search for: setdt

...Sorry - this may be trivial, but I am struggling here for this. For two dataframes (A and B), I wish to identify (based on a primary key-column present in both A & B) - 1. Which records (rows) of A did not match with B, and 2. Which records of B did not match with A ? I came across a setdt function while browsing, but when I tried it, it says - Could not find function "setdt". Overall, if there is any way of doing it (preferably in some simplified way), please advise. Many thanks in advance. regards, Tito [[alternative HTML version deleted]]

...> > For two dataframes (A and B), I wish to identify (based on a primary > key-column present in both A & B) - > > 1. Which records (rows) of A did not match with B, and > > > > 2. Which records of B did not match with A ? > > > > I came across a setdt function while browsing, but when I tried it, it says > - Could not find function "setdt". > > > > Overall, if there is any way of doing it (preferably in some simplified > way), please advise. > > > Many thanks in advance. > > > regards, > &...

...d B), I wish to identify (based on a primary >> key-column present in both A & B) - >> >> 1. Which records (rows) of A did not match with B, and >> >> >> >> 2. Which records of B did not match with A ? >> >> >> >> I came across a setdt function while browsing, but when I tried it, it >> says >> - Could not find function "setdt". >> >> >> >> Overall, if there is any way of doing it (preferably in some simplified >> way), please advise. >> >> >> Many thanks in ad...

...stackoverflow.com/questions/33990760/converting-factors-to-binary-in-r df <-data.frame(a = c(1,2,3), b = c(1,1,2), c = c("Rose","Pink","Red"), d = c(2,3,4)) cbind(df[1:2], sapply(levels(df$c), function(x) as.integer(x == df$c)), df[4]) o así library(data.table) setDT(df)[, c(levels(df$c), "c") := c(lapply(levels(c), function(x) as.integer(x == c)), .(NULL))] Pero no me resuelve el tener que hacerlo algunos cientos de veces, que es lo que querría evitar. Sé que es evidente cómo se tiene que hacer, pero soy ciego a esa evidencia :-( Muchas gracia...

...>>> key-column present in both A & B) - >>> >>> 1. Which records (rows) of A did not match with B, and >>> >>> >>> >>> 2. Which records of B did not match with A ? >>> >>> >>> >>> I came across a setdt function while browsing, but when I tried it, it >>> says >>> - Could not find function "setdt". >>> >>> >>> >>> Overall, if there is any way of doing it (preferably in some simplified >>> way), please advise. >>> &gt...

Data.frame is returned by SQL query. It does have column names. In the function, I make small changes to some columns. Something like: Myquery <- ?SELECT date, price, stock FROM stocktab WHERE stock = ?ABC? AND date > ?2025-01-01?;? Prices <- dbGetQuery(con, myquery) SetDT(Prices) Prices[, date = as.Date(date)] R CMD check say ?no visible binding for global variable ?date?? Sent from my iPhone On Jan 28, 2025, at 1:24?AM, Sorkin, John <jsorkin at som.umaryland.edu> wrote: ? There you go, once again helping strengthen ;) John Get Outlook for iOS<https:...

...rame(a = c(1,2,3), b = c(1,1,2), c = > > c("Rose","Pink","Red"), d = c(2,3,4)) > > > > cbind(df[1:2], sapply(levels(df$c), function(x) as.integer(x == df$c)), > > df[4]) > > > > o así > > > > library(data.table) > > setDT(df)[, c(levels(df$c), "c") := > > c(lapply(levels(c), function(x) as.integer(x == c)), .(NULL))] > > > > > > Pero no me resuelve el tener que hacerlo algunos cientos de veces, que es > > lo que querría evitar. Sé que es evidente cómo se tiene que hacer, p...

...am struggling here for this. >> For two dataframes (A and B), I wish to identify (based on a primary >> key-column present in both A & B) - >> 1. Which records (rows) of A did not match with B, and >> 2. Which records of B did not match with A ? >> I came across a setdt function while browsing, but when I tried it, it says >> - Could not find function "setdt". >> Overall, if there is any way of doing it (preferably in some simplified >> way), please advise. >> Many thanks in advance. >> regards, >> Tito >> [...

...`data.frame`. When it detects a `tbl_df`, it tries to redirect to `reshape2::dcast()`, but since that appears to be deprecated, it will fail in future versions at some point. To avoid this, consider converting?`keeptabs` into a `data.table` directly before calling `dcast()`. For example: >?setDT(keeptabs) > out1 <- dcast(keeptabs, ID_Key ~ column1, fun.aggregate = length, value.var = "column1") > out2 <- dcast(keeptabs, ID_Key ~ column2, fun.aggregate = length, value.var = "column2") If?`keeptabs` is a `data.table` at the time of calling `dcas...

