I have the following data table:
library(data.table)
set.seed(1)
DT <- data.table(ind=1:100,x=sample(100),y=sample(100),group=c(rep("A",50),rep("B",50)))
Now the problem I have is that I need to take every value in column "x" (that is, each given ID), and add all the existing values in column "y" to it. I also need to do it separately per column "group". Let's assume we start with ID = 1. This element has the value: x_1 = 68, and y_1 = 76. We also see y_2 = 39, y_3 = 24, etc. So what I want to compute is the sums x_1 + y_1, x_1 + y2, x_1 + y_3, etc. But not only for x_1, but also for x_2, x_3, etc. So for x_2 it would look like: x_2 + y_1, x_2 + y_2, x_2 + y_3, etc. This should also be done separately per column "group" (in this regard the dataset should simple be split by group).
Edit: Exemplary code to do this only for X_1 and group A:
current_X <- DT[1,x] # not needed, just to illustrate
vector_current_X <- rep(DT[1,x],nrow(DT[group == "A"]))
DT[group == "A",copy_current_X := vector_current_X]
DT[,sum_current_X_Y := copy_current_X + y]
DT
One apparent issue with this approach is that if it were applied to all x, then a lot of columns would be added to the final DT. So I am not sure if it is the best approach. In the end, I am just looking for the lowest sum (per element x) with each element y, and per group.
I know how to do operations per group, and I also know the lapply functions. The issue is that from my understanding, I need to include a row-wise loop. And next, the structure of the result will be different from the original data table, because we have many additional observations. I have seen before that you can save lists inside a data.table, but I am unsure if that is the best approach. My dataset is much larger, so efficiency is important.
Thanks for any hints how to approach this.
You can do this:
DT[, .(.BY$x+DT[group==.BY$group,y]), by=.(x,group)]
This returns N rows per x, where N is the size of x's group. We leverage the special (.BY), which is available in j when utilizing by. Basically, .BY is a named list, containing the values of the grouping variables. Here, I'm adding the value of x (.BY$x) to the vector of y values from the subset of DT where the group is equal to the current group value (.BY$group)
Output:
x group V1
<int> <char> <int>
1: 68 A 144
2: 68 A 107
3: 68 A 92
4: 68 A 121
5: 68 A 160
---
4996: 4 B 25
4997: 4 B 66
4998: 4 B 83
4999: 4 B 27
5000: 4 B 68
You can also accomplish this via a join:
DT[,!c("y")][DT[, .(y,group)], on=.(group), allow.cartesian=T][, total:=x+y][order(ind)]
Output:
ind x group y total
<int> <int> <char> <int> <int>
1: 1 68 A 76 144
2: 1 68 A 39 107
3: 1 68 A 24 92
4: 1 68 A 53 121
5: 1 68 A 92 160
---
4996: 100 4 B 21 25
4997: 100 4 B 62 66
4998: 100 4 B 79 83
4999: 100 4 B 23 27
5000: 100 4 B 64 68
If I understand correctly, the requested result requires a cross join where each element of x is combined with each element of y (within each group).
This can be accomplished easily using the CJ() function:
DT[, CJ(x, y, sorted = FALSE), by = group][, sum_x_y := x + y][]
group x y sum_x_y
1: A 68 76 144
2: A 68 39 107
3: A 68 24 92
4: A 68 53 121
5: A 68 92 160
---
4996: B 4 21 25
4997: B 4 62 66
4998: B 4 79 83
4999: B 4 23 27
5000: B 4 64 68
Related
I have a data.table in R, and I'm looking to create a vector based on .SDcols row by row.
library("data.table")
dt = data.table(
id=1:6,
A1=sample(100,6),
A2=sample(100,6),
A3=sample(100,6),
B1=sample(100,6),
B2=sample(100,6),
B3=sample(100,6)
)
dt[,x1:=paste(.SD,collapse = ","),.SDcols=A1:B3,by=id]
dt[,x2:=strsplit(x1,",")] # x2 vector of characters
now, I got x2 with a vector of characters.
however, I expected x2 with a vector of integers.
