I am trying to call different columns of a data.table inside a loop, to get unique values of each column.
Consider the simple data.table below.
> df <- data.table(var_a = rep(1:10, 2),
+ var_b = 1:20)
> df
var_a var_b
1: 1 1
2: 2 2
3: 3 3
4: 4 4
5: 5 5
6: 6 6
7: 7 7
8: 8 8
9: 9 9
10: 10 10
11: 1 11
12: 2 12
13: 3 13
14: 4 14
15: 5 15
16: 6 16
17: 7 17
18: 8 18
19: 9 19
20: 10 20
My code works when I call for a specific column outside a loop,
> unique(df$var_a)
[1] 1 2 3 4 5 6 7 8 9 10
> unique(df[, var_a])
[1] 1 2 3 4 5 6 7 8 9 10
> unique(df[, "var_a"])
var_a
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 8
9: 9
10: 10
but not when I do so within a loop that goes through different columns of the data.table.
> for(v in c("var_a","var_b")){
+ print(v)
+ df$v
+ unique(df[, .v])
+ unique(df[, "v"])
+ }
[1] "var_a"
Error in `[.data.table`(df, , .v) :
j (the 2nd argument inside [...]) is a single symbol but column name '.v' is not found. Perhaps you intended DT[, ...v]. This difference to data.frame is deliberate and explained in FAQ 1.1.
>
> unique(df[, ..var_a])
Error in `[.data.table`(df, , ..var_a) :
Variable 'var_a' is not found in calling scope. Looking in calling scope because you used the .. prefix.
For the first problem, when you're referencing a column name indirectly, you can either use double-dot ..v syntax, or add with=FALSE in the data.table::[ construct:
for (v in c("var_a", "var_b")) {
print(v)
print(df$v)
### either one of these will work:
print(unique(df[, ..v]))
# print(unique(df[, v, with = FALSE]))
}
# [1] "var_a"
# NULL
# var_a
# <int>
# 1: 1
# 2: 2
# 3: 3
# 4: 4
# 5: 5
# 6: 6
# 7: 7
# 8: 8
# 9: 9
# 10: 10
# [1] "var_b"
# NULL
# var_b
# <int>
# 1: 1
# 2: 2
# 3: 3
# 4: 4
# 5: 5
# 6: 6
# 7: 7
# 8: 8
# 9: 9
# 10: 10
# 11: 11
# 12: 12
# 13: 13
# 14: 14
# 15: 15
# 16: 16
# 17: 17
# 18: 18
# 19: 19
# 20: 20
# var_b
But this just prints it without changing anything. If all you want to do is look at unique values within each column (and not change the underlying frame), then I'd likely go with
lapply(df[,.(var_a, var_b)], unique)
# $var_a
# [1] 1 2 3 4 5 6 7 8 9 10
# $var_b
# [1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
which shows the name and unique values. The use of lapply (whether on df as a whole or a subset of columns) is also preferable to another recommendation to use apply(df, 2, unique), though in this case it returns the same results.
Use .subset2 to refer to a column by its name:
for(v in c("var_a","var_b")) {
print(unique(.subset2(df, v)))
}
following the information on the first error, this would be the correct way to call in a loop:
for(v in c("var_a","var_b")){
print(unique(df[, ..v]))
}
# won't print all the lines
as for the second error you have not declared a variable called "var_a", it looks like you want to select by name.
# works as you have shown
unique(df[, "var_a"])
# works once the variable is declared
var_a <- "var_a"
unique(df[, ..var_a])
You may also be interested in the env param of data.table (see development version); here is an illustration below, but you could use this in a loop too.
v="var_a"
df[, v, env=list(v=v)]
Output:
[1] 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10
Related
On my projects I usually do the data prepping with a few functions, so my code usually look like this:
readAndClean("directory") %>%
processing() %>%
readyForModelling()
Where I'm passing a data.table object from one function to another.
