Re: [R] Removing display of R row names from list.
Beats me. You can print a single data frame with
print(dat1, rownames = FALSE) but it is not clear to me how to do it within a
function.
I am sure someone who actually know what they are doing will be along in a
moment. Sorry not to have been of more help.
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 18:11:36 +0300
To: jrkrid...@inbox.com
Subject: Re: [R] Removing display of R row names from list.
Here is a small example. it is not the real data I am working on but this can
help
#Chick weight example
names(ChickWeight) - tolower(names(ChickWeight))
chick_m - melt(ChickWeight, id=2:4, na.rm=TRUE)
class(chick_m)
test = function(data){
i = 1
table = list()
table [[i]] - dcast(chick_m, time ~ variable, mean)
i = i + 1
print(table)
}
test(chick_m )
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[https://sites.google.com/a/aims.ac.za/fredo/]
On Wed, Jul 29, 2015 at 5:12 PM, John Kane jrkrid...@inbox.com wrote:
HI Frederic,
Can you supply a small example of the problem?
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 14:15:58 +0300
To: r-help@r-project.org
Subject: [R] Removing display of R row names from list.
Dear All,
Is there a way to not include the row names when creating the list?
Thanks!
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[https://sites.google.com/a/aims.ac.za/fredo/]
[[alternative HTML version deleted]]
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[https://stat.ethz.ch/mailman/listinfo/r-help]
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[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] Imbalanced random forest
This might help: http://bit.ly/1MUP0Lj On Wed, Jul 29, 2015 at 11:00 AM, jpara3 j.para.fernan...@hotmail.com wrote: ¿How can i set up a study with random forest where the response is highly imbalanced? - Guided Tours Basque Country Guided tours in the three capitals of the Basque Country: Bilbao, Vitoria-Gasteiz and San Sebastian, as well as in their provinces. Available languages. Travel planners for groups and design of tourist routes across the Basque Country. -- View this message in context: http://r.789695.n4.nabble.com/Imbalanced-random-forest-tp4710524.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] indices of mismatch element in two vector with missing values
Thank you everyone for taking the time to reply. Thanks Martin and Hervé for your solutions. Now I just need to remember to check for NA match. -- View this message in context: http://r.789695.n4.nabble.com/indices-of-mismatch-element-in-two-vector-with-missing-values-tp4710497p4710518.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Removing display of R row names from list.
1. Please post in plain text, not HTML.
2. Please do your homework. Data frames (the class of the result of
dcast) **always** have row names. ?data.frame for details.
3. Read up on S3 methods if you have not done so already (any good R
tutorial will tell you about them, including the Intro to R tutorial
that ships with R). Then see the row.names argument of
?print.data.frame .
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Wed, Jul 29, 2015 at 8:11 AM, Frederic Ntirenganya ntfr...@gmail.com wrote:
Here is a small example. it is not the real data I am working on but this
can help
#Chick weight example
names(ChickWeight) - tolower(names(ChickWeight))
chick_m - melt(ChickWeight, id=2:4, na.rm=TRUE)
class(chick_m)
test = function(data){
i = 1
table = list()
table [[i]] - dcast(chick_m, time ~ variable, mean)
i = i + 1
print(table)
}
test(chick_m )
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
On Wed, Jul 29, 2015 at 5:12 PM, John Kane jrkrid...@inbox.com wrote:
HI Frederic,
Can you supply a small example of the problem?
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 14:15:58 +0300
To: r-help@r-project.org
Subject: [R] Removing display of R row names from list.
Dear All,
Is there a way to not include the row names when creating the list?
Thanks!
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[[alternative HTML version deleted]]
__
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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.
FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop!
Check it out at http://www.inbox.com/earth
[[alternative HTML version deleted]]
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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-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] How to simulate informative censoring in a Cox PH model?
As models become more complex it becomes harder to distinguish
different parts and their effects. Even for a straight forward linear
regression model if X1 and X2 are correlated with each other then it
becomes difficult to distinguish between the effects of X1^2, X2^2,
and X1*X2. In your case the informative censoring and model
misspecification are becoming hard to distinguish (and it could be
argued that having informative censoring is really just a form of
model misspecification). So I don't think so much that you are doing
things wrong, just that you are learning that the models are complex.
Another approach to simulation that you could try is to simulate the
event time and censoring time using copulas (and therefore they can be
correlated to give informative censoring, but without relying on a
term that you could have included in the model) then consider the
event censored if the censoring time is shorter.
On Fri, Jul 24, 2015 at 10:14 AM, Daniel Meddings dpmeddi...@gmail.com wrote:
Hi Greg
Many thanks for taking the time to respond to my query. You are right about
pointing out the distinction between what variables are and are not included
in the event times process and in the censoring process. I clearly forgot
this important aspect. I amended my code to do as you advise and now I am
indeed getting biased estimates when using the informatively censored
responses. The problem is now that the estimates from the independently
censored responses are the same - i.e. they are just as biased. Thus the
bias seems to be due entirely to model mis-specification and not the
informative censoring.
