I would like to convert variable to z-scores. How to do that for each factor cell level separately using a loop?
Example DATA:
df = data.frame(Cell = c(rep("13a",5),rep("1b",5),rep("5b",5)),
condition = rep(c("a","b","c","d","e"),3),
variable = c(58,55,36,29,53,57,53,54,52,52,45,49,48,46,45))
Is this a good start?... Maybe the loop is not necessary buy I would like to learn how to write loops...
# Final data frame containing the results of all loops
df_z = data.frame()
# Loop through by cell
for (i in 1:unique(df$Cell)) {
df_z$myZ <- scale(variable)
}
It can be done with a group_by operation
library(dplyr)
df %>%
group_by(Cell) %>%
mutate(myZ = as.numeric(scale(variable)))
Or with data.table
library(data.table)
setDT(df)[, myZ := as.numeric(scale(variable)), by = Cell][]
In the case for for loop, we can subset the in each of the iteration and assign the scaled values to the created 'myZ' variable
un1 <- unique(df$Cell)
df$myZ <- NA
for(un in un1) {
i1 <- df$Cell == un
df$myZ[i1] <- as.numeric(scale(df$variable[i1]))
}
Or with split
df$myZ <- unsplit(lapply(split(df$variable, df$Cell), scale), df$Cell)
We can use ave in base R :
df$myZ <- with(df, ave(variable, Cell, FUN = scale))
df
# Cell condition variable myZ
#1 13a a 58 0.917
#2 13a b 55 0.684
#3 13a c 36 -0.792
#4 13a d 29 -1.336
#5 13a e 53 0.528
#6 1b a 57 1.640
#7 1b b 53 -0.289
#8 1b c 54 0.193
#9 1b d 52 -0.772
#10 1b e 52 -0.772
#11 5b a 45 -0.881
#12 5b b 49 1.321
#13 5b c 48 0.771
#14 5b d 46 -0.330
#15 5b e 45 -0.881
Related
I am working on a problem set and absolutely cannot figure this one out. I think I've fried my brain to the point where it doesn't even make sense anymore.
Here is a look at the data ...
sex age chol tg ht wt sbp dbp vldl hdl ldl bmi
<chr> <int> <int> <int> <dbl> <dbl> <int> <int> <int> <int> <int> <dbl>
1 M 60 137 50 68.2 112. 110 70 10 53 74 2.40
2 M 26 154 202 82.8 185. 88 64 34 31 92 2.70
3 M 33 198 108 64.2 147 120 80 22 34 132 3.56
4 F 27 154 47 63.2 129 110 76 9 57 88 3.22
5 M 36 212 79 67.5 176. 130 100 16 37 159 3.87
6 F 31 197 90 64.5 121 122 78 18 58 111 2.91
7 M 28 178 163 66.5 167 118 68 19 30 135 3.78
8 F 28 146 60 63 105. 120 80 12 46 88 2.64
9 F 25 231 165 64 126 130 72 23 70 137 3.08
10 M 22 163 30 68.8 173 112 70 6 50 107 3.66
# … with 182 more rows
I must write a function, myTtest, to perform the following task:
Perform a two-sample t-tests to compare the differences of a series of numeric variables between each level of a classification variable
The first argument, dat, is a data frame
The second argument, classVar, is a character vector of length 1. It is the name of the classification variable, such as 'sex.'
The third argument, numVar, is a character vector that contains the name of the numeric variables, such as c("age", "chol", "tg"). This means I need to perform three t-tests to compare the difference of those between males and females.
The function should return a data frame with the following variables: Varname, F.mean, M.mean, t (for t-statistics), df (for degrees of freedom), and p (for p-value).
I should be able to run this ...
myTtest(dat = chol, classVar = "sex", numVar = c("age", "chol", "tg")
... and then get the data frame to appear.
Any help is greatly appreciated. I am pulling my hair out over this one! As well, as noted in my comment below, this has to be done without Tidyverse ... which is why I'm having so much trouble to begin with.
The intuition for this solution is that you can loop over your dependent variables, and call t.test() in each loop. Then save the results from each DV and stack them together in one big data frame.
