ake a suggestion, if you really want a tibble that allows you
> to know what each row is for, consider one of many methods for saving the
> previous row names as a new column. I used that to take the data.frame
> version I made above and got:
>
> > temp <- as_tibble(result.
I work with a list of crypto assets daily closing prices in a xts
class. Here is a limited example:
asset.xts.lst <- list(BTCUSDT = structure(c(26759.63, 26862, 26852.48, 27154.15,
27973.45), dim = c(5L, 1L), index = structure(c(1697068800, 1697155200,
1697241600, 1697328000, 1697414400), tzone =
or your answer.
Why not use
> factors? Are you trying to test contrasts maybe? I would be surprised if
> the function for the statistical test you are trying to use does not
> deal with that already on its own.
>
> HTH,
> Ivan
>
>
> On 27/09/2023 13:01, arnaud gaboury
I have two data.frames:
mydf1 <- structure(list(symbol = "ETHUSDT", cummulative_quote_qty =
1999.9122, side = "BUY", time = structure(1695656875.805, tzone = "", class
= c("POSIXct", "POSIXt"))), row.names = c(NA, -1L), class = c("data.table",
"data.frame"))
mydf2 <- structure(list(symbol = c("ET
I have a list of 9 lists called my.list. Each one of these 9 lists is
itself a list of 6 data.frames. Most of these data.frames have 0 rows and 0
columns.
I want to return all data.frames from the list with row numbers different
from 0.
I first created the following function:
non_empty_df <- funct
I need to download basic OHLCV (Open, High, Low, Close, Volume) in a
hourly format from various cryptocurrency exchanges.
There is the crypto package[0] but it has been removed from CRAN. Then
there is the coinmarketcapr[1] package on CRAN, but it is limited to a
paid service, coinmarketcap, which
On Thu, Dec 10, 2015 at 1:47 PM, Giorgio Garziano <
giorgio.garzi...@ericsson.com> wrote:
> my_convert <- function(col) {
> v <- grep("[0-9]{2}.[0-9]{2}.[0-9]{4}", col);
> w <- grep("[0-9]+,[0-9]+", col)
> col2 <- col
> if (length(v) == length(col)){
> col2 <- as.Date(col, format="%d.%
On Thu, Dec 10, 2015 at 12:54 PM, Duncan Murdoch
wrote:
> On 10/12/2015 6:12 AM, arnaud gaboury wrote:
>
>> Here is a sample of my data frame, obtained with read_csv2 from readr
>> package.
>>
>> myDf <- structure(list(X15 = c("30.09.2015", &q
Here is a sample of my data frame, obtained with read_csv2 from readr package.
myDf <- structure(list(X15 = c("30.09.2015", "05.10.2015", "30.09.2015",
"29.09.2015", "10.10.2015"), X16 = c("02.10.2015", "06.10.2015",
"01.10.2015", "01.10.2015", "13.10.2015"), X17 = c("Grains",
"Grains", "Grains",
I was doing some cleaning ony my linux machine and, among others, I
try to clean my R environment variables accordingly [0] and [1].
I am not really sure how to declare in a clean manner these startup variables.
Here is my setup:
1- my home folder
-$ ls ~/.config/R
env/ helper/ Renviron Rprofi
---
Sent from my phone. Please excuse my brevity.
On August 22, 2015 7:51:39 AM PDT, arnaud gaboury
wrote:
>I want to build R with an optimized BLAS library.
>My OS: Fedora 22
>Hardware: CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian
>CPU(s)
I want to build R with an optimized BLAS library.
My OS: Fedora 22
Hardware: CPU op-mode(s): 32-bit, 64-bit Byte Order: Little Endian
CPU(s): 8 Thread(s) per core: 2 Vendor ID: GenuineIntel Model name:
Intel(R) Core(TM) i7-2600K CPU @ 3.40GHz
I am a little confused when it comes to choose a method
On Sun, Aug 16, 2015, 2:29 PM Swagato Chatterjee
wrote:
Hello,
I have written a R script which runs a regression of a dataset and saves
the result in a csv file.