I've gotten a habit to always start this functions with:
processing <- function(data_init){
data <- copy(data_init)
}
to avoid making changes to the DT on the global environment, as the following example will:
test <- data.table(cars[1:10,])
processing <- function(data_init){
data_init[, id := 1:.N]
return("done")
}
test
# speed dist
# 1: 4 2
# 2: 4 10
# 3: 7 4
# 4: 7 22
# 5: 8 16
# 6: 9 10
# 7: 10 18
# 8: 10 26
# 9: 10 34
# 10: 11 17
processing(test)
# [1] "done"
test
# speed dist id
# 1: 4 2 1
# 2: 4 10 2
# 3: 7 4 3
# 4: 7 22 4
# 5: 8 16 5
# 6: 9 10 6
# 7: 10 18 7
# 8: 10 26 8
# 9: 10 34 9
# 10: 11 17 10
But this always seems a little ugly to me.
Is it the correct way of handling data.tables inside functions?
Recently I saw a question (can't find the link) that was something like this
I want to add a column on a data.frame that computes the variance of a different column while removing the current observation.
dt = data.table(
id = c(1:13),
v = c(9,5,8,1,25,14,7,87,98,63,32,12,15)
)
So, with a for() loop:
res = NULL
for(i in 1:13){
res[i] = var(dt[-i,v])
}
I tried doing this in data.table, using negative indexing with .I, but to my surprise none of the following works:
#1
dt[,var := var(dt[,v][-.I])]
#2
dt[,var := var(dt$v[-.I])]
#3
fun = function(x){
v = c(9,5,8,1,25,14,7,87,98,63,32,12,15)
var(v[-x])
}
dt[,var := fun(.I)]
#4
fun = function(x){
var(dt[-x,v])
}
dt[,var := fun(.I)]
All of those gives the same output:
id v var
1: 1 9 NA
2: 2 5 NA
3: 3 8 NA
4: 4 1 NA
5: 5 25 NA
6: 6 14 NA
7: 7 7 NA
8: 8 87 NA
9: 9 98 NA
10: 10 63 NA
11: 11 32 NA
12: 12 12 NA
13: 13 15 NA
What am I missing? I thought it was a problem with .I being passed to functions, but a dummy example:
fun = function(x,c){
x*c
}
dt[,dummy := fun(.I,2)]
id v var
1: 1 9 2
2: 2 5 4
3: 3 8 6
4: 4 1 8
5: 5 25 10
6: 6 14 12
7: 7 7 14
8: 8 87 16
9: 9 98 18
10: 10 63 20
11: 11 32 22
12: 12 12 24
13: 13 15 26
works fine.
Why can't I use .I in this specific scenario?
You may use .BY:
a list containing a length 1 vector for each item in by
dt[ , var_v := dt[id != .BY$id, var(v)], by = id]
Variance is calculated once per row (by = id). In each calculation, the current row is excluded using id != .BY$id in the 'inner' i.
all.equal(dt$var_v, res)
# [1] TRUE
Why doesn't your code work? Because...
.I is an integer vector equal to seq_len(nrow(x)),
...your -.I not only removes current observation, it removes all rows in one go from 'v'.
