The data comes from another question I was playing around with:
dt <- data.table(user=c(rep(3, 5), rep(4, 5)),
country=c(rep(1,4),rep(2,6)),
event=1:10, key="user")
# user country event
#1: 3 1 1
#2: 3 1 2
#3: 3 1 3
#4: 3 1 4
#5: 3 2 5
#6: 4 2 6
#7: 4 2 7
#8: 4 2 8
#9: 4 2 9
#10: 4 2 10
And here's the surprising behavior:
dt[user == 3, as.data.frame(table(country))]
# country Freq
#1 1 4
#2 2 1
dt[user == 4, as.data.frame(table(country))]
# country Freq
#1 2 5
dt[, as.data.frame(table(country)), by = user]
# user country Freq
#1: 3 1 4
#2: 3 2 1
#3: 4 1 5
# ^^^ - why is this 1 instead of 2?!
Thanks mnel and Victor K. The natural follow-up is - shouldn't it be 2, i.e. is this a bug? I expected
dt[, blah, by = user]
to return identical result to
rbind(dt[user == 3, blah], dt[user == 4, blah])
Is that expectation incorrect?
The idiomatic data.table approach is to use .N
dt[ , .N, by = list(user, country)]
This will be far quicker and it will also retain country as the same class as in the original.
As mnel noted in comments, as.data.frame(table(...)) produces a data frame where the first variable is a factor. For user == 4, there is only one level in the factor, which is stored internally as 1.
What you want is factor levels, but what you get is how factors are stored internally (as integers, starting from 1). The following provides the expected result:
> dt[, lapply(as.data.frame(table(country)), as.character), by = user]
user country Freq
1: 3 1 4
2: 3 2 1
3: 4 2 5
Update. Regarding your second question: no, I think data.table behaviour is correct. Same thing happens in plain R when you join two factors with different levels:
> a <- factor(3:5)
> b <- factor(6:8)
> a
[1] 3 4 5
Levels: 3 4 5
> b
[1] 6 7 8
Levels: 6 7 8
> c(a,b)
[1] 1 2 3 1 2 3
Related
Given two data.table:
dt1 <- data.table(id = c(1,-99,2,2,-99), a = c(2,1,-99,-99,3), b = c(5,3,3,2,5), c = c(-99,-99,-99,2,5))
dt2 <- data.table(id = c(2,3,1,4,3),a = c(6,4,3,2,6), b = c(3,7,8,8,3), c = c(2,2,4,3,2))
> dt1
id a b c
1: 1 2 5 -99
2: -99 1 3 -99
3: 2 -99 3 -99
4: 2 -99 2 2
5: -99 3 5 5
> dt2
id a b c
1: 2 6 3 2
2: 3 4 7 2
3: 1 3 8 4
4: 4 2 8 3
5: 3 6 3 2
How can one replace the -99 of dt1 with the values of dt2?
Wanted results should be dt3:
> dt3
id a b c
1: 1 2 5 2
2: 3 1 3 2
3: 2 3 3 4
4: 2 2 2 2
5: 3 3 5 5
You can do the following:
dt3 <- as.data.frame(dt1)
dt2 <- as.data.frame(dt2)
dt3[dt3 == -99] <- dt2[dt3 == -99]
dt3
# id a b c
# 1 1 2 5 2
# 2 3 1 3 2
# 3 2 3 3 4
# 4 2 2 2 2
# 5 3 3 5 5
If your data is all of the same type (as in your example) then transforming them to matrix is a lot faster and transparent:
dt1a <- as.matrix(dt1) ## convert to matrix
dt2a <- as.matrix(dt2)
# make a matrix of the same shape to access the right entries
missing_idx <- dt1a == -99
dt1a[missing_idx] <- dt2a[missing_idx] ## replace by reference
This is a vectorized operation, so it should be fast.
Note: If you do this make sure the two data sources match exactly in shape and order of rows/columns. If they don't then you need to join by the relevant keys and pick the correct columns.
EDIT: The conversion to matrix may be unnecessary. See kath's answer for a more terse solution.
Simple way could be to use setDF function to convert to data.frame and use data frame sub-setting methods. Restore to data.table at the end.
