[R] iplots
Hi guys, I am trying to plot a scatterplot with lines instead of with points using function iplot in library iplots. The best I can come up with is x =(1:100) y = rnorm(100) iplot(x,y,ptDiam=0) ilines(x,y) Do you guys know of any paramter to set the type in iplot or iplot.opt? Thanks, Fernando Álvarez __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Populate a matrix
This last solution is what I was looking for, I was trying to avoid loops. Thanks! -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Rainer Schuermann Sent: 5. oktober 2011 18:29 To: r-help@r-project.org Subject: Re: [R] Populate a matrix m - matrix( rep( y, length( x ) ), length( y ), length( x ) ) On Wednesday 05 October 2011 18:11:18 fernando.cabr...@nordea.com wrote: Hi guys I have vectors x - c(1,2,3,4) and y - c(4,3,9) and would like to generate a matrix which has 3 rows (length(y)) and 4 columns (length(x)), and each row is the corresponding y element repeated length(x) times. 4,4,4,4 3,3,3,3 9,9,9,9 Thanks. Fernando Álvarez __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Populate a matrix
Hi guys I have vectors x - c(1,2,3,4) and y - c(4,3,9) and would like to generate a matrix which has 3 rows (length(y)) and 4 columns (length(x)), and each row is the corresponding y element repeated length(x) times. 4,4,4,4 3,3,3,3 9,9,9,9 Thanks. Fernando Álvarez __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix/Vector manipulation
Stylish, but ifelse only includes a cumsum less or equal than v and ignores the
remainder, if v does not fit equally in say the first two weight buckets.
R - c(1.2, 1.3, 1.5)
W - c(3,2,5)
my_cumsum(4, R, W) # should take 3*1.2 + 1*1.3
[1] 4.0
sum(ifelse(cumsum(W) = 4, W, 0) * R) # ignores the 1*1.3 part because 3+2 4
[1] 3.6
Cheers,
Fer
-Original Message-
From: David Reiner [mailto:david.rei...@xrtrading.com]
Sent: 3. oktober 2011 17:57
To: Cabrera, Fernando Álvarez; r-help@r-project.org
Subject: RE: [R] Matrix/Vector manipulation
sum(ifelse(cumsum(W)=v, W, 0) * R)
HTH,
David L. Reiner
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of fernando.cabr...@nordea.com
Sent: Monday, October 03, 2011 9:50 AM
To: r-help@r-project.org
Subject: [SPAM] - [R] Matrix/Vector manipulation - Bayesian Filter detected spam
Hi guys,
Have the following problem computing vectors with pure vector algebra and end
up reverting to recursion or for-looping.
Function my_cumsum calculates a weighted average (W) of ratios (R), but only up
to the given size/volume (v). Now I recurse into the vector (from left to
right) with what you have left from the difference of volume minus current
weight, and stop when the difference is less than or equal to the current
weight.
Vectors W and R have the same length, and v is always a positive integer.
W: {w_1 w_2 .. w_m}
R: {r_1 r_2 .. r_m}
my_cumsum - function(v, R, W) {
if (v = W[1]) # check the head
v*R[1]
else
W[1]*R[1] + my_cumsum(v - W[1], R[2:length(R)], W[2:length(W)])
# recurse the tail }
Any help is greatly appreciated!
Fernando Alvarez
Great ideas originate in the muscles. ~ Thomas A. Edison
__
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and proprietary to XR Trading, LLC (XR) and/or its affiliates, and are
protected by intellectual property laws. Without the prior written consent of
XR, the XR Content may not (i) be disclosed to any third party or (ii) be
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its affiliates, on behalf of XR or its affiliates.
THE XR CONTENT IS PROVIDED AS IS, WITHOUT REPRESENTATIONS OR WARRANTIES OF ANY
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DISCLAIMS ANY AND ALL WARRANTIES, EXPRESS AND IMPLIED, RELATING TO THE XR
CONTENT, AND NEITHER XR NOR ANY OF ITS AFFILIATES SHALL IN ANY EVENT BE LIABLE
FOR ANY DAMAGES OF ANY NATURE WHATSOEVER, INCLUDING, BUT NOT LIMITED TO,
DIRECT, INDIRECT, CONSEQUENTIAL, SPECIAL AND PUNITIVE DAMAGES, LOSS OF PROFITS
AND TRADING LOSSES, RESULTING FROM ANY PERSON'S USE OR RELIANCE UPON, OR
INABILITY TO USE, ANY XR CONTENT, EVEN IF XR IS ADVISED OF THE POSSIBILITY OF
SUCH DAMAGES OR IF SUCH DAMAGES WERE FORESEEABLE.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix/Vector manipulation
Correction to my previous mail: my_cumsum(4,R,W) does not return 4.0, it
returns 4.9!