A small illustration which starts with your attempt (just without the assignment :=) and simplifies it step by step:
# your attempt
dt[ , var(dt[, v][-.I])]
# [1] NA
# without the `var`, indexing only
dt[ , dt[ , v][-.I]]
# numeric(0)
# an empty vector
# same indexing written in a simpler way
dt[ , v[-.I]]
# numeric(0)
# even more simplified, with a vector of values
# and its corresponding indexes (equivalent to .I)
v <- as.numeric(11:14)
i <- 1:4
v[i]
# [1] 11 12 13 14
x[-i]
# numeric(0)
Here's a brute-force thought:
exvar <- function(x, na.rm = FALSE) sapply(seq_len(length(x)), function(i) var(x[-i], na.rm = na.rm))
dt[,var := exvar(v)]
dt
# id v var
# 1: 1 9 1115.538
# 2: 2 5 1098.265
# 3: 3 8 1111.515
# 4: 4 1 1077.841
# 5: 5 25 1153.114
# 6: 6 14 1132.697
# 7: 7 7 1107.295
# 8: 8 87 822.447
# 9: 9 98 684.697
# 10: 10 63 1040.265
# 11: 11 32 1153.697
# 12: 12 12 1126.424
# 13: 13 15 1135.538
Finding the last position of a vector that is less than a given value is fairly straightforward (see e.g. this question
But, doing this line by line for a column in a data.frame or data.table is horribly slow. For example, we can do it like this (which is ok on small data, but not good on big data)
library(data.table)
set.seed(123)
x = sort(sample(20,5))
# [1] 6 8 15 16 17
y = data.table(V1 = 1:20)
y[, last.x := tail(which(x <= V1), 1), by = 1:nrow(y)]
# V1 last.x
# 1: 1 NA
# 2: 2 NA
# 3: 3 NA
# 4: 4 NA
# 5: 5 NA
# 6: 6 1
# 7: 7 1
# 8: 8 2
# 9: 9 2
# 10: 10 2
# 11: 11 2
# 12: 12 2
# 13: 13 2
# 14: 14 2
# 15: 15 3
# 16: 16 4
# 17: 17 5
# 18: 18 5
# 19: 19 5
# 20: 20 5
Is there a fast, vectorised way to get the same thing? Preferably using data.table or base R.
You may use findInterval
y[ , last.x := findInterval(V1, x)]
Slightly more convoluted using cut. But on the other hand, you get the NAs right away:
y[ , last.x := as.numeric(cut(V1, c(x, Inf), right = FALSE))]
Pretty simple in base R
x<-c(6L, 8L, 15L, 16L, 17L)
y<-1:20
cumsum(y %in% x)
[1] 0 0 0 0 0 1 1 2 2 2 2 2 2 2 3 4 5 5 5 5
i hope this one isn´t stupid.
I have two dataframes with Variables ID and gender/sex. In df1, there are NAs. In df2, the variable is complete. I want to complete the column in df1 with the values from df2.
(In df1 the variable is called "gender". In df2 it is called "sex".)
Here is what i tried so far:
#example-data
ID<-seq(1,30,by=1)
df1<-as.data.frame(ID)
df2<-df1
df1$gender<-c(NA,"2","1",NA,"2","2","2","2","2","2",NA,"2","1","1",NA,"2","2","2","2","2","1","2","2",NA,"2","2","2","2","2",NA)
df2$sex<-c("2","2","1","2","2","2","2","2","2","2","2","2","1","1","2","2","2","2","2","2","1","2","2","2","2","2","2","2","2","2")
#Approach 1:
NAs.a <- is.na(df1$gender)
df1$gender[NAs.a] <- df2[match(df1$ID[NAs.a], df2$ID),]$sex
#Approach 2 (i like dplyr a lot, perhaps there´s a way to use it):
library("dplyr")
temp<-df2 %>% select(ID,gender)
#EDIT:
#df<-left_join(df1$gender,df2$gender, by="ID")
df<-left_join(df1,df2, by="ID")
Thank you very much.
Here's a quick solution using data.tables binary join this will join only gender with sex and leave all the rest of the columns untouched
library(data.table)
setkey(setDT(df1), ID)
df1[df2, gender := i.sex][]
# ID gender
# 1: 1 2
# 2: 2 2
# 3: 3 1
# 4: 4 2
# 5: 5 2
# 6: 6 2
# 7: 7 2
# 8: 8 2
# 9: 9 2
# 10: 10 2
# 11: 11 2
# 12: 12 2
# 13: 13 1
# 14: 14 1
# 15: 15 2
# 16: 16 2
# 17: 17 2
# 18: 18 2
# 19: 19 2
# 20: 20 2
# 21: 21 1
# 22: 22 2
# 23: 23 2
# 24: 24 2
# 25: 25 2
# 26: 26 2
# 27: 27 2
# 28: 28 2
# 29: 29 2
# 30: 30 2
This would probably be the simplest with base R.
