49 30.70872 44.65144 128.39415
We could do this with the devel version of data.table i.e. v1.9.5. Instructions to install the devel version are here.
We convert the 'data.frame' to 'data.table' (setDT), create a column 'rn' with the option keep.rownames=TRUE. We melt the dataset by specifying the patterns in the measure to convert from 'wide' to 'long' format. Grouped by 'rn', we do the lm and get the coef. This can be assigned as a new column in the original dataset ('exp') while removing the unwanted 'rn' column by assigning (:=) it to NULL.
library(data.table)#v1.9.5+
modelh <- melt(setDT(exp, keep.rownames=TRUE), measure=patterns('^age', '^h'),
value.name=c('age', 'h'))[, {model <- lm(age ~h)
coef(model)[1] + 100 * coef(model)[2]},rn]$V1
exp[, modelh:= modelh][, rn := NULL]
exp
# exp re age1 age2 h h2 modelh
# 1: A 1 10 30 19.23298 46.67906 68.85506
# 2: A 2 11 31 17.73018 47.55402 66.17050
# 3: A 3 12 32 26.56967 46.69174 84.98486
# 4: A 4 13 33 11.69149 47.74486 61.98766
# 5: A 5 14 34 24.05648 46.10051 82.90167
# 6: A 6 15 35 24.51312 44.85710 89.21053
# 7: A 7 16 36 34.37208 47.85151 113.37492
# 8: A 8 17 37 21.10962 48.40977 74.79483
# 9: A 9 18 38 26.39676 46.74548 90.34187
#10: A 10 19 39 15.10786 45.38862 75.07002
#11: B 1 20 40 28.74989 46.44153 100.54666
#12: B 2 21 41 36.46497 48.64253 125.34773
#13: B 3 22 42 18.41062 45.74346 81.70062
#14: B 4 23 43 21.95464 48.77079 81.20773
#15: B 5 24 44 32.87653 47.47637 115.95097
#16: B 6 25 45 30.07065 48.44727 101.10688
#17: B 7 26 46 16.13836 44.90204 84.31080
#18: B 8 27 47 20.72575 47.14695 87.00805
#19: B 9 28 48 20.78425 48.94782 84.25406
#20: B 10 29 49 30.70872 44.65144 128.39415
Great (double) answer from #akrun.
Just a suggestion for your future analysis as you mentioned "it's an example of a bigger problem". Obviously, if you are really interested in building models rowwise then you'll create more and more columns as your age and h observations increase. If you get N observations you'll have to use 2xN columns for those 2 variables only.
I'd suggest to use a long data format in order to increase your rows instead of your columns.
Something like:
exp[1,] # how your first row (model building info) looks like
# exp re age1 age2 h h2
# 1 A 1 10 30 19.23298 46.67906
reshape(exp[1,], # how your model building info is transformed
varying = list(c("age1","age2"),
c("h","h2")),
v.names = c("age_value","h_value"),
direction = "long")
# exp re time age_value h_value id
# 1.1 A 1 1 10 19.23298 1
# 1.2 A 1 2 30 46.67906 1
Apologies if the "bigger problem" refers to something else and this answer is irrelevant.
With base R, the function sprintf can help us create formulas. And lapply carries out the calculation.
strings <- sprintf("c(%f,%f) ~ c(%f,%f)", exp$age1, exp$age2, exp$h, exp$h2)
lst <- lapply(strings, function(x) {model <- lm(as.formula(x));coef(model)[1] + 100 * coef(model)[2]})
exp$modelh <- unlist(lst)
exp
# exp re age1 age2 h h2 modelh
# 1 A 1 10 30 19.23298 46.67906 68.85506
# 2 A 2 11 31 17.73018 47.55402 66.17050
# 3 A 3 12 32 26.56967 46.69174 84.98486
# 4 A 4 13 33 11.69149 47.74486 61.98766
# 5 A 5 14 34 24.05648 46.10051 82.90167
# 6 A 6 15 35 24.51312 44.85710 89.21053
# 7 A 7 16 36 34.37208 47.85151 113.37493
# 8 A 8 17 37 21.10962 48.40977 74.79483
# 9 A 9 18 38 26.39676 46.74548 90.34187
# 10 A 10 19 39 15.10786 45.38862 75.07002
# 11 B 1 20 40 28.74989 46.44153 100.54666
# 12 B 2 21 41 36.46497 48.64253 125.34773
# 13 B 3 22 42 18.41062 45.74346 81.70062
# 14 B 4 23 43 21.95464 48.77079 81.20773
# 15 B 5 24 44 32.87653 47.47637 115.95097
# 16 B 6 25 45 30.07065 48.44727 101.10688
# 17 B 7 26 46 16.13836 44.90204 84.31080
# 18 B 8 27 47 20.72575 47.14695 87.00805
# 19 B 9 28 48 20.78425 48.94782 84.25406
# 20 B 10 29 49 30.70872 44.65144 128.39416
In the lapply function the expression as.formula(x) is what converts the formulas created in the first line into a format usable by the lm function.