Now this dataset has to be edited periodically which is done in a server. I
need to run the R script in a server so that the results
adley Wickham "
> to report problems in a user-contributed package.
>
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Mon, Apr 20, 2015 at 12:10 AM, arnaud gaboury > wrote:
>
>> R 3.2.0 on Linux
>>
>&
--
extract_numeric
function (x)
{
as.numeric(gsub("[^0-9.-]+", "", as.character(x)))
}
-
Is there any particular reason for the hyphen in gsub() ? Why not
remove it thus ?
TY much Jim
>
> Jim
R 3.2.0 on Linux
library(tidyr)
playerStats <- c("LVL 10", "5,671,448 AP l6,000,000 AP", "Unique
Portals Visited 1,038",
"XM Collected 15,327,123 XM", "Hacks 14,268", "Resonators Deployed 11,126",
"Links Created 1,744", "Control Fields Created 294", "Mind Units
Ca
On Mon, Apr 20, 2015 at 9:10 AM, arnaud gaboury
wrote:
> R 3.2.0 on Linux
>
>
> library(tidyr)
>
> playerStats <- c("LVL 10", "5,671,448 AP l6,000,000 AP", "Unique
> Portals Visited 1,038",
> "XM
On Thu, Apr 16, 2015 at 3:25 PM, arnaud gaboury
wrote:
> On a Linux 64 bits, R.3.1.2, with tidyr() loaded.
>
> gabx@hortensia [R] separate(rawStats, 'toto')
> Error in dyn.load(file, DLLpath = DLLpath, ...) :
> unable to load shared object
> '/developemen
On a Linux 64 bits, R.3.1.2, with tidyr() loaded.
gabx@hortensia [R] separate(rawStats, 'toto')
Error in dyn.load(file, DLLpath = DLLpath, ...) :
unable to load shared object
'/developement/language/r/library/stringi/libs/stringi.so':
libicui18n.so.54: cannot open shared object file: No such f
On Tue, Apr 14, 2015 at 10:09 PM, John Sorkin
wrote:
> I suggest that you investigate installing RStudio server on the Linux
> Box. If you do this, you can logon to RStudio (on the Linux server), and
> it will look exactly like RStudio running on a windows box. You may need
> some help configurin
On Tue, Apr 14, 2015 at 7:09 PM, Sarah Goslee wrote:
>
> Hi Michael,
>
> On Tue, Apr 14, 2015 at 12:57 PM, Michael Haenlein
> wrote:
> > Dear all,
> >
> > I am used to running R locally on my Windows-based PC. Since some of my
> > computations are taking a lot of time I am now trying to move to a
This is the wrong part of my code.
>
>> idName=users[users$id %in% ext]
> idname
> 1: U03AEKWTL agreenmamba
> 2: U032FHV3S poisonivy
> 3: U03AEKYL4 vairis
>
Best is to use:
idNames <- users[pmatch(ext, users$id, duplicates.ok = T)]. This leave
me with an ordered and dupl
I have been searching for a while now, but can't put all pieces of the
puzzle together.
Goal : I want to replace all these kinds of patterns <@U032FHV3S> by
this <@agreenmamba>. In a more generic way, it is replacing 'id' by
user 'name'.
I have two df:
The first, 'history', is some message histor
On Thu, Feb 12, 2015 at 7:12 PM, arnaud gaboury
wrote:
> On Thu, Feb 12, 2015 at 3:40 PM, arnaud gaboury
> wrote:
>> I have two df (and dt):
>>
>> df1
>> structure(list(name = c("poisonivy", "poisonivy", "poisonivy",
>&
On Thu, Feb 12, 2015 at 3:40 PM, arnaud gaboury
wrote:
> I have two df (and dt):
>
> df1
> structure(list(name = c("poisonivy", "poisonivy", "poisonivy",
> "poisonivy", "poisonivy", "poisonivy", "poisonivy&
I have two df (and dt):
df1
structure(list(name = c("poisonivy", "poisonivy", "poisonivy",
"poisonivy", "poisonivy", "poisonivy", "poisonivy", "poisonivy",
"cruzecontrol", "agreenmamba", "agreenmamba", "vairis", "vairis",
"vairis", "vairis", "vairis", "vairis", "xaeth"), text = c("ok",
"need items
I am little lost between all the possibilities to apply a function to
a data.frame or data.table.
