Displaying 5 results from an estimated 5 matches for "12e6".
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2005 May 02
2
RMySQL query: why result takes so much memory in R ?
...ll 12 millions of records I want to import 3 fields only.
The fields are specified as:id int(11), group char(15), measurement
float(4,2).
Why does this take > 1G RAM? I run R on suse linux, with 1G RAM and with
the code below it even fills the whole 1G of swap. I just don't
understand how 12e6 * 3 can fill such a huge range of RAM? Thanks for
clarification and potential solutions.
## my code
library(RMySQL)
drv <- dbDriver("MySQL")
ch <- dbConnect(drv,dbname="testdb",
user="root",password="mysql")
testdb <- dbGetQuery(c...
2018 Feb 20
0
Take the maximum of every 12 columns
Ista, et. al: efficiency?
(Note: I needed to correct my previous post: do.call() is required for
pmax() over the data frame)
> x <- data.frame(matrix(runif(12e6), ncol=12))
> system.time(r1 <- do.call(pmax,x))
user system elapsed
0.049 0.000 0.049
> identical(r1,r2)
[1] FALSE
> system.time(r2 <- apply(x,1,max))
user system elapsed
2.162 0.045 2.207
## 150 times slower!
> identical(r1,r2)
[1] TRUE
pmax() is there f...
2018 Feb 20
2
Take the maximum of every 12 columns
On Tue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com>
wrote:
> Ista, et. al: efficiency?
> (Note: I needed to correct my previous post: do.call() is required for
> pmax() over the data frame)
>
> > x <- data.frame(matrix(runif(12e6), ncol=12))
>
> > system.time(r1 <- do.call(pmax,x))
> user system elapsed
> 0.049 0.000 0.049
>
> > identical(r1,r2)
> [1] FALSE
> > system.time(r2 <- apply(x,1,max))
> user system elapsed
> 2.162 0.045 2.207
>
> ## 150 times s...
2018 Feb 20
3
Take the maximum of every 12 columns
This is what I was looking for. Thank you everyone!
Sincerely,
Milu
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2018 Feb 20
0
Take the maximum of every 12 columns
...ue, Feb 20, 2018 at 11:58 AM, Bert Gunter <bgunter.4567 at gmail.com>
> wrote:
>
>> Ista, et. al: efficiency?
>> (Note: I needed to correct my previous post: do.call() is required for
>> pmax() over the data frame)
>>
>> > x <- data.frame(matrix(runif(12e6), ncol=12))
>>
>> > system.time(r1 <- do.call(pmax,x))
>> user system elapsed
>> 0.049 0.000 0.049
>>
>> > identical(r1,r2)
>> [1] FALSE
>> > system.time(r2 <- apply(x,1,max))
>> user system elapsed
>> 2.162...