I've been getting used to data.tables and just cannot seem to find the answer to something that feels so simple (or at least is with data frames).
I want to use data.table to aggregate, however, I don't always know which column to aggregate ahead of time (it takes input from the user). I want to define what column to use based off of a character vector. Here's a short example of what I want to do:
require(data.table)
myDT <- data.table(a = 1:10, b = 11:20, n1 = c("first", "second"))
aggWith <- "a"
Now I want to use the aggWith object to define what column to sum on. This does not work:
> myDT.Agg <- myDT[, .(Agg = sum(aggWith)), by = .(n1)]
Error in sum(aggWith) : invalid 'type' (character) of argument
Nor does this:
> myDT.Agg <- myDT[, .(Agg = sum(aggWith)), by = .(n1), with = FALSE]
Error in sum(aggWith) : invalid 'type' (character) of argument
This does:
myDT.Agg <- myDT[, .(Agg = sum(a)), by = .(n1)]
However, I want to be able to define which column "a" is arbitrarily based off a character vector. I've looking through ?data.table, but am just not seeing what I need. Sorry in advance if this is really simple and I'm just overlooking something.
We could specify the 'aggWith' as .SDcols and then get the sum of .SD
myDT[, list(Agg= sum(.SD[[1L]] )), by = n1, .SDcols=aggWith]
If there are multiple columns, then loop with lapply
myDT[, lapply(.SD, sum), by = n1, .SDcols= aggWith]
Another option would be to use eval(as.name
myDT[, list(Agg= sum(eval(as.name(aggWith)))), by = n1]
Related
my data.table contain K columns called claims, among other 30 columns. I want to subset the data.table, such that only rows remain which do not have 0 claims.
So, firstly i get all the column names i need for filtering. For the purpose of this example, i have chosen K = 2
> claimsCols = c("claimsnext", paste0("claims" , 1:K))
> claimsCols
[1] "claimsnext" "claims1" "claims2"
i have tried subsetting like:
for(i in claimsCols){
BTplan <- BTplan[ claimsCols[i] == 0, ]
i+1
}
this doent work:
Error in i + 1 : non-numeric argument to binary operator
I am sure there is a better way to do this?
I would basically do what akrun does
idx = BTplan[ , Reduce(`&`, .SD), .SDcols = patterns('claims')]
BTplan = BTplan[idx]
The innovations are:
Use patterns in .SDcols to specify the columns to include by pattern
& automatically converts numeric to logical, i.e. 1.1 & 2.2 is TRUE, and becomes FALSE as soon as there's a 0 anywhere (hence filtering the corresponding row)
In a future version of data.table this will be slightly more efficient and concise (and hopefully more readable):
idx = BTplan[ , pall(.SD), .SDcols = patterns('claims')]
BTplan = BTplan[idx]
Keep an eye on this pull request:
https://github.com/Rdatatable/data.table/pull/4448
In the OP's code, the i is each of the elements of 'claimsCols' which is character, so i +1 won't work and in fact, it is not needed
for(colnm in claimsCols) {
BTplan <- BTplan[BTplan[[colnm]] != 0,]
}
Or using data.table syntax
library(data.table)
setDT(BTplan)
BTplan[BTplan[, Reduce(`&`, lapply(.SD, `!=`, 0)),.SDcols = claimsCols]]
How can I program a loop so that all eight tables are calculated one after the other?
The code:
dt_M1_I <- M1_I
dt_M1_I <- data.table(dt_M1_I)
dt_M1_I[,I:=as.numeric(gsub(",",".",I))]
dt_M1_I[,day:=substr(t,1,10)]
dt_M1_I[,hour:=substr(t,12,16)]
dt_M1_I_median <- dt_M1_I[,list(median_I=median(I,na.rm = TRUE)),by=.(day,hour)]
This should be calculated for:
M1_I
M2_I
M3_I
M4_I
M1_U
M2_U
M3_U
M4_U
Thank you very much for your help!