Benchmark
library(dplyr)
library(microbenchmark)
set.seed(100)
big.exp <- data.frame(age1=sample(30, 1e4, T),
age2=sample(30:50, 1e4, T),
h=runif(1e4, 10, 40),
h2= 40 + runif(1e4,4,9))
microbenchmark(
plafort = {strings <- sprintf("c(%f,%f) ~ c(%f,%f)", big.exp$age1, big.exp$age2, big.exp$h, big.exp$h2)
lst <- lapply(strings, function(x) {model <- lm(as.formula(x));coef(model)[1] + 100 * coef(model)[2]})
big.exp$modelh <- unlist(lst)},
akdplyr = {big.exp %>%
rowwise() %>%
do({
age <- c(.$age1, .$age2)
h <- c(.$h, .$h2)
model <- lm(age ~ h)
data.frame(., modelh = coef(model)[1] + 100*coef(model)[2])
} )}
,times=5)
t: seconds
expr min lq mean median uq max neval cld
plafort 13.00605 13.41113 13.92165 13.56927 14.53814 15.08366 5 a
akdplyr 26.95064 27.64240 29.40892 27.86258 31.02955 33.55940 5 b
(Note: I downloaded the newest 1.9.5 devel version of data.table today, but continued to receive errors when trying to test it.
The results also differ fractionally (1.93 x 10^-8). Rounding likely accounts for the difference.)
all.equal(pl, ak)
[1] "Attributes: < Component “class”: Lengths (1, 3) differ (string compare on first 1) >"
[2] "Attributes: < Component “class”: 1 string mismatch >"
[3] "Component “modelh”: Mean relative difference: 1.933893e-08"
Conclusion
The lapply approach seems to perform well compared to dplyr with respect to speed, but it's 5 digit rounding may be an issue. Improvements may be possible. Perhaps using apply after converting to matrix to increase speed and efficiency.
I want to add many new columns simultaneously to a data.table based on by-group computations. A working example of my data would look something like this:
Time Stock x1 x2 x3
1: 2014-08-22 A 15 27 34
2: 2014-08-23 A 39 44 29
3: 2014-08-24 A 20 50 5
4: 2014-08-22 B 42 22 43
5: 2014-08-23 B 44 45 12
6: 2014-08-24 B 3 21 2
Now I want to scale and sum many of the variables to get an output like:
Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
1: 2014-08-22 A 15 27 34 -1.1175975 0.7310560 121 68
2: 2014-08-23 A 39 44 29 0.3073393 0.4085313 121 68
3: 2014-08-24 A 20 50 5 0.8102582 -1.1395873 121 68
4: 2014-08-22 B 42 22 43 -0.5401315 1.1226726 88 57
5: 2014-08-23 B 44 45 12 1.1539172 -0.3274462 88 57
6: 2014-08-24 B 3 21 2 -0.6137858 -0.7952265 88 57
A brute force implementation of my problem would be:
library(data.table)
set.seed(123)
d <- data.table(Time = rep(seq.Date( Sys.Date(), length=3, by="day" )),
Stock = rep(LETTERS[1:2], each=3 ),
x1 = sample(1:50, 6),
x2 = sample(1:50, 6),
x3 = sample(1:50, 6))
d[,x2_scale:=scale(x2),by=Stock]
d[,x3_scale:=scale(x3),by=Stock]
d[,x2_sum:=sum(x2),by=Stock]
d[,x3_sum:=sum(x3),by=Stock]
Other posts describing a similar issue (Add multiple columns to R data.table in one function call? and Assign multiple columns using := in data.table, by group) suggest the following solution:
d[, c("x2_scale","x3_scale"):=list(scale(x2),scale(x3)), by=Stock]
d[, c("x2_sum","x3_sum"):=list(sum(x2),sum(x3)), by=Stock]
But again, this would get very messy with a lot of variables and also this brings up an error message with scale (but not with sum since this isn't returning a vector).
Is there a more efficient way to achieve the required result (keeping in mind that my actual data set is quite large)?