R > dt
id A1 A2 A3 B1 B2 B3 x2
1: 1 72 23 76 10 35 14 c(72,23,76,10,35,14)
2: 2 44 28 77 29 20 63 c(44,28,77,29,20,63)
3: 3 18 34 43 77 76 100 c(18,34,43,77,76,100)
4: 4 15 33 50 87 86 86 c(15,33,50,87,86,86)
5: 5 71 71 41 75 8 3 c(71,71,41,75,8,3)
6: 6 11 89 98 42 72 27 c(11,89,98,42,72,27)
I tried with several solutions, all failed.
dt[,x2:=.(list(.SD)),.SDcols=A1:B3,by=id] #x2 is <data.table>
dt[,x2:=.(lapply(.SD,c)),.SDcols=A1:B3,by=id]
dt[,x2:=.(c(.SD)), .SDcols=A1:B3,by=id] #RHS 1 is length 6 (greater than the size (1) of group 1). The last 5 element(s) will be discarded.
dt[,x2:=c(.SD),.SDcols=A1:B3,by=id] # x2 equals A1
dt[,x2:=lapply(.SD,c),.SDcols=A1:B3,by=id] # x2 equals A1
dt[,x2:=sapply(.SD,c),.SDcols=A1:B3,by=id] # x2 equals A1
Any suggestion?
Thanks in advance
=====================================================================
edit: thanks Jaap,
dt[, x2 := lapply(strsplit(x1, ","), as.integer)] # it works
Still, I wonder any beautiful solution?
=====================================================================
edit2:
new solutions, base function is much more useful than I thought.
dt[,ABC0:=apply(rbind(.SD), 1, list),.SDcols=A1:B3,by=id]
dt[,ABC1:=apply(cbind(.SD), 1, list),.SDcols=A1:B3,by=id]
or more simple
dt[,ABC2:=lapply(.SD,rbind),.SDcols=A1:B3]
I'm new to R and still getting to grips with how it handles data (my background is spreadsheets and databases). the problem I have is as follows. My data looks like this (it is held in CSV):
RecNo Var1 Var2 Var3
41 800 201.8 Y
43 140 39 N
47 60 20.24 N
49 687 77 Y
54 570 135 Y
58 1250 467 N
61 211 52 N
64 96 117.3 N
68 687 77 Y
Column 1 (RecNo) is my observation number; while it is a number, it is not required for my analysis. Column 4 (Var3) is a Yes/No column which, again, I do not currently need for the analysis but will need later in the process to add information in the output.
I need to normalise the numeric data in my dataframe to values between 0 and 1 without losing the other information. I have the following function:
normalize <- function(x) {
x <- sweep(x, 2, apply(x, 2, min))
sweep(x, 2, apply(x, 2, max), "/")
}
However, when I apply it to my above data by calling
myResult <- normalize(myData)
it returns an error because of the text in Column 4. If I set the text in this column to binary values it runs fine, but then also normalises my case numbers, which I don't want.
So, my question is: How can I change my normalize function above to accept the names of the columns to transform, while outputting the full dataset (i.e. without losing columns)?
I could not get TUSHAr's suggestion to work, but I have found two solutions that work fine:
1. akrun's suggestion above:
myData2 <- myData1 %>% mutate_at(2:3, funs((.-min(.))/max(.-min(.))))
This produces the following:
RecNo Var1 Var2 Var3
1 41 0.62184874 0.40601834 Y
2 43 0.06722689 0.04195255 N
3 47 0.00000000 0.00000000 N
4 49 0.52689076 0.12693105 Y
5 54 0.42857143 0.25663508 Y
6 58 1.00000000 1.00000000 N
7 61 0.12689076 0.07102414 N
8 64 0.03025210 0.21718329 N
9 68 0.52689076 0.12693105 Y
Alternatively, there is the package BBmisc which allowed me the following after transforming my record numbers to factors:
> myData <- myData %>% mutate(RecNo = factor(RecNo))
> myNorm <- normalize(myData2, method="range", range = c(0,1), margin = 1)
> myNorm
RecNo Var1 Var2 Var3
1 41 0.62184874 0.40601834 Y
2 43 0.06722689 0.04195255 N
3 47 0.00000000 0.00000000 N
4 49 0.52689076 0.12693105 Y
5 54 0.42857143 0.25663508 Y
6 58 1.00000000 1.00000000 N
7 61 0.12689076 0.07102414 N
8 64 0.03025210 0.21718329 N
9 68 0.52689076 0.12693105 Y
EDIT: For completion I include TUSHAr's solution as well, showing as always that there are many ways around a single problem:
normalize<-function(x){
minval=apply(x[,c(2,3)],2,min)
maxval=apply(x[,c(2,3)],2,max)
#print(minval)
#print(maxval)
y=sweep(x[,c(2,3)],2,minval)
#print(y)
sweep(y,2,(maxval-minval),"/")
}
df[,c(2,3)]=normalize(df)
Thank you for your help!