Whenever you have several variables of the same kind, especially when you find yourself numbering them, as you did, step back and replace them with a single list variable. I do not recommend doing what the other answer suggested.
That is, instead of M1_I…M4_I and M1_U…M4_U, have two variables m_i and m_u (using lower case in variable names is conventional), which are each lists of four data.tables.
Alternatively, you might want to use a single variable, m, which contains nested lists of data.tables (m = list(list(i = …, u = …), …)).
Assuming the first, you can then iterate over them as follows:
give_this_a_meaningful_name = function (df) {
dt <- data.table(df)
dt[, I := as.numeric(gsub(",", ".", I))]
dt[, day := substr(t, 1, 10)]
dt[, hour := substr(t, 12, 16)]
dt[, list(median_I = median(I, na.rm = TRUE)), by = .(day, hour)]
}
m_i_median = lapply(m_i, give_this_a_meaningful_name)
(Note also the introduction of consistent spacing around operators; good readability is paramount for writing bug-free code.)
You can use a combination of a for loop and the get/assign functions like this:
# create a vector of the data.frame names
dts <- c('M1_I', 'M2_I', 'M3_I', 'M4_I', 'M1_U', 'M2_U', 'M3_U', 'M4_U')
# iterate over each dataframe
for (dt in dts){
# get the actual dataframe (not the string name of it)
tmp <- get(dt)
tmp <- data.table(tmp)
tmp[, I:=as.numeric(gsub(",",".",I))]
tmp[, day:=substr(t,1,10)]
tmp[, hour:=substr(t,12,16)]
tmp <- tmp[,list(median_I=median(I,na.rm = TRUE)),by=.(day,hour)]
# assign the modified dataframe to the name you want (the paste adds the 'dt_' to the front)
assign(paste0('dt_', dt), tmp)
}
I'm trying to manipulate a number of data.tables in similar ways, and would like to write a function to accomplish this. I would like to pass in a parameter containing a list of columns that would have the operations performed. This works fine when the vector declaration of columns is the left hand side of the := operator, but not if it is declared earlier (or passed into the function). The follow code shows the issue.
dt = data.table(a = letters, b = 1:2, c=1:13)
colsToDelete = c('b', 'c')
dt[,colsToDelete := NULL] # doesn't work but I don't understand why not.
dt[,c('b', 'c') := NULL] # works fine, but doesn't allow passing in of columns
The error is "Adding new column 'colsToDelete' then assigning NULL (deleting it)." So clearly, it's interpreting 'colsToDelete' as a new column name.
The same issue occurs when doing something along these lines
dt[, colNames := lapply(.SD, adjustValue, y=factor), .SDcols = colNames]
I new to R, but rather more experienced with some other languages, so this may be a silly question.
It's basically because we allow symbols on LHS of := to add new columns, for convenience: ex: DT[, col := val]. So, in order to distinguish col itself being the name from whatever is stored in col being the column names, we check if the LHS is a name or an expression.
If it's a name, it adds the column with the name as such on the LHS, and if expression, then it gets evaluated.
DT[, col := val] # col is the column name.
DT[, (col) := val] # col gets evaluated and replaced with its value
DT[, c(col) := val] # same as above
The preferred idiom is: dt[, (colsToDelete) := NULL]
HTH
I am surprised no answer provided uses the set() function.
set(DT, , colsToDelete, NULL)
This should be the easiest.
To extend on previous answer, you can delete columns by reference doing:
# delete columns 10 to 15
dt[ , (10:15) := NULL ]
or
# delete columns 3, 5 and 10 to 15
dt[ , (c(3,5,10:15)) := NULL ]
This code did the job for me. you need to have the position of the columns to be deleted e.g., posvec as mentioned in the ?set
j: Column name(s) (character) or number(s) (integer) to be assigned value
when column(s) already exist, and only column name(s) if they are to
be created.