I think with a small modification to your last code you can easily do both for as many variables you want
vars <- c("x2", "x3") # <- Choose the variable you want to operate on
d[, paste0(vars, "_", "scale") := lapply(.SD, function(x) scale(x)[, 1]), .SDcols = vars, by = Stock]
d[, paste0(vars, "_", "sum") := lapply(.SD, sum), .SDcols = vars, by = Stock]
## Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
## 1: 2014-08-22 A 13 14 32 -1.1338934 1.1323092 87 44
## 2: 2014-08-23 A 25 39 9 0.7559289 -0.3701780 87 44
## 3: 2014-08-24 A 18 34 3 0.3779645 -0.7621312 87 44
## 4: 2014-08-22 B 44 8 6 -0.4730162 -0.7258662 59 32
## 5: 2014-08-23 B 49 3 18 -0.6757374 1.1406469 59 32
## 6: 2014-08-24 B 15 48 8 1.1487535 -0.4147807 59 32
For simple functions (that don't need special treatment like scale) you could easily do something like
vars <- c("x2", "x3") # <- Define the variable you want to operate on
funs <- c("min", "max", "mean", "sum") # <- define your function
for(i in funs){
d[, paste0(vars, "_", i) := lapply(.SD, eval(i)), .SDcols = vars, by = Stock]
}
Another variation using data.table
vars <- c("x2", "x3")
d[, paste0(rep(vars, each=2), "_", c("scale", "sum")) := do.call(`cbind`,
lapply(.SD, function(x) list(scale(x)[,1], sum(x)))), .SDcols=vars, by=Stock]
d
# Time Stock x1 x2 x3 x2_scale x2_sum x3_scale x3_sum
#1: 2014-08-22 A 15 27 34 -1.1175975 121 0.7310560 68
#2: 2014-08-23 A 39 44 29 0.3073393 121 0.4085313 68
#3: 2014-08-24 A 20 50 5 0.8102582 121 -1.1395873 68
#4: 2014-08-22 B 42 22 43 -0.5401315 88 1.1226726 57
#5: 2014-08-23 B 44 45 12 1.1539172 88 -0.3274462 57
#6: 2014-08-24 B 3 21 2 -0.6137858 88 -0.7952265 57
Based on comments from #Arun, you could also do:
cols <- paste0(rep(vars, each=2), "_", c("scale", "sum"))
d[,(cols):= unlist(lapply(.SD, function(x) list(scale(x)[,1L], sum(x))),
rec=F), by=Stock, .SDcols=vars]
You're probably looking for a pure data.table solution, but you could also consider using dplyr here since it works with data.tables as well (no need for conversion). Then, from dplyr you could use the function mutate_all as I do in this example here (with the first data set you showed in your question):
library(dplyr)
dt %>%
group_by(Stock) %>%
mutate_all(funs(sum, scale), x2, x3)
#Source: local data table [6 x 9]
#Groups: Stock
#
# Time Stock x1 x2 x3 x2_sum x3_sum x2_scale x3_scale
#1 2014-08-22 A 15 27 34 121 68 -1.1175975 0.7310560
#2 2014-08-23 A 39 44 29 121 68 0.3073393 0.4085313
#3 2014-08-24 A 20 50 5 121 68 0.8102582 -1.1395873
#4 2014-08-22 B 42 22 43 88 57 -0.5401315 1.1226726
#5 2014-08-23 B 44 45 12 88 57 1.1539172 -0.3274462
#6 2014-08-24 B 3 21 2 88 57 -0.6137858 -0.7952265
You can easily add more functions to be calculated which will create more columns for you. Note that mutate_all applies the function to each column except the grouping variable (Stock) by default. But you can either specify the columns you only want to apply the functions to (which I did in this example) or you can specify which columns you don't want to apply the functions to (that would be, e.g. -c(x2,x3) instead of where I wrote x2, x3).
EDIT: replaced mutate_each above with mutate_all as mutate_each will be deprecated in the near future.
EDIT: cleaner version using functional. I think this is the closest to the dplyr answer.
library(functional)
funs <- list(scale=Compose(scale, c), sum=sum) # See data.table issue #783 on github for the need for this
cols <- paste0("x", 2:3)
cols.all <- outer(cols, names(funs), paste, sep="_")
d[,
c(cols.all) := unlist(lapply(funs, Curry(lapply, X=.SD)), rec=F),
.SDcols=cols,
by=Stock
]
Produces:
Time Stock x1 x2 x3 x2_scale x3_scale x2_sum x3_sum
1: 2014-08-22 A 15 27 34 -1.1175975 0.7310560 121 68
2: 2014-08-23 A 39 44 29 0.3073393 0.4085313 121 68
3: 2014-08-24 A 20 50 5 0.8102582 -1.1395873 121 68
4: 2014-08-22 B 42 22 43 -0.5401315 1.1226726 88 57
5: 2014-08-23 B 44 45 12 1.1539172 -0.3274462 88 57
6: 2014-08-24 B 3 21 2 -0.6137858 -0.7952265 88 57