#Change to data.frmae
setDF(dt1)
setDF(dt2)
# Perform assignment
dt1[dt1==-99] = dt2[dt1==-99]
# Restore back to data.table
setDT(dt1)
setDT(dt2)
dt1
# id a b c
# 1 1 2 5 2
# 2 3 1 3 2
# 3 2 3 3 4
# 4 2 2 2 2
# 5 3 3 5 5
This simple trick would work efficiently.
dt1<-as.matrix(dt1)
dt2<-as.matrix(dt2)
index.replace = dt1==-99
dt1[index.replace] = dt2[index.replace]
as.data.table(dt1)
as.data.table(dt2)
This should work, a simple approach:
for (i in 1:nrow(dt1)){
for (j in 1:ncol(dt1)){
if (dt1[i,j] == -99) dt1[i,j] = dt2[i,j]
}
}
I have a hard time to understand the "Vectorised between" example in data.table packages document V1.10.4?
X = data.table(a=1:5, b=6:10, c=c(5:1))
> X
a b c
1: 1 6 5
2: 2 7 4
3: 3 8 3
4: 4 9 2
5: 5 10 1
# NEW feature in v1.9.8, vectorised between
> X[c %between% list(a,b)]
a b c
1: 1 6 5
2: 2 7 4
3: 3 8 3
X[between(c, a, b)] # same as above
Can someone please explain it to me how dose it work? why only 5,4,3 from c was selected? Thanks.
-----As posted in comments----
In row 4, 2 is not between 4 and 9....between(c=2,a=4,b=9).
between uses >= and <= (rather than > and <). That's why in row 3, it returns 3 (since its TRUE)
I have a large data set - around 32mil rows. I have information on the telephone number, the origin of the call, and the destination.
For each telephone number, I want to count the number of times it appeared either as Origin or as Destination.
An example data table is as follows:
library(data.table)
dt <- data.table(Tel=seq(1,5,1), Origin=seq(1,5,1), Destination=seq(3,7,1))
Tel Origin Destination
1: 1 1 3
2: 2 2 4
3: 3 3 5
4: 4 4 6
5: 5 5 7
I have working code, but it takes too long for my data since it involves a for loop. How can I optimize it?
Here it is:
for (i in unique(dt$Tel)){
index <- (dt$Origin == i | dt$Destination == i)
dt[dt$Tel ==i, "N"] <- sum(index)
}
Result:
Tel Origin Destination N
1: 1 1 3 1
2: 2 2 4 1
3: 3 3 5 2
4: 4 4 6 2
5: 5 5 7 2
Where N tells that Tel=1 appears 1, Tel=2 appears 1, Tel=3,4 and 5 each appear 2 times.
We can do a melt and match
dt[, N := melt(dt, id.var = "Tel")[, tabulate(match(value, Tel))]]
Or another option is to loop through the columns 2 and 3, use %in% to check whether the values in 'Tel' are present, then with Reduce and + get the sum of logical elements for each 'Tel', assign (:=) the values to 'N'
dt[, N := Reduce(`+`, lapply(.SD, function(x) Tel %in% x)), .SDcols = 2:3]
dt
# Tel Origin Destination N
#1: 1 1 3 1
#2: 2 2 4 1
#3: 3 3 5 2
#4: 4 4 6 2
#5: 5 5 7 2
A second method constructs a temporary data.table which is then joins to the original. This is longer and likely less efficient than #akrun's, but can be useful to see.
# get temporary data.table as the sum of origin and destination frequencies
temp <- setnames(data.table(table(unlist(dt[, .(Origin, Destination)], use.names=FALSE))),
c("Tel", "N"))
# turn the variables into integers (Tel is the name of the table above, and thus character)
temp <- temp[, lapply(temp, as.integer)]
Now, join the original table on
dt <- temp[dt, on="Tel"]
dt
Tel N Origin Destination
1: 1 1 1 3
2: 2 1 2 4
3: 3 2 3 5
4: 4 2 4 6
5: 5 2 5 7
You can get the desired column order using setcolorder
setcolorder(dt, c("Tel", "Origin", "Destination", "N"))
I need to create (with R) a rolling index of pairs from a data set that includes groups. Consider the following data set:
times <- c(4,3,2)
V1 <- unlist(lapply(times, function(x) seq(1, x)))
df <- data.frame(group = rep(1:length(times), times = times),
V1 = V1,
rolling_index = c(1,1,2,2,3,3,4,5,5))
df
group V1 rolling_index
1 1 1 1
2 1 2 1
3 1 3 2
4 1 4 2
5 2 1 3
6 2 2 3
7 2 3 4
8 3 1 5
9 3 2 5
The data frame I have includes the variables group and V1. Within each group V1 designates a running index (that may or may not start at 1).