normalize<-function(x){
minval=apply(x[,c(2,3)],2,min)
maxval=apply(x[,c(2,3)],2,max)
#print(minval)
#print(maxval)
y=sweep(x[,c(2,3)],2,minval)
#print(y)
sweep(y,2,(maxval-minval),"/")
}
df[,c(2,3)]=normalize(df)
I have a data frame which looks like this:
sub = c("X001","X001", "X001","X002","X002","X001","X002","X001","X002","X002","X002","X002")
revenue = c(20, 15, -10,-25,20,-20, 17,9,14,12, -9, 11)
df = data.frame(sub, revenue)
I want to aggregate it in such a way that the second column should show the sum of all revenue for the sub, the third column should show the sum on absolute value, the fourth column should show the sum of all positive values and the fifth column should show the sum of all negative values.
The result should look like this:
Sub All Sum Absolute Sum Positive Sum Negative Sum
X001 14 74 44 -30
X002 40 108 74 -34
I've written code that calculates the All sum:
y<-aggregate(df$revenue, by=list(Feature=x$Sub), FUN=sum)
I would really appreciate it if someone more knowledgable in R would help me in calculating the other three columns.
Here's how to do that with dplyr:
library(dplyr)
df%>%
group_by(sub)%>%
summarise(All_Sum=sum(revenue),Absolute_Sum=sum(abs(revenue)),
Positive_Sum=(sum(revenue[revenue>0])),Negative_Sum=(sum(revenue[revenue<0])))
sub All_Sum Absolute_Sum Positive_Sum Negative_Sum
<fctr> <dbl> <dbl> <dbl> <dbl>
1 X001 14 74 44 -30
2 X002 40 108 74 -34
In base R using aggregate:
aggregate(.~sub, df, function(a) c(sum(a), sum(abs(a)), sum(a[a>0]), sum(a[a<0])))
# sub revenue.1 revenue.2 revenue.3 revenue.4
#1 X001 14 74 44 -30
#2 X002 40 108 74 -34
We can also use data.table
library(data.table)
setDT(df)[, .(All_Sum = sum(revenue), Absolute_Sum = sum(abs(revenue)),
Positive_Sum = sum(revenue[revenue>0]), Negative_Sum = sum(revenue[revenue<0])), by = sub]
# sub All_Sum Absolute_Sum Positive_Sum Negative_Sum
#1: X001 14 74 44 -30
#2: X002 40 108 74 -34
I have a binomail dataset that looks like this:
df <- data.frame(replicate(4,sample(1:200,1000,rep=TRUE)))
addme <- data.frame(replicate(1,sample(0:1,1000,rep=TRUE)))
df <- cbind(df,addme)
df <-df[order(df$replicate.1..sample.0.1..1000..rep...TRUE..),]
The data is currently soreted in a way to show the instances belonging to 0 group then the ones belonging to the 1 group. Is there a way I can sort the data in a 0-1-0-1-0... fashion? I mean to show a row that belongs to the 0 group, the row after belonging to the 1 group then the zero group and so on...
All I can think about is complex functions. I hope there's a simple way around it.
Thank you,
Here's an attempt, which will add any extra 1's at the end:
First make some example data:
set.seed(2)
df <- data.frame(replicate(4,sample(1:200,10,rep=TRUE)),
addme=sample(0:1,10,rep=TRUE))
Then order:
with(df, df[unique(as.vector(rbind(which(addme==0),which(addme==1)))),])
# X1 X2 X3 X4 addme
#2 141 48 78 33 0
#1 37 111 133 3 1
#3 115 153 168 163 0
#5 189 82 70 103 1
#4 34 37 31 174 0
#6 189 171 98 126 1
#8 167 46 72 57 0
#7 26 196 30 169 1
#9 94 89 193 134 1
#10 110 15 27 31 1
#Warning message:
#In rbind(which(addme == 0), which(addme == 1)) :
# number of columns of result is not a multiple of vector length (arg 1)
Here's another way using dplyr, which would make it suitable for within-group ordering. It's also probably pretty quick. If there's unbalanced numbers of 0's and 1's, it will leave them at the end.
library(dplyr)
df %>%
arrange(addme) %>%
mutate(n0 = sum(addme == 0),
orderme = seq_along(addme) - (n0 * addme) + (0.5 * addme)) %>%
arrange(orderme) %>%
select(-n0, -orderme)
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