idx <- is.na(df1$gender)
df1$gender[idx] = df2$sex[idx]
You could do
df1 %>% select(ID) %>% left_join(df2, by = "ID")
# ID sex
#1 1 2
#2 2 2
#3 3 1
#4 4 2
#5 5 2
#6 6 2
#.. ..
This assumes - as in the example - that all ID's from df1 are also present in df2 and have a sex/gender information there.
If you have other columns in your data you could also try this instead:
df1 %>% select(-gender) %>% left_join(df2[c("ID", "sex")], by = "ID")
I have a data.table
dt2 <- data.table(urn=1:10,freq=0, freqband="")
dt2$freqband = NA
dt2$freq <- 1:7 #does give a warning message
## urn freq freqband
## 1: 1 1 NA
## 2: 2 2 NA
## 3: 3 3 NA
## 4: 4 4 NA
## 5: 5 5 NA
## 6: 6 6 NA
## 7: 7 7 NA
## 8: 8 1 NA
## 9: 9 2 NA
##10: 10 3 NA
i also have a function that I am wanting to use to group my freq column
fn_GetFrequency <- function(numgifts) {
if (numgifts <5) return("<5")
if (numgifts >=5) return("5+")
return("ERROR")
}
I am wanting to set the freqband column based on this function. In some cases it will be all records, in some cases it will be a subset. My current approach is (for a subset):
dt2[dt2$urn < 9, freqband := fn_GetFrequency(freq)]
using this approach I get the warning:
Warning message:
In if (numgifts < 5) return("<5") :
the condition has length > 1 and only the first element will be used
then it sets all the records to have a value of "<5" rather than the correct value. I'm figuring that I need to use some sort of lapply/sapply/etc function, however I still haven't been able to quite grasp how they work in order to use them to solve my problem.
Any help would be greatly appreciated.
EDIT: How might you do this if you use a function that requires 2 parameters?
UPDATED: to include the output of dt2 after my attempted update
urn freq freqband
1: 1 1 <5
2: 2 2 <5
3: 3 3 <5
4: 4 4 <5
5: 5 5 <5
6: 6 6 <5
7: 7 7 <5
8: 8 1 <5
9: 9 2 NA
10: 10 3 NA
UPDATE: I tried this code to and it worked to deliver the desired output, and it allows me to have a function I can call in other places of code too.
dt2[dt2$urn < 9, freqband := sapply(freq, fn_GetFrequency)]
> fn_GetFrequency <- function(numgifts) {
+ ifelse (numgifts <5, "<5", "5+")
+ }
> dt2[dt2$urn < 9, freqband := fn_GetFrequency(freq)]
> dt2
urn freq freqband
1: 1 1 <5
2: 2 2 <5
3: 3 3 <5
4: 4 4 <5
5: 5 5 5+
6: 6 6 5+
7: 7 7 5+
8: 8 1 <5
9: 9 2 NA
10: 10 3 NA
For multiple bands (which I'm sure has been asked before) you should use the findInterval function. And I'm doing it the data.table way reather than the dataframe way:
dt2[ urn==8, freq := -1 ] # and something to test the <0 condition
dt2[ urn <= 8, freqband := c("ERROR", "<5", "5+")[
findInterval(freq,c(-Inf, 0, 5 ,Inf))] ]
dt2
urn freq freqband
1: 1 1 <5
2: 2 2 <5
3: 3 3 <5
4: 4 4 <5
5: 5 5 5+
6: 6 6 5+
7: 7 7 5+
8: 8 -1 ERROR
9: 9 2 NA
10: 10 3 NA