Here is mine:
structure(list(name = c("poisonivy", "poisonivy", "poisonivy",
"poisonivy", "poisonivy", "poisonivy", "poisonivy", "poisonivy",
"cruzecontrol", "agreenmamba", "agreenmamba", "vairis", "
gabx@hortensia [R] sessionInfo()
R version 3.1.2 (2014-10-31)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
LC_MONETARY=en_US.UTF-8
[6] LC_MESSAGES=en_US.UTF-8LC_PAPER=en_US.UTF-8 LC_NAM
On Thu, Oct 2, 2014 at 4:54 AM, Jason Eyerly wrote:
> A lot of excellent suggestions! Thank you everyone for the input. I’ve
> purchased via Amazon:
>
> "A Beginner's Guide to R" by Zuur
> "Data Manipulation with R" by Spector
> “Introductory Statistics with R.” by Peter Dalgaard
>
> Are there an
>> I got the benchmark script, which I've attached, from Texas Advanced
>> Computing Center. Here are my results (elapsed times, in secs):
Where can we get the benchmark script?
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> The best that I can see for R would be if someone were to post a "how
> to use MKL for compiling R" type document.
I build R with MKL and ICC on my Archlinux box[1][2].
If I can help in anything, I will do it.
[1]https://wiki.archlinux.org/index.php/R
[2]https://aur.archlinux.org/packages/r-mk
On Thu, Aug 28, 2014 at 11:45 AM, Martin Spindler
wrote:
> Dear all,
> I would like to us openBLAS in R under Linux / Unix.
> Which steps do I have to undertake? Does someone know a detailed
> description? (I found some sources on the web, but none was really
> helpful for me.)
> Thanks and best,
On Fri, Aug 22, 2014 at 2:03 PM, S Ellison wrote:
>> We are currently trying to migrate 3 users of "R" to a citrix based
>> environment, but are coming across major issues trying to install the
>> packages to the relevant image.
>
Please be more precised in your issue.
__
>
> R> as.vector(sapply(my.cache.list, function(x)strsplit(x, "\\.")[[1]][2]))
> [1] "subject_test" "subject_train" "y_test""y_train"
>
>
> R> gsub("df\\.(.*)\\.RData", "\\1", my.cache.list)
> [1] "subject_test" "subject_train" "y_test""y_train"
>
>
> Note that "." will match any
A directory is full of data.frames cache files. All these files have
the same pattern:
df.some_name.RData
my.cache.list <- c("df.subject_test.RData", "df.subject_train.RData",
"df.y_test.RData",
"df.y_train.RData")
I want to keep only the part inside the two points. After lots of
headache using
TY very much for your setdiffDF(). It does the job perfectly.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Petr Savicky
Sent: lundi 27 février 2012 20:41
To: r-help@r-project.org
Subject: Re: [R
vance which of my two df will have the greates dimension (I can add some
lines to deal with it, but again, seems very heavy).
I hoped I could find a better solution.
A2CT2 Ltd.
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: lundi 27 février 2012 18:42
To: A
ot;Coffee C", "GC", "Sugar No 11",
"ZS", "ZS"), Price = c(2331, 2356, 2440, 204.55, 205.45, 17792,
24.81, 1273.5, 1276.25, 2331, 2356, 2440, 2450, 204.55, 205.45,
17792, 24.81, 1273.5, 1276.25), Nbr.Lots = c(-61, -61, 6, 40,
40, -1, -1, -1, 1, -61, -
I don't understand where is your problem. Are you looking for "jaunty" as a
package on R Crans?? Are you looking to install R package on your Ubuntu box?
As you know, Jaunty is no more supported by Ubuntu, so packets have been moved
in Archives.
