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 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
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
--
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
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
>>
>&
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
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", "
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
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&
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",
>&
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
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
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
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 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
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
>
> 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
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.
__
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,
> 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
>> 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?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinf
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
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
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
---
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 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
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",
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
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.%
I have a data frame called "pose":
DESCRIPTION QUANITY CLOSING.PRICE
1 WHEAT May/101467.75
2 WHEAT May/102467.75
3 WHEAT May/101467.75
4 WHEAT May/101467.75
5 COTTON NO.2 May/101 78.1
CLOSING.PRICE POSITION
1 COTTON NO.2 May/10 78.133
2 WHEAT May/10 467.754
Op is a data.frame object.The trick is done!
*******
Arnaud Gaboury
Mobile: +41 79 392 79 56
BBM: 255B488F
__
R-help@
Dear group,
Here is a data.frame, "lme":
> lme
DESCRIPTION CLOSING.PRICE POSITION
4PRIMARY NICKEL USD 04/06/1025,755.71000
5PRIMARY NICKEL USD 10/06/1025,760.86000
6 PRM HGH GD ALUMINIUM USD 09/07/10 2,415.90000
8 SPC
0
16 SPCL HIGH GRAD2,393.0
If someone has a better idea, it is welcomed.
-Original Message-
From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
Sent: Thursday, April 15, 2010 9:50 AM
To: 'r-help@r-project.org'
Subject: sum rows in a data.frame
Dear group,
Her
Dear group,
I have this following data.frame:
> c
DESCRIPTION CREATED.DATE QUANITY CLOSING.PRICE
26 PRM HGH GD ALU 2010-04-09 -12,415.9000
27 PRM HGH GD ALU 2010-04-09 12,415.9000
28 PRIMARY NICKEL 2010-03-04 1 25,755.7100
29 PRIMARY NICKEL 2010-03-05
Why don't you try something like :
Xd$x=as.date(xd$x,format="%y/%m/%d").
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Christian Raschke
> Sent: Thursday, April 15, 2010 8:28 PM
> To: r-help@r-project.org
> Subject: Re:
Dear group,
Here is my data.frame :
df <-
structure(list(DESCRIPTION = c("PRM HGH GD ALU", "PRM HGH GD ALU",
"PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL",
"STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ",
"STANDARD LEAD ", "STANDARD LEAD ", "STA
the CREATED.DATE
*******
Arnaud Gaboury
Mobile: +41 79 392 79 56
BBM: 255B488F
***
From: Ista Zahn [mailto:istaz...@gmail.com]
Sent: Friday, April 16, 2010 1:05 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation
Hi,
I'm not sure I understand
Excellent!! You saved me hours and hours of turning around and around.
TY so much.
From: Ista Zahn [mailto:istaz...@gmail.com]
Sent: Friday, April 16, 2010 1:37 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation
It works for me...
> DF <-
+ str
class = "data.frame")
Here is my code :
>opfut=ddply(futures, c("CONTRAT","SETTLEMENT"), summarise, POSITION=
sum(QUANTITY))
Here is the output:
> opfut
CONTRAT SETTLEMENT POSITION
1 SUGAR NO.11 May/10 16.54005
2
oo.com]
> Sent: Friday, April 16, 2010 4:02 PM
> To: arnaud Gaboury; r-help@r-project.org
> Subject: Re: [R] data.frame and ddply
>
> You can do something like this after the output from opfut
> opfut <- data.frame(opfut$CONTRAT,opfut$POSITION,opfut$SETTLEMENT)
> name
.", "1,386.", "1,388.", "1,389.", "1,389."