I want to create a new indexing variable that looks like rolling_index. This variable groups rows within the same group and consecutive V1 value, thus creating a new rolling index. This new index must be consecutive over groups. If there is an uneven amount of rows within a group (e.g. group 2), then the last, single row gets its own rolling index value.
You can try
library(data.table)
setDT(df)[, gr:=as.numeric(gl(.N, 2, .N)), group][,
rollindex:=cumsum(c(TRUE,abs(diff(gr))>0))][,gr:= NULL]
# group V1 rolling_index rollindex
#1: 1 1 1 1
#2: 1 2 1 1
#3: 1 3 2 2
#4: 1 4 2 2
#5: 2 1 3 3
#6: 2 2 3 3
#7: 2 3 4 4
#8: 3 1 5 5
#9: 3 2 5 5
Or using base R
indx1 <- !duplicated(df$group)
indx2 <- with(df, ave(group, group, FUN=function(x)
gl(length(x), 2, length(x))))
cumsum(c(TRUE,diff(indx2)>0)|indx1)
#[1] 1 1 2 2 3 3 4 5 5
Update
The above methods are based on the 'group' column. Suppose you already have a sequence column ('V1') by group as showed in the example, creation of rolling index is easier
cumsum(!!df$V1 %%2)
#[1] 1 1 2 2 3 3 4 5 5
As mentioned in the post, if the 'V1' column do not start at '1' for some groups, we can get the sequence from the 'group' and then do the cumsum as above
cumsum(!!with(df, ave(seq_along(group), group, FUN=seq_along))%%2)
#[1] 1 1 2 2 3 3 4 5 5
There is probably a simpler way but you can do:
rep_each <- unlist(mapply(function(q,r) {c(rep(2, q),rep(1, r))},
q=table(df$group)%/%2,
r=table(df$group)%%2))
df$rolling_index <- inverse.rle(x=list(lengths=rep_each, values=seq(rep_each)))
df$rolling_index
#[1] 1 1 2 2 3 3 4 5 5
I have monthly data in one data.table and annual data in another data.table and now I want to match the annual data to the respective observation in the monthly data.
My approach is as follows: Duplicating the annual data for every month and then join the monthly and annual data. And now I have a question regarding the duplication of rows. I know how to do it, but I'm not sure if it is the best way to do it, so some opinions would be great.
Here is an exemplatory data.table DT for my annual data and how I currently duplicate:
library(data.table)
DT <- data.table(ID = paste(rep(c("a", "b"), each=3), c(1:3, 1:3), sep="_"),
values = 10:15,
startMonth = seq(from=1, by=2, length=6),
endMonth = seq(from=3, by=3, length=6))
DT
ID values startMonth endMonth
[1,] a_1 10 1 3
[2,] a_2 11 3 6
[3,] a_3 12 5 9
[4,] b_1 13 7 12
[5,] b_2 14 9 15
[6,] b_3 15 11 18
#1. Alternative
DT1 <- DT[, list(MONTH=startMonth:endMonth), by="ID"]
setkey(DT, ID)
setkey(DT1, ID)
DT1[DT]
ID MONTH values startMonth endMonth
a_1 1 10 1 3
a_1 2 10 1 3
a_1 3 10 1 3
a_2 3 11 3 6
[...]
The last join is exactly what I want. However, DT[, list(MONTH=startMonth:endMonth), by="ID"] already does everything I want except adding the other columns to DT, so I was wondering if I could get rid of the last three rows in my code, i.e. the setkey and join operations. It turns out, you can, just do the following:
#2. Alternative: More intuitiv and just one line of code
DT[, list(MONTH=startMonth:endMonth, values, startMonth, endMonth), by="ID"]
ID MONTH values startMonth endMonth
a_1 1 10 1 3
a_1 2 10 1 3
a_1 3 10 1 3
a_2 3 11 3 6
...