I'll leave out some bits for you to fill in, but here's the gist:
First, some example data:
set.seed(123)
n <- 20
grp <- sample(c("m", "f"), n, replace = TRUE)
df <- data.frame(grp = grp, age = rnorm(n), chol = rnorm(n), tg = rnorm(n))
df
grp age chol tg
1 m 1.2240818 0.42646422 0.25331851
2 m 0.3598138 -0.29507148 -0.02854676
3 m 0.4007715 0.89512566 -0.04287046
4 f 0.1106827 0.87813349 1.36860228
5 m -0.5558411 0.82158108 -0.22577099
6 f 1.7869131 0.68864025 1.51647060
7 f 0.4978505 0.55391765 -1.54875280
8 f -1.9666172 -0.06191171 0.58461375
9 m 0.7013559 -0.30596266 0.12385424
10 m -0.4727914 -0.38047100 0.21594157
Now make a container that each of the model outputs will go into:
fits_df <- data.frame()
Loop over each DV and append the model output to fits_df each time with rbind:
for (dv in c("age", "chol", "tg")) {
frml <- as.formula(paste0(dv, " ~ grp")) # make a model formula: dv ~ grp
fit <- t.test(frml, two.sided = TRUE, data = df) # perform the t-test
# hint: use str(fit) to figure out how to pull out each value you care about
fit_df <- data.frame(
dv = col,
f_mean = xxx,
m_mean = xxx,
t = xxx,
df = xxx,
p = xxx
)
fits_df <- rbind(fits_df, fit_df)
}
Your output will look like this:
fits_df
dv f_mean m_mean t df p
1 age -0.18558068 -0.04446755 -0.297 15.679 0.7704954
2 chol 0.07731514 0.22158672 -0.375 17.828 0.7119400
3 tg 0.09349567 0.23693052 -0.345 14.284 0.7352112
One note: When you're pulling out values from fit, you may get odd row names in your output data frame. This is due to the names property of the various fit attributes. You can get rid of these by using as.numeric() or as.character() wrappers around the values you pull from fit (for example, fit$statistic can be cleaned up with as.character(round(fit$statistic, 3))).
I have a list containing two data frames:
sample_list <- list("tables" = data.frame(weight = sample(1:50, 20, replace = T)),
"chairs" = data.frame(height = sample(1:50, 20, replace = T)))
I would like to use lapply to run a function over all the data frames in this list. In the output of each function, I need to create another column with the name of the source data frame (see mutate):
lapply(sample_list, function(x) {
x %>%
filter(x >= 20) %>%
mutate(groupName = names(x))
})
For some reason, I can't figure out how to make this work. How do I pass the name of the data frame into mutate? Right now it is returning the name of the first column in that data frame, rather than the name of the data frame itself.
Thanks!
We can loop through names of sample_list instead of looping through the list
lapply(names(sample_list), function(x) {
sample_list[[x]] %>%
filter_at(vars(1),~. >= 20) %>%
mutate(groupName = x)
})
Update Sep-2021
cleaner way using purrr::map
purrr::map(names(sample_list), ~sample_list[[.x]] %>%
filter_at(vars(1),~. >= 20) %>%
mutate(groupName = .x)
)
You can try purrr::imap() to map over both elements and elements' name.
# purrr::imap
purrr::imap(sample_list, function(element,name){
head(mutate(element,groupName = name))
})
# or mapply, but you need to specify names of the list
myfun <- function(element,name){
head(mutate(element,groupName = name))
}
mapply(myfun,sample_list,names(sample_list),SIMPLIFY = FALSE)
$tables
weight groupName
1 42 tables
2 24 tables
3 13 tables
4 31 tables
5 9 tables
6 27 tables
$chairs
height groupName
1 18 chairs
2 6 chairs
3 34 chairs
4 37 chairs
5 36 chairs
6 49 chairs
Using Map from base R
Map(function(dat, grp) cbind(dat, group_name = grp)[dat[[1]] > 20,],
sample_list, names(sample_list))
You can use Map with the function data.frame to add the names.
Map(`data.frame`, sample_list, groupName = names(sample_list))
#Map(`[<-`, sample_list, "groupName", value = names(sample_list)) #Alternative
#$tables
# weight groupName
#1 22 tables
#2 12 tables
#3 9 tables
#4 26 tables
#5 39 tables
#6 6 tables
#7 31 tables
#8 9 tables
#9 39 tables
#10 4 tables
#11 37 tables
#12 30 tables
#13 20 tables
#14 35 tables
#15 31 tables
#16 46 tables
#17 44 tables
#18 30 tables
#19 12 tables
#20 46 tables
#
#$chairs
# height groupName
#1 12 chairs
#2 17 chairs
#3 35 chairs
#4 40 chairs
#5 23 chairs
#6 21 chairs
#7 48 chairs
#8 24 chairs
#9 20 chairs
#10 41 chairs
#11 43 chairs
#12 45 chairs
#13 47 chairs
#14 13 chairs
#15 35 chairs
#16 32 chairs
#17 26 chairs
#18 34 chairs
#19 33 chairs
#20 8 chairs
In case it should also be subseted to those >= 20:
lapply(sample_list, function(x) x[x[,1] >= 20,, drop = FALSE])
When it should be done in one step I would use the way already posted by #akrun.