), Net.Charges..sum. = c(-4.32, -4.32, -21.6, -4.32, -4.32, -4.32,
-4.32, -4.32, -4.32, -4.32, -4.32, -12.5, -12.5, -12.5, -12.5,
-12.5, -3.3, -3.3, -3.3, -3.3, -3.3, -3.3, -3.3, -3.3)
06/102020-03-041
25,755.71
In the DESCRIPTION column, I want to get rid of the date (09/07/10 .). I
know the function substr(x, start, stop), but in my case, I need to indicate
I want to start from the end of the vector, and I have no idea how to pass
this arg
23, 2010 6:11 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] substract start from the end of the vector
>
> Hi,
>
> On Fri, Apr 23, 2010 at 11:57 AM, arnaud Gaboury
> wrote:
> > Dear group,
> >
> > Here is my df :
> >
>
o have the 2 Dataframes "PositionFut" & "PositionLme" in my
environment, as I will need them later for another function. If I do a
save(), it will be saved in my directory, but won't see it when I do ls(),
even if I do a get() before. What is the process to save thes
2010 11:44 AM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] save a data frame in environment
>
> Don't print them but return() them from your function position() within
> a list. Then you will have a list of two data.frames for later usage
> a
Dear group,
Here is my data.frame :
avprix <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN May/10",
"ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SPCL HIGH GRADE ZINC USD
Jul/10",
"STANDARD LEAD USD Jul/10"), prix = c(-1.5, -1082, 11084, 1983.5,
-2464, -118), quantity = c(0, -3, 8,
Dear group,
I need to modify some characters in a data frame. I want to use gsub and the
regex functionalities to do this.
Here is the data frame (df):
DESCRIPTIONprix quantity
1 CORN Jul/10-1.50
2 CORN May/10 -1082.0
, -1082, 11084, 1983.5,
-2464, -118), quantity = c(0, -3, 8, 2, -1, 0)), .Names = c("DESCRIPTION",
"prix", "quantity"), row.names = c(NA, -6L), class = "data.frame")
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Wednesday, April 28, 2010 1:31
TY so much david. We are getting close. But I need to keep "USD" in my
object name (i.e "STANDARD LEAD USD")
*******
Arnaud Gaboury
Mobile: +41 79 392 79 56
BBM: 255B488F
***
> -Original Message-
> From: David
Dear group,
Here is my df:
avprix <-
structure(list(DESCRIPTION = c("CORN Jul/10", "CORN May/10",
"ROBUSTA COFFEE (10) Jul/10", "SOYBEANS Jul/10", "SPCL HIGH GRADE ZINC USD",
"STANDARD LEAD USD"), prix = c(-1.5, -1082, 11084, 1983.5, -2464,
-118), quantity = c(0, -3, 8, 2, -1, 0), fees = c(-8.
I want the 0s to be at the end or beginning.
> -Original Message-
> From: Erik Iverson [mailto:er...@ccbr.umn.edu]
> Sent: Wednesday, April 28, 2010 5:21 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] data frame move columns
>
>
> >
TY david. Here we go with a correct code!
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Wednesday, April 28, 2010 5:28 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] data frame move columns
>
>
>
Dear group,
I know this issue has been already covered, and before you reply I must say
I have read the R-FAQ and search the mailing list archive.
I still can't manage to change my factor to numeric as I couldn't find any
clear answer.
Here is my df :
Pose1 <-
structure(list(DESCRIPTION = struct
TY petr, I was just trying something like that few mn ago :-)
as.numeric(gsub(",", "", S)) does exactly what I want.
> -Original Message-
> From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
> Sent: Thursday, April 29, 2010 1:28 PM
> To: arnaud Ga
Dear group,
Here is my data frame:
position100415 <-
structure(list(DESCRIPTION = structure(1:9, .Label = c(" SUGAR NO.11 Jul/10
",
" SUGAR NO.11 May/10 ", "CORN May/10 ", "COTTON NO.2 Jul/10 ",
"CRUDE OIL miNY May/10 ", "ROBUSTA COFFEE (10) Jul/10 ", "SILVER May/10 ",
"SOYBEANS Jul/10 ", "WHE
TY so much, I ignored about this argument. Adding stripe.white=T as an
argument in my read.csv does the job.