Arnaud Gaboury
A2CT2 Ltd.
s = "data.frame")
As you can see, they have same column names.
My idea was to merge these two df when passing as argument "not to take into
account duplicate rows", so I will get one df with rows which are not in both
df.
Is it possible? How to do it?
TY for any help.
Ar
TY Elai for your answer. One solution has been given earlier in this list by
Sarah Goslee and William Dunlap.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of ilai
Sent: vendredi 24 février 2012 20:14
could try: add the openJDK path in your
environment.
Hope this help.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of John Kane
Sent: vendredi 24 février 2012 19:32
To: r-help@r-project.org
Subject: [
B","CC","DD","DD"), y = 1:8)
>df2$y <- df2$y * mult[as.character(df2$x)]
> df2
x y
1 AA 2
2 AA 4
3 BB 15
4 BB 20
5 BB 25
6 CC 6
7 DD 14
8 DD 16
WORKING
Ty both of you and have a good weekend.
Arnaud Gaboury
A2CT2 Ltd.
-Original Mes
, 3L, -3L, -89L, -1L, -1L, -51L,
-51L)), .Names = c("Product", "reported.Price", "reported.Nbr.Lots"
), row.names = c(7L, 4L, 5L, 6L, 13L, 14L, 15L, 16L, 17L, 18L,
19L, 8L, 9L, 10L, 11L, 12L, 20L, 21L, 22L, 23L, 35L, 36L, 37L,
38L, 39L, 40L, 41L, 42L, 31L, 32L, 24L, 2
e")
Row will change. I am looking to multiply reported.Price by 100 IF Product=CL,
multiply by 10 IF product=GC, multiply by 100 IF product=HG, multiply by 1000
IF Product=NG, multiply by 100 IF product=RB.
I hope I am clear enough, and YES I have tried many workarounds myself before
pos
e longer than the chosen example shown here. It seems
your tip didn't do the job.
I am expecting this as result :
> df
x y
1 AA 10 > if df$x==AA, df$y<-1*10
2 BB 50 > if df$x==BB, df$y<-2*25
3 CC 3 NOTHING
4 AA 40> if df$x==AA, df$y<-4*10
TY Uwe,
So I will have to write a line for each condition? Right?
In fact I was trying to do something with apply in one line, but couldn't
achieve any result. In fact, all my transformation will be multiplying one
object by a specific number according to the value of df$x.
Arnaud Ga
en df$y=df$y*10
if df$x=="BB" then df$y=df$y*25
and so on with other conditions.
TY for any help.
Trading
A2CT2 Ltd.
Arnaud Gaboury
A2CT2 Ltd.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE
sum the Filled.Qty column and renamed it Nbr.Lots. This line works.
What I would like is changing the col names in the same line, thus avoiding
another line with
>colnames(exportfile)<-c("Contract","Price","Nbr.Lots")
Is there a possibility to change my
Ah, I feel stupid!
OK, it works for me.
TY for your help
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: jeudi 9 février 2012 20:02
To: Arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] loading packages in a function
TY for your answer, but here what i did :
#load needed lybrary
suppressPackageStartupMessages(library(plyr))
suppressPackageStartupMessages(library(car))
library(plyr)
library(car)
But I still get :
Loading required package: MASS
Loading required package: nnet
Arnaud Gaboury
A2CT2 Ltd
red package: MASS
Loading required package: nnet
YOU DID A GOOD JOB,SEND EMAIL
Last line is the supposed result of my function, so it ok.
How to get rid of the first two lines, only for esthetic purpose?
TY for your time.
Arnaud Gaboury
A2CT2 Ltd.
_
TY much.
Works for me.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: jeudi 9 février 2012 17:47
To: Arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] ifelse
cat() returns a null value so it's proble
ot;), cat("NO\n")) :
replacement has length zero
I have the correct answer, YES, but with an Error.
Why?
TY for any help
Arnaud Gaboury
A2CT2 Ltd.
__
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Thanks so much Peter. You are the man.
Easy way and working for me.