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of fernando.cabr...@nordea.com
Sent: 4. oktober 2011 08:37
To: r-help@r-project.org
Subject: Re: [R] Matrix/Vector manipulation
Stylish, but ifelse only includes a cumsum less or equal than v and ignores the
remainder, if v does not fit equally in say the first two weight buckets.
R - c(1.2, 1.3, 1.5)
W - c(3,2,5)
my_cumsum(4, R, W) # should take 3*1.2 + 1*1.3
[1] 4.0
sum(ifelse(cumsum(W) = 4, W, 0) * R) # ignores the 1*1.3 part because 3+2 4
[1] 3.6
Cheers,
Fer
-Original Message-
From: David Reiner [mailto:david.rei...@xrtrading.com]
Sent: 3. oktober 2011 17:57
To: Cabrera, Fernando Álvarez; r-help@r-project.org
Subject: RE: [R] Matrix/Vector manipulation
sum(ifelse(cumsum(W)=v, W, 0) * R)
HTH,
David L. Reiner
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of fernando.cabr...@nordea.com
Sent: Monday, October 03, 2011 9:50 AM
To: r-help@r-project.org
Subject: [SPAM] - [R] Matrix/Vector manipulation - Bayesian Filter detected spam
Hi guys,
Have the following problem computing vectors with pure vector algebra and end
up reverting to recursion or for-looping.
Function my_cumsum calculates a weighted average (W) of ratios (R), but only up
to the given size/volume (v). Now I recurse into the vector (from left to
right) with what you have left from the difference of volume minus current
weight, and stop when the difference is less than or equal to the current
weight.
Vectors W and R have the same length, and v is always a positive integer.
W: {w_1 w_2 .. w_m}
R: {r_1 r_2 .. r_m}
my_cumsum - function(v, R, W) {
if (v = W[1]) # check the head
v*R[1]
else
W[1]*R[1] + my_cumsum(v - W[1], R[2:length(R)], W[2:length(W)])
# recurse the tail }
Any help is greatly appreciated!
Fernando Alvarez
Great ideas originate in the muscles. ~ Thomas A. Edison
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
This e-mail and any materials attached hereto, including, without limitation,
all content hereof and thereof (collectively, XR Content) are confidential
and proprietary to XR Trading, LLC (XR) and/or its affiliates, and are
protected by intellectual property laws. Without the prior written consent of
XR, the XR Content may not (i) be disclosed to any third party or (ii) be
reproduced or otherwise used by anyone other than current employees of XR or
its affiliates, on behalf of XR or its affiliates.
THE XR CONTENT IS PROVIDED AS IS, WITHOUT REPRESENTATIONS OR WARRANTIES OF ANY
KIND. TO THE MAXIMUM EXTENT PERMISSIBLE UNDER APPLICABLE LAW, XR HEREBY
DISCLAIMS ANY AND ALL WARRANTIES, EXPRESS AND IMPLIED, RELATING TO THE XR
CONTENT, AND NEITHER XR NOR ANY OF ITS AFFILIATES SHALL IN ANY EVENT BE LIABLE
FOR ANY DAMAGES OF ANY NATURE WHATSOEVER, INCLUDING, BUT NOT LIMITED TO,
DIRECT, INDIRECT, CONSEQUENTIAL, SPECIAL AND PUNITIVE DAMAGES, LOSS OF PROFITS
AND TRADING LOSSES, RESULTING FROM ANY PERSON'S USE OR RELIANCE UPON, OR
INABILITY TO USE, ANY XR CONTENT, EVEN IF XR IS ADVISED OF THE POSSIBILITY OF
SUCH DAMAGES OR IF SUCH DAMAGES WERE FORESEEABLE.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Matrix/Vector manipulation
Hi guys,
Have the following problem computing vectors with pure vector algebra and end
up reverting to recursion or for-looping.