This, however, only works because I hardcoded the column names into the list expression. In my real data, I do not know the names of all columns in advance, so I was wondering if I could just tell data.table to return the column MONTH that I compute as shown above and all the other columns of DT. .SD seemed to be able to do the trick, but:
DT[, list(MONTH=startMonth:endMonth, .SD), by="ID"]
Error in `[.data.table`(DT, , list(YEAR = startMonth:endMonth, .SD), by = "ID") :
maxn (4) is not exact multiple of this j column's length (3)
So to summarize, I know how it's been done, but I was just wondering if this is the best way to do it because I'm still struggling a little bit with the syntax of data.table and often read in posts and on the wiki that there are good and bads ways of doing things. Also, I don't quite get why I get an error when using .SD. I thought it is just any easy way to tell data.table that you want all columns. What do I miss?
Looking at this I realized that the answer was only possible because ID was a unique key (without duplicates). Here is another answer with duplicates. But, by the way, some NA seem to creep in. Could this be a bug? I'm using v1.8.7 (commit 796).
library(data.table)
DT <- data.table(x=c(1,1,1,1,2,2,3),y=c(1,1,2,3,1,1,2))
DT[,rep:=1L][c(2,7),rep:=c(2L,3L)] # duplicate row 2 and triple row 7
DT[,num:=1:.N] # to group each row by itself
DT
x y rep num
1: 1 1 1 1
2: 1 1 2 2
3: 1 2 1 3
4: 1 3 1 4
5: 2 1 1 5
6: 2 1 1 6
7: 3 2 3 7
DT[,cbind(.SD,dup=1:rep),by="num"]
num x y rep dup
1: 1 1 1 1 1
2: 2 1 1 1 NA # why these NA?
3: 2 1 1 2 NA
4: 3 1 2 1 1
5: 4 1 3 1 1
6: 5 2 1 1 1
7: 6 2 1 1 1
8: 7 3 2 3 1
9: 7 3 2 3 2
10: 7 3 2 3 3
Just for completeness, a faster way is to rep the row numbers and then take the subset in one step (no grouping and no use of cbind or .SD) :
DT[rep(num,rep)]
x y rep num
1: 1 1 1 1
2: 1 1 2 2
3: 1 1 2 2
4: 1 2 1 3
5: 1 3 1 4
6: 2 1 1 5
7: 2 1 1 6
8: 3 2 3 7
9: 3 2 3 7
10: 3 2 3 7
where in this example data the column rep happens to be the same name as the rep() base function.
Great question. What you tried was very reasonable. Assuming you're using v1.7.1 it's now easier to make list columns. In this case it's trying to make one list column out of .SD (3 items) alongside the MONTH column of the 2nd group (4 items). I'll raise it as a bug [EDIT: now fixed in v1.7.5], thanks.
In the meantime, try :
DT[, cbind(MONTH=startMonth:endMonth, .SD), by="ID"]
ID MONTH values startMonth endMonth
a_1 1 10 1 3
a_1 2 10 1 3
a_1 3 10 1 3
a_2 3 11 3 6
...
Also, just to check you've seen roll=TRUE? Typically you'd have just one startMonth column (irregular with gaps) and then just roll join to it. Your example data has overlapping month ranges though, so that complicates it.
Here is a function I wrote which mimics disaggregate (I needed something that handled complex data). It might be useful for you, if it isn't overkill. To expand only rows, set the argument fact to c(1,12) where 12 would be for 12 'month' rows for each 'year' row.
zexpand<-function(inarray, fact=2, interp=FALSE, ...) {
fact<-as.integer(round(fact))
switch(as.character(length(fact)),
'1' = xfact<-yfact<-fact,
'2'= {xfact<-fact[1]; yfact<-fact[2]},
{xfact<-fact[1]; yfact<-fact[2];warning(' fact is too long. First two values used.')})
if (xfact < 1) { stop('fact[1] must be > 0') }
if (yfact < 1) { stop('fact[2] must be > 0') }
# new nonloop method, seems to work just ducky
bigtmp <- matrix(rep(t(inarray), each=xfact), nrow(inarray), ncol(inarray)*xfact, byr=T)
#does column expansion
bigx <- t(matrix(rep((bigtmp),each=yfact),ncol(bigtmp),nrow(bigtmp)*yfact,byr=T))
return(invisible(bigx))
}
The fastest and most succinct way of doing it:
DT[rep(1:nrow(DT), endMonth - startMonth)]
We can also enumerate by group by:
dd <- DT[rep(1:nrow(DT), endMonth - startMonth)]
dd[, nn := 1:.N, by = ID]
dd