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: peter dalgaard [mailto:pda...@gmail.com]
Sent: mercredi 8 février 2012 16:15
To: David Reiner
Cc: Arnaud Gaboury; jim holtman; r-help@r-project.org
Subject: Re: [R
David,
You are not far indeed, as I trade commodities, and prices are the ones from
grain market: Corn, wheat and Soybeans.
They are quoted in 1/4, and my trading platform displays them in 2,4,6 and my
statements are in 25,50,75.
TY
Arnaud Gaboury
A2CT2 Ltd.
-Original Message
TY Jim,
It do the trick.
I was trying to play without success with the format() options.
No simplest way so?
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: jim holtman [mailto:jholt...@gmail.com]
Sent: mercredi 8 février 2012 15:36
To: Arnaud Gaboury
Cc: r-help@r-project.org
.
All numbers are in fact 2.1/4, 2.1/2, 2.3/4.
How can I tell R 2.2 is 2.25, 2.4 is 2.50 and 2.6 is 2.75 ?
TY for any help.
Arnaud Gaboury
A2CT2 Ltd.
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
> names1<-recode(df$names,"'BO'='BOO';'CL'='CLR';'C'='CC'")
> df1<-data.frame(names1,price)
> df1
names1 price
1BOO10
2 CC25
3CLR20
TY for the tip.
Any possibility to write this in o
I used in fact recode() from epilcac package, not the one you mentioned!
Arnaud Gaboury
A2CT2 Ltd.
-Original Message-
From: Arnaud Gaboury
Sent: mardi 7 février 2012 20:25
To: Jorge I Velez
Cc: r-help@r-project.org; Arnaud Gaboury
Subject: RE: [R] replace elements of a data frame
I
df
names price
1 BOO10
2 C25
3CL20
As you can see, "BO" has been replaced by "BOO", but with a warning!
Arnaud Gaboury
A2CT2 Ltd.
Trade: +41 22 849 88 63
Fax: +41 22 849 88 66
arnaud.gabo...@a2ct2.com
This email and any files transmitted with it are
y "BOB", "C" by "CR", "CL" by "CLO", and the list is
more long.
I can do that for each element:
>df[df=="BO"]<-"BOB"
But my df is bigger indeed with other elements.
I was thinking using replace(), but can't get a
;> 978, 0, 0, 0)), .Names = c("DESCRIPTION", "pl", "PL", "POSITION",
>> "SETTLEMENT"), row.names = c(NA, -13L), class = "data.frame")
>>
>> I am looking for one data frame with the column $PL=zz$PL+av$PL.
>> I have
Dear list,
here are my two data frames:
av <-
structure(list(DESCRIPTION = c("COFFEE C Sep/10", "COPPER Sep/10",
"CORN Dec/10", "CRUDE OIL miNY Sep/10", "GOLD Aug/10", "HENRY HUB
NATURAL GAS Sep/10",
"PALLADIUM Sep/10", "SILVER Sep/10", "SOYBEANS Nov/10", "SPCL HIGH
GRADE ZINC USD",
"SUGAR NO.11
Jorge,
Your line works and give the desired result. Now I need to be able to work
with i instead of 100419..., as I need to be able to change these numbers.
TY for your help
From: Jorge Ivan Velez [mailto:jorgeivanve...@gmail.com]
Sent: Wednesday, June 02, 2010 5:09 PM
To: arnaud Gaboury
Cc
Here we go :
dd<-data.frame(do.call(rbind,
mget(paste("DailyPL",sel[-1],sep=""),envir=.GlobalEnv)),row.names=NULL)
TY so much Joshua and Jorge.
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Wednesday, June 02, 2010 5
et(paste("DailyPL", sel[-1], sep = ""), envir =
> .GlobalEnv), :
unused argument(s) (row.names = NULL)
Why this error?