Function my_cumsum calculates a weighted average (W) of ratios (R), but only up
to the given size/volume (v). Now I recurse into the vector (from left to
right) with what you have left from the difference of volume minus current
weight, and stop when the difference is less than or equal to the current
weight.
Vectors W and R have the same length, and v is always a positive integer.
W: {w_1 w_2 .. w_m}
R: {r_1 r_2 .. r_m}
my_cumsum - function(v, R, W) {
if (v = W[1]) # check the head
v*R[1]
else
W[1]*R[1] + my_cumsum(v - W[1], R[2:length(R)], W[2:length(W)])
# recurse the tail
}
Any help is greatly appreciated!
Fernando Alvarez
Great ideas originate in the muscles. ~ Thomas A. Edison
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Matrix and list indices
Thanks Michael it works!
Have to say it is amazing what you can do in R with a few lines (a line in this
case) of code.
Fernando
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: 27. september 2011 15:43
To: Cabrera, Fernando Álvarez
Subject: Re: [R] Matrix and list indices
Untested, I believe this should work, though you might need to modify for
floating point funny business in testing the equalities:
my_list - list( earth=array(c(0,0,45,0,0,45,0,45),dim=c(2,2,2)),
mars=array(c(8:1),dim=c(2,2,2)))
my_list$earth[my_list$earth==0] - my_list$mars[my_list$earth==0]
Hope this helps,
Michael Weylandt
On Tue, Sep 27, 2011 at 8:49 AM,
fernando.cabr...@nordea.commailto:fernando.cabr...@nordea.com wrote:
Hi guys,
I am trying to replace all elements of earth that are equal to zero with their
corresponding elements in mars. I can do the replace with a bunch of for-loops,
but I don't think this is the R way of doing things.
my_list - list( earth=array(c(0,0,45,0,0,45,0,45),dim=c(2,2,2)),
mars=array(c(8:1),dim=c(2,2,2)))
my_list
for (i in c(1:2)) {
for (j in c(1:2)) {
for (k in c(1:2)) {
if (my_list$earth[i,j,k] == 0) {
my_list$earth[i,j,k] - my_list$mars[i,j,k]
}
}
}
}
my_list
Do you guys have any suggestions for getting rid of the ugly for-loops?
Many thanks,
Fernando Álvarez
Nordea e-Markets
Strandgade 3
PO Box 850
DK-0900 Copenhagen C
Denmark
Tel.: +45 33 33 32 67tel:%2B45%2033%2033%2032%2067
Mobile: +45 61 55 27 54tel:%2B45%2061%2055%2027%2054
This transmission is intended solely for the person or entity to whom it is
addressed. It may contain privileged and confidential information. If you are
not the intended recipient, please be notified that any dissemination,
distribution or copying is strictly prohibited. If you have received this
transmission by mistake, please let us know and then delete it from your system.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Matrix and list indices
Hi guys,
I am trying to replace all elements of earth that are equal to zero with their
corresponding elements in mars. I can do the replace with a bunch of for-loops,
but I don't think this is the R way of doing things.
my_list - list( earth=array(c(0,0,45,0,0,45,0,45),dim=c(2,2,2)),
mars=array(c(8:1),dim=c(2,2,2)))
my_list
for (i in c(1:2)) {
for (j in c(1:2)) {
for (k in c(1:2)) {
if (my_list$earth[i,j,k] == 0) {
my_list$earth[i,j,k] - my_list$mars[i,j,k]
}
}
}
}
my_list
Do you guys have any suggestions for getting rid of the ugly for-loops?
Many thanks,
Fernando Álvarez
Nordea e-Markets
Strandgade 3
PO Box 850
DK-0900 Copenhagen C
Denmark
Tel.: +45 33 33 32 67
Mobile: +45 61 55 27 54
This transmission is intended solely for the person or entity to whom it is
addressed. It may contain privileged and confidential information. If you are
not the intended recipient, please be notified that any dissemination,
distribution or copying is strictly prohibited. If you have received this
transmission by mistake, please let us know and then delete it from your system.
P Please consider the impact on the environment before printing this e-mail.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