> -Original Message-
> From: Uwe Ligges [mailto:lig...@statistik.tu-dortmund.de]
> Sent: Friday, April 30, 2010 1:08 PM
> To: arnaud Gaboury
> Cc: r-hel
Dear group,
I am losing my mind with a simple question. Sorry if obvious, but I maybe
start to be confused after days and days of reading documentations.
Df :
df <-
structure(list(a = 1:3, b = 4:6, c = structure(c(1L, 1L, 1L), class =
"factor", .Label = "w")), .Names = c("a",
"b", "c"), row.na
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
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 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
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 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 =
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.
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
;> 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
quot;,
"POSITION", "prix"), row.names = c(NA, -17L), class = "data.frame")
I want to remove all the dates in $DESCRIPTION. I was thinking using
something like this:
>df3$DESCRIPTION<- gsub("here is a regex expression","",df3$DESCRIPTION).
Can&#
TY for the two answers. Both work.
> -Original Message-
> From: Phil Spector [mailto:spec...@stat.berkeley.edu]
> Sent: Tuesday, May 11, 2010 11:21 AM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] Regex and gsub
>
> df3$DESCRIPTION =
>
Dear group,
I have 3 data frames I would like to merge.
Here they are:
pose16 <-
structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L,
6L, 7L, 8L, 9L, 12L), .Label = c("COPPER May/10", "COTTON NO.2 Jul/10",
"CRUDE OIL miNY May/10", "GOLD Jun/10", "ROBUSTA COFFEE (10) Jul/10",
"SOYBEANS
In fact, that is one solution. I can then do a
ddply(x,"DESCRIPTION",summarise, pl=sum(prix),quantity=sum(POSITION)). It
does work. I was just wandering if it exists another solution.
From: jim holtman [mailto:jholt...@gmail.com]
Sent: Tuesday, May 11, 2010 2:14 PM
To: arnaud Gab
Dear group,
I have these 2 following data frame:
allcon <-
structure(list(DESCRIPTION = structure(1:17, .Label = c("COFFEE C Jul/10",
"COPPER May/10", "CORN Jul/10", "CORN May/10", "COTTON NO.2 Jul/10",
"CRUDE OIL miNY May/10", "GOLD Jun/10", "HENRY HUB NATURAL GAS May/10",
"ROBUSTA COFFEE (10
vice, I would SYSTEMATICALLY add the argument
strip.white=T when I read a csv file, so all blanks are removed.
> -Original Message-
> From: arnaud Gaboury [mailto:arnaud.gabo...@gmail.com]
> Sent: Tuesday, May 11, 2010 4:25 PM
> To: r-help@r-project.org
> Cc: 'arnaud Gaboury&
Dear group,
Here is my df :
pose16 <-
structure(list(DESCRIPTION = structure(c(1L, 2L, 3L, 4L, 5L,
6L, 7L, 8L, 9L, 12L), .Label = c("COPPER May/10", "COTTON NO.2 Jul/10",
"CRUDE OIL miNY May/10", "GOLD Jun/10", "ROBUSTA COFFEE (10) Jul/10",
"SOYBEANS Jul/10", "SUGAR NO.11 Jul/10", "SUGAR NO.11
t; > c(702.9, -244.77, -166.48, -2273.8, -24876, -2985, -80.9, 15.95,
> > -2512.5,
> > 2423)
> >
> >
> > On Wed, May 12, 2010 at 11:53 AM, arnaud Gaboury
> > wrote:
> >
> >> The result is different with your code.
> >>
> >> M
M HGH GD ALUMINIUM USD",
"SPCL HIGH GRADE ZINC USD", "STANDARD LEAD USD"), class = "factor"),
POSITION = c(2, -3, -2, -2, -18, -3, -5, 1, -5, 1), prix = c(-702.9,
244.77, 166.48, 2273.8, 24876, 2985, 80.9, -15.95, 2512.5,
-2423)), .Names = c("DES
TY. Didn't know this function.