TY for your help
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Wednesday, June 02, 2010 5:08 PM
;, "100421"),
"dd" is only equal to "DailyPL100421"
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Wednesday, June 02, 2010 4:48 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re
lyPL100421)
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Wednesday, June 02, 2010 4:48 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] bind select data frames
>
> Hello,
>
> Does this do what you are look
Dear group,
Here is my environment:
> ls()
[1] "DailyPL100419" "DailyPL100420" "DailyPL100421" "dd""i"
"l" "PLglobal" "Pos100416" "Pos100419" "Pos100420"
"Pos100421" "position"
[13] "result""sel" "Trad100416""Trad100419"
"Trad10
I do really think it is a very good idea.
TY
> -Original Message-
> From: h.wick...@gmail.com [mailto:h.wick...@gmail.com] On Behalf Of
> Hadley Wickham
> Sent: Wednesday, June 02, 2010 3:31 PM
> To: arnaud Gaboury
> Cc: Peter Ehlers; r-help@r-project.org; Prof Bria
The correct line is :
> l[(which(100420==l)-1):which(100420==l)]
[1] "100419" "100420"
> -Original Message-----
> From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
> Sent: Wednesday, June 02, 2010 9:41 AM
> To: r-help@r-project.org
> Cc:
Dear group,
Here is a list of elements :
l <-
c("100415", "100416", "100419", "100420", "100421", "100422",
"100423", "100426", "100427", "100428", "100429", "100430", "100503",
"100504", "100505", "100506", "100507", "100510", "100511", "100512",
"100513", "100514", "100517", "100518", "1005
Maybe not the cleanest way, but I create a fake data frame with one row so
ddply() is happy!!
> if (nrow(futures)==0) futures<-data.frame(...)
> -Original Message-
> From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
> Sent: Tuesday, June 01, 2010 12:07 PM
> To: a
TY for the tip. The lower case is in fact the culprit.
> -Original Message-
> From: Erik Iverson [mailto:er...@ccbr.umn.edu]
> Sent: Tuesday, June 01, 2010 6:05 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] as.date
>
>
> > Whe
Dear group,
Here is my df (obtained with a read.csv2()):
df <-
structure(list(DESCRIPTION = c("COTTON NO.2 Jul/10", "COTTON NO.2 Jul/10",
"PALLADIUM Jun/10", "PALLADIUM Jun/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10"), CREATED.DATE = c("13/05/2010", "13/05/2010",
"14/05/2010", "14/05/2010
It is indeed ddply() from package plyr.
> -Original Message-
> From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
> Sent: Tuesday, June 01, 2010 12:24 PM
> To: Peter Ehlers
> Cc: arnaud Gaboury; r-help@r-project.org
> Subject: Re: [R] data frame manipulation with
to aggregate
> -Original Message-
> From: Patrick Hausmann [mailto:patrick.hausm...@uni-bremen.de]
> Sent: Tuesday, June 01, 2010 11:38 AM
> To: arnaud Gaboury
> Subject: Re: [R] data frame manipulation ddply
>
> Hi Arnaud,
>
> maybe "agg
Dear group,
Here is my data frame:
futures <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
"CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
"SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO
0 9:47 AM
> To: arnaud Gaboury
> Subject: Re: [R] data frame manipulation with zero rows
>
> On Tue, 1 Jun 2010, arnaud Gaboury wrote:
>
> > Dear group,
> >
> > Here is the kind of data.frame I obtain every day with my function :
> >
> > futures <
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN Jul/10",
"CORN Jul/10", "CORN Jul/10", "CORN Jul/10", "LIVE CATTLE Aug/10",
"LIVE CATTLE Aug/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11 Jul/10",
"SUGAR N
> From: jim holtman [mailto:jholt...@gmail.com]
> Sent: Thursday, May 27, 2010 2:34 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] switch function
>
> try this:
>
> > toBuy <- trades$Trade.Status == "DEL" & trades$Buy.Sel
, 2L, 1L), Price = c("15.2500", "368.", "368.5000"),
Net.Charges..sum. = c(4.01, -8.64, -4.32)), .Names = c("Trade.Status",
"Instrument.Long.Name", "Delivery.Prompt.Date", "Buy.Sell..Cleared.",