...t;,"column2"))) } keeptabs <- bind_rows(keeptabs) out1 <- dcast(keeptabs, ID_Key ~ column1, fun.aggregate = length, value.var = "column1") out2 <- dcast(keeptabs, ID_Key ~ column2, fun.aggregate = length, value.var = "column2") out1 <- setDT(out1 |> rename_with(~ paste0("column1_name_", .x, recycle0 = TRUE), -ID_Key)) out2 <- setDT(out1 |> rename_with(~ paste0("column2_name_", .x, recycle0 = TRUE), -ID_Key)) all_cols <- unique(c(names(out1), names(out2))) out1_missing <- setdiff(all_cols,...

...; > c("Rose","Pink","Red"), d = c(2,3,4)) > > > > > > cbind(df[1:2], sapply(levels(df$c), function(x) as.integer(x == df$c)), > > > df[4]) > > > > > > o as? > > > > > > library(data.table) > > > setDT(df)[, c(levels(df$c), "c") := > > > c(lapply(levels(c), function(x) as.integer(x == c)), .(NULL))] > > > > > > > > > Pero no me resuelve el tener que hacerlo algunos cientos de veces, que > es > > > lo que querr?a evitar. S? que es evide...

...ses. My data is organized in a data.table.? My goal is to perform analyses according to some groups.? The results of analysis are objects.? If these objects could be stored as elements of a data.table, this would help downstream summarizing of results. Let me try another example. carsdt <- setDT(copy(mtcars)) carsdt[, unique(cyl) |> length()] #[1] 3 carsreg <- carsdt[, .(fit = lm(mpg ~ disp + hp + wt)), by = .(cyl)] #I would like a data.table with three rows, one each for "lm" object corresponding to cyl value carsreg[, .N] #[1] 36 #Here each component of "lm&quo...

...urned by SQL query. It does have column names. In the function, I make small changes to some columns. > > Something like: > > Myquery <- ?SELECT date, price, stock FROM stocktab WHERE stock = ?ABC? AND date > ?2025-01-01?;? > > Prices <- dbGetQuery(con, myquery) > SetDT(Prices) > Prices[, date = as.Date(date)] If Prices were a regular dataframe at this point, then the message would be correct. You can't calculate `as.Date(date)` without telling R where to look for the `date` variable. However, you have set it to be a data.table instead. They use nonst...

...eturned by SQL query. It does have column names. In the function, I make small changes to some columns. >> Something like: >> Myquery <- ?SELECT date, price, stock FROM stocktab WHERE stock = ?ABC? AND date > ?2025-01-01?;? >> Prices <- dbGetQuery(con, myquery) >> SetDT(Prices) >> Prices[, date = as.Date(date)] > > If Prices were a regular dataframe at this point, then the message would be correct. You can't calculate `as.Date(date)` without telling R where to look for the `date` variable. > > However, you have set it to be a data.table ins...

...eturned by SQL query. It does have column names. In the function, I make small changes to some columns. >> Something like: >> Myquery <- ?SELECT date, price, stock FROM stocktab WHERE stock = ?ABC? AND date > ?2025-01-01?;? >> Prices <- dbGetQuery(con, myquery) >> SetDT(Prices) >> Prices[, date = as.Date(date)] > > If Prices were a regular dataframe at this point, then the message would be correct. You can't calculate `as.Date(date)` without telling R where to look for the `date` variable. > > However, you have set it to be a data.table ins...

Well, you may have good reasons to do things this way -- and you certainly do not have to explain them here. But you might wish to consider using R's poly() function and a basic nested list structure to do something quite similar that seems much simpler to me, anyway: x <- rnorm(20) df <- data.frame(x = x, y = x + .1*x^2 + rnorm(20, sd = .2)) result <- with(df,

How is the server configured to handle memory distribution for individual users. I see it has over 700GB of total system memory, but how much can be assigned it each individual user? AAgain - just curious, and wondering how much memory was assigned to your instance when you were running R. regards, Gregg On Wednesday, December 11th, 2024 at 9:49 AM, Deramus, Thomas Patrick <tderamus at

There you go, once again helping strengthen ;) John Get Outlook for iOS<https://aka.ms/o0ukef> ________________________________ From: R-help <r-help-bounces at r-project.org> on behalf of avi.e.gross at gmail.com <avi.e.gross at gmail.com> Sent: Tuesday, January 28, 2025 12:01:25 AM To: 'Naresh Gurbuxani' <naresh_gurbuxani at hotmail.com>; r-help at r-project.org

I am an old, long time SAS programmer. I need to produce R code that processes a dataframe in a manner that is equivalent to that produced by using a by statement in SAS and an if first.day statement and a retain statement: I want to take data (olddata) that looks like this ID Day 1 1 1 1 1 2 1 2 1 3 1 3 1 4 1 4 1 5 1 5 2 5 2 5 2 5 2 6 2 6 2 6 3 10 3 10 and make it look like this: (withing each