DT_removed_slected_col = set(DT, j = posvec, value = NULL)
Also if you want to get the posvec you can try this:
selected_col = c('col_a','col_b',...)
selected_col = unlist(sapply(selected_col, function(x) grep(x,names(DT))))
namvec = names(selected_col) #col names
posvec = unname(selected_col) #col positions
I'm trying to use data.table rather data.frame(for a faster code). Despite the syntax difference between than, I'm having problems when I need to extract a specific character column and use it as character vector. When I call:
library(data.table)
DT <- fread("file.txt")
vect <- as.character(DT[, 1, with = FALSE])
class(vect)
###[1] "character"
head(vect)
It returns:
[1] "c(\"uc003hzj.4\", \"uc021ofx.1\", \"uc021olu.1\", \"uc021ome.1\", \"uc021oov.1\", \"uc021opl.1\", \"uc021osl.1\", \"uc021ovd.1\", \"uc021ovp.1\", \"uc021pdq.1\", \"uc021pdv.1\", \"uc021pdw.1\")
Any ideas of how to avoid these "\" in the output?
The as.character works on vectors and not on data.frame/data.table objects in the way the OP expected. So, if we need to get the first column as character class, subset with .SD[[1L]] and apply the as.character
DT[, as.character(.SD[[1L]])]
If there are multiple columns, we can specify the column index with .SDcols and loop over the .SD to convert to character and assign (:=) the output back to the particular columns.
DT[, (1:2) := lapply(.SD, as.character), .SDcols= 1:2]
data
DT <- data.table(Col1 = 1:5, Col2= 6:10, Col3= LETTERS[1:5])
I have a data frame with two columns. I want to add an additional two columns to the data set with counts based on aggregates.
df <- structure(list(ID = c(1045937900, 1045937900),
SMS.Type = c("DF1", "WCB14"),
SMS.Date = c("12/02/2015 19:51", "13/02/2015 08:38"),
Reply.Date = c("", "13/02/2015 09:52")
), row.names = 4286:4287, class = "data.frame")
I want to simply count the number of Instances of SMS.Type and Reply.Date where there is no null. So in the toy example below, i will generate the 2 for SMS.Type and 1 for Reply.Date
I then want to add this to the data frame as total counts (Im aware they will duplicate out for the number of rows in the original dataset but thats ok)
I have been playing around with aggregate and count function but to no avail
mytempdf <-aggregate(cbind(testtrain$SMS.Type,testtrain$Response.option)~testtrain$ID,
train,
function(x) length(unique(which(!is.na(x)))))
mytempdf <- aggregate(testtrain$Reply.Date~testtrain$ID,
testtrain,
function(x) length(which(!is.na(x))))
Can anyone help?
Thank you for your time
Using data.table you could do (I've added a real NA to your original data).
I'm also not sure if you really looking for length(unique()) or just length?
library(data.table)
cols <- c("SMS.Type", "Reply.Date")
setDT(df)[, paste0(cols, ".count") :=
lapply(.SD, function(x) length(unique(na.omit(x)))),
.SDcols = cols,
by = ID]
# ID SMS.Type SMS.Date Reply.Date SMS.Type.count Reply.Date.count
# 1: 1045937900 DF1 12/02/2015 19:51 NA 2 1
# 2: 1045937900 WCB14 13/02/2015 08:38 13/02/2015 09:52 2 1
In the devel version (v >= 1.9.5) you also could use uniqueN function
Explanation
This is a general solution which will work on any number of desired columns. All you need to do is to put the columns names into cols.
lapply(.SD, is calling a certain function over the columns specified in .SDcols = cols
paste0(cols, ".count") creates new column names while adding count to the column names specified in cols
:= performs assignment by reference, meaning, updates the newly created columns with the output of lapply(.SD, in place
by argument is specifying the aggregator columns
After converting your empty strings to NAs:
library(dplyr)
mutate(df, SMS.Type.count = sum(!is.na(SMS.Type)),
Reply.Date.count = sum(!is.na(Reply.Date)))