Here is a code to generate a data.frame :
ref_variables=LETTERS[1:10]
row=100
d0=seq(1:100)
for (i in seq_along(ref_variables)){
dtemp=sample(seq(1:row),row,TRUE)
d0=data.frame(d0,dtemp)
}
d0[,1]=NULL
names(d0)=ref_variables
I have a dataset, data.frame or data.table, whatever.
Let's say I want to modify the columns 2 to 4 by dividing each of them by the first one. Of Course, I can make a loop like this :
columns_name_to_divide=c("B","C","H")
column_divisor="A"
for (i in seq_along(columns_name_to_divide)){
ds[columns_name_to_divide[i]] = ds[columns_name_to_divide[i]] / ds[column_divisor]
}
But is there a way more elegant to do it?
> d0[2:4] <- d0[,2:4]/d0[,1]
This will substitute your original values with result you get after dividing column 2,3,4 by column 1. Rest will remain the same.
If you want to create 3 new columns in d0 with new values after dividing column 2,3,4 by column 1 This will not replace the original values in column 2,3, and 4. The calculated values would be in column 11,12 and 13 respectively.
> dim(d0)
# [1] 100 10
> d0[11:13] <- d0[,2:4]/d0[,1]
> dim(d0)
# [1] 100 13
To round up the new values, you can simply add round() function to 2 decimal places like below:
> d0[2:4] <- round(d0[,2:4]/d0[,1],2) # Original values subtituted at 2,3,4
# OR
> d0[11:13] <- round(d0[,2:4]/d0[,1],2) # New columns added, original columns are untouched.
We can use set from data.table which would be make this more efficient as the overhead of .[data.table is avoided when called multiple times (though not in this case).
library(data.table)
setDT(d0)
for(j in columns_name_to_divide){
set(d0, i = NULL, j = j, value = d0[[j]]/d0[[column_divisor]])
}
Or using lapply
setDT(d0)[, (columns_name_to_divide) := lapply(.SD, `/`,
d0[[column_divisor]]), .SDcols = columns_name_to_divide]
Or an elegant option using dplyr
library(dplyr)
library(magrittr)
d0 %<>%
mutate_each_(funs(./d0[[column_divisor]]), columns_name_to_divide)
head(d0)
# A B C D E F G H I J
#1 60 0.4000000 1.1500000 6 86 27 19 0.150000 94 97
#2 11 0.6363636 0.3636364 25 52 44 82 8.818182 84 68
#3 80 0.8750000 1.1375000 72 34 56 69 0.125000 34 17
#4 77 0.3116883 1.0259740 9 44 87 61 1.064935 79 40
#5 18 0.3333333 5.0555556 60 69 62 89 2.166667 21 34
#6 42 1.3333333 2.3095238 61 20 87 95 1.428571 78 63
Benchmarks
set.seed(42)
d1 <- as.data.frame(matrix(sample(1:9, 1e7*7, replace=TRUE), ncol=7))
d2 <- copy(d1)
d3 <- copy(d1)
system.time({
d2 %<>%
mutate_each(funs(./d2[["V2"]]), V4:V7)
})
# user system elapsed
# 0.52 0.39 0.91
system.time({
d1[,4:7] <- d1[,4:7]/d1[,2]
})
# user system elapsed
# 1.72 0.72 2.44
system.time({
setDT(d3)
for(j in 4:7){
set(d3, i = NULL, j = j, value = d3[[j]]/d3[["V2"]])
}
})
# user system elapsed
# 0.32 0.16 0.47
You can do this:
library(data.table)
cols <- names(df)[2:4]
col1 <- names(df)[1]
setDT(df)[, (cols) := lapply (cols, function(x) get(x) / get(col1) )]
# sample data for reproducible example:
df <- data.frame(V1=rep(10,5),
V2=rep(20,5),
V3=rep(30,5),
V4=rep(40,5),
V5=rep(50,5))
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