"Volume", "Price
..@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want to apply the switch command on every entry of the
,
2L, 1L), Price = c("15.2500", "368.", "368.5000"), Net.Charges..sum. =
c(4.01,
-8.64, -4.32)), .Names = c("Trade.Status", "Instrument.Long.Name",
"Delivery.Prompt.Date", "Buy.Sell..Cleared.", "Volume", "Price"
[1] "Buy"
$Volume
[1] "Buy"
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulat
] "Buy"
$Volume
[1] "Buy"
$Price
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change el
Dear group,
Here is my df :
trades <-
structure(list(Trade.Status = c("DEL", "INS", "INS"), Instrument.Long.Name =
c("SUGAR NO.11",
"CORN", "CORN"), Delivery.Prompt.Date = c("Jul/10", "Jul/10",
"Jul/10"), Buy.Sell..Cleared. = c("Sell", "Buy", "Buy"), Volume = c(1L,
2L, 1L), Price = c("15.2500", "
<- X
for (i in which(Y=="DEL")){
new.vect[i]<-switch(
EXPR = X[i],
Sell="Buy",
Buy="Sell",
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury
wrote:
Dear group,
Here is my df :
trade <-
structure(list(Tr
Dear group,
Here is my df :
trade <-
structure(list(Trade.Status = c("DEL", "INS", "INS"), Instrument.Long.Name =
c("SUGAR NO.11",
"CORN", "CORN"), Delivery.Prompt.Date = c("Jul/10", "Jul/10",
"Jul/10"), Buy.Sell..Cleared. = c("Sell", "Buy", "Buy"), Volume = c(1L,
2L, 1L), Price = c("15.2500",
se in fact (minimum of one element different from zero).
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 2:48 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] (no subject)
What exactly are you trying to do? If you want to know which position is
w
Oops, forgot to give a subject
> -Original Message-
> From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
> Sent: Wednesday, May 26, 2010 2:31 PM
> To: r-help@r-project.org
> Cc: 'arnaud Gaboury'
> Subject:
>
> Dear group,
>
> Here is my dat
ge if one of the element of the POSITION
column is different from zero.
I tried using mapply with some line like this :
> mapply(if,u$POSITION,==0,print("WARNING:POSITIONS ARE WRONG",quote=F))
But it seems it is not the correct way to pass the various arguments.
Any help is appreci
Dear group,
Here is my function:
#return the daily PL for day y
PLDaily<-function(x,y)
{
#find elements in my directory with "LSCPos" in the name, keep the numeric
part in the name and
#create a list
l<-gsub("\\D","",dir()[grep("LSCPos",dir())])
#select in the list the desired elements
Thank you so much. You are totally right.
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Monday, May 24, 2010 6:33 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] writing function
>
> My guess is that ei
Dear group,
Here is my environment after I run a function, myfun()
>myfun()
> ls()
[1] "allcon""avprix16" "DailyPL100416" "DailyPL100419"
"DailyPL100420" "l" "ll""myl" "PL"
"PLdaily" "PLglobal" "PLmonthly"
[13] "Pos100415" "Pos100
Do you think there is a way to add somewhere the argument row.names=NULL ? Or
should I have to write another line to remove the row.names?
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Monday, May 24, 2010 5:07 PM
> To: arnaud Gaboury
Maybe is there a neater solution, but the function mget() does the trick. So
until further advice, I will work with your solution.
Thank you
> -Original Message-
> From: Joshua Wiley [mailto:jwiley.ps...@gmail.com]
> Sent: Monday, May 24, 2010 5:07 PM
> To: arnaud Gab
[,2] [,3]
[1,] "DailyPL100416" "DailyPL100419" "DailyPL100420"
That's not what I want! I expect "DF" to be a data.frame binded by row.
I suspect there is an issue with get() or assign(), or something like that.
Any help is
Dear group,
I have a function, let's call it myfun, wich give me a list of result:
R1,R2,R3...
There is a loop in this function to get my results. Here is the structure of
my function:
Myfun<-function()
{
For (i in X ){
---instructions-
Ri
{
{
All Results (R1,R2...) are Data.
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