From: Henrique Dallazuanna [mailto:www...@gmail.com]
Sent: Wednesday, May 12, 2010 3:32 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame subscription
Try this:
transform(pose16, prix = POSITION * SETTLEMENT, SETTLEMENT =
Dear group,
Here are my objects in my environment:
> ls()
[1] "Pos100415" "Pos100416" "posA" "pose15""pose16""pose16t"
"position" "trade" "x"
I need to pass the object "Pos100415" to a function. This element is a
data.frame, obtained through a function: Pos(x)<-myfun(x) with
Wonderful!!
TY so much for the help.
> -Original Message-
> From: Patrizio Frederic [mailto:frederic.patri...@gmail.com]
> Sent: Thursday, May 13, 2010 10:47 AM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] access objects in my environment
>
Dear group,
I have this command line in a function:
zz<-merge(transform(merge(value,allcon,all.y=T),SHORTDESCRIPTION=NULL,VALUE=
NULL,PL=-VALUE*pl,quantity=NULL),PosB,all.x=T,sort=F)
I want to cut it in 2 lines (just for convenient purpose). Here is what I
did :
zz<-merge(transform(merge(value
Removing the + sign resolves the problem.
TY
> -Original Message-
> From: Muhammad Rahiz [mailto:muhammad.ra...@ouce.ox.ac.uk]
> Sent: Thursday, May 13, 2010 3:15 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] long command line cut in 2 part
Dear group,
Here is an object:
z <-
c("LSCPos100415.csv", "LSCPos100416.csv", "LSCPos100419.csv",
"LSCPos100420.csv", "LSCPos100421.csv", "LSCPos100422.csv",
"LSCPos100423.csv",
"LSCPos100426.csv", "LSCPos100427.csv", "LSCPos100428.csv",
"LSCPos100429.csv",
"LSCPos100430.csv", "LSCPos100503.cs
Of course this helps!
I didn't noticed the difference between sub and gsub.
TY
> -Original Message-
> From: Bernd Weiss [mailto:bernd.we...@uni-koeln.de]
> Sent: Friday, May 14, 2010 12:23 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R
Dear group,
I am trying to write functions, but as a beginner, everything is not so
obvious.
Let's say I want the results in a list of elemts like this :
tot1, tot2, etc
Here is a function:
toto <-
function(x,y)
{
for(i in x:y){
paste(c("tot",i),collapse="")<-(i*2)
}
}
If I type this :
>to
want is 100415,100416,100419,100420.
The object "l" is a list, but can be of any other kind if it can help.
I tried something like this :
>select<-l[c(x:y)]
But it doesn't return the needed result
Any help would be appreciated
***
Arnaud Gaboury
Mobile
See ?which for more info.
>
> HTH,
> Ivan
>
> Le 5/21/2010 11:07, arnaud Gaboury a écrit :
> > Dear group,
> >
> > Here is a list :
> >
> > l<-
> > list("100415", "100416", "100419", "100420"
Dear group,
Here is my environment :
> ls()
[1] "l" "PLglobal" "Pos100415" "Pos100416" "Pos100419"
"Pos100420" "position" "select" "Trad100415" "Trad100416"
"Trad100419" "Trad100420" "trade" "y"
With objects :
> l
[1] "100415" "100416" "100419" "100420" "100421" "10
ip about "select" not to use as an object name
> -Original Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Friday, May 21, 2010 4:00 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] indexing problem
>
>
> On Ma
t]
> Sent: Friday, May 21, 2010 4:00 PM
> To: arnaud Gaboury
> Cc: r-help@r-project.org
> Subject: Re: [R] indexing problem
>
>
> On May 21, 2010, at 9:19 AM, arnaud Gaboury wrote:
>
> > Dear group,
> >
> > Here is my environment :
> >
> >
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.
[,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
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
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
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
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 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
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
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
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
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",
<- 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 :
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", "
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