To give a concrete example I simulate event times T_i and censoring times
C_i from the following models;
T_i~ Weibull(lambda_t(x),v_t),lambda_t(x)=lambda_t*exp( beta_t_0 +
(beta_t_1*Treat_i) + (beta_t_2*Z_i) + (beta_t_3*Treat_i*Z_i) )
C_i~ Weibull(lambda_c(x),v_c),lambda_c(x)=lambda_c*exp( beta_c_0 +
(beta_c_1*Treat_i) + (beta_c_2*Z_i) + (beta_c_3*Treat_i*Z_i) )
D_i~Weibull(lambda_d(x),v_D), lambda_d(x)=lamda_d*exp( beta_d_0)
where ;
beta_t_0 = 1, beta_t_1 = -1, beta_t_2 = 1, beta_t_3 = -2, lambda_t=0.5
beta_c_0 = 0.2, beta_c_1 = -2, beta_c_2 = 2, beta_c_3 = -2,
lambda_c=0.5
beta_d_0 = -0.7, lambda_d=0.1
When I fit the cox model to both the informatively censored responses and
the independent censored responses I include only the Treatment covariate in
the model.
I simulate Treatment from a Bernoulli distribution with p=0.5 and Z_i from a
beta distribution so that Z ranges from 0 to 1 (I like to think of Z as a
poor prognosis probability where Z_i=1 means a subject is 100% certain to
have a poor prognosis and Z_i=0 means zero chance). These parameter choices
give approximately 27% and 25% censoring for the informatively censored
responses (using C_i) and the independent censored responses (using D_i)
respectively. I use N=2000 subjects and 2000 simulation replications.
The above simulation I get estimates of beta_t_2 of -1.526 and -1.537 for
the informatively censored responses and the independent censored responses
respectively.
Furthermore when I fit a cox model to the full responses (no censoring at
all) I get an estimate of beta_t_2 of -1.542. This represents the best that
can possibly be done given that Z and Treat*Z are not in the model. Clearly
censoring is not making much of a difference here - model mis-specification
dominates.
I still must be doing something wrong but I cannot figure this one out.
Thanks
Dan
On Thu, Jul 23, 2015 at 12:33 AM, Greg Snow 538...@gmail.com wrote:
I think that the Cox model still works well when the only information
in the censoring is conditional on variables in the model. What you
describe could be called non-informative conditional on x.
To really see the difference you need informative censoring that
depends on something not included in the model. One option would be
to use copulas to generate dependent data and then transform the
values using your Weibul. Or you could generate your event times and
censoring times based on x1 and x2, but then only include x1 in the
model.
On Wed, Jul 22, 2015 at 2:20 AM, Daniel Meddings dpmeddi...@gmail.com
wrote:
I wish to simulate event times where the censoring is informative, and
to
compare parameter estimator quality from a Cox PH model with estimates
obtained from event times generated with non-informative censoring.
However
I am struggling to do this, and I conclude rather than a technical flaw
in
my code I instead do not understand what is meant by informative and
un-informative censoring.
My approach is to simulate an event time T dependent on a vector of
covariates x having hazard function h(t|x)=lambda*exp(beta'*x)v*t^{v-1}.
This corresponds to T~ Weibull(lambda(x),v), where the scale parameter
lambda(x)=lambda*exp(beta'*x) depends on x and the shape parameter v is
fixed. I have N subjects where
[R] PythonInR. Python script in R with parameters required. Download satellite images from NASA
Hello, I'm trying to execute a python script within R (3.2.1 x 64) with the PythonInR package. I would like to download an order of satellite images from Nasa using a python script ( http://landsat.usgs.gov/documents/espa_bulk_downloader_v1.0.0.zip) but I have no success. I first run the pyExecfile command with the *feedparser.py* script and then the *download_espa_order.py* giving the required parameters (my mail acount and the order number), here is the code: setwd(C:/Python27) install.packages(PythonInR) library(PythonInR) pyConnect(pythonExePath=C:/Python27/python.exe) pyIsConnected() # autodetectPython(C:/Python27/python.exe) pyExecfile(C:/Landsat/feedparser.py) pyExecfile(C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com -o magifranqu...@gmail.com-07222015-120911 -d C:/Landsat/ESPA) and I get this error: Error: unexpected string constant in pyExecfile(C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com The code C:/Landsat/download_espa_order.py -e magifranqu...@gmail.com -o magifranqu...@gmail.com-07222015-120911 -d C:/Landsat/ESPA runs ok when I use it within system console. I appreciate if someone could help me to solve this problem. Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Removing display of R row names from list.
Here is a small example. it is not the real data I am working on but this
can help
#Chick weight example
names(ChickWeight) - tolower(names(ChickWeight))
chick_m - melt(ChickWeight, id=2:4, na.rm=TRUE)
class(chick_m)
test = function(data){
i = 1
table = list()
table [[i]] - dcast(chick_m, time ~ variable, mean)
i = i + 1
print(table)
}
test(chick_m )
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
On Wed, Jul 29, 2015 at 5:12 PM, John Kane jrkrid...@inbox.com wrote:
HI Frederic,
Can you supply a small example of the problem?
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 14:15:58 +0300
To: r-help@r-project.org
Subject: [R] Removing display of R row names from list.
Dear All,
Is there a way to not include the row names when creating the list?
Thanks!
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop!
Check it out at http://www.inbox.com/earth
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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] vectorized sub, gsub, grep, etc.
There is confusion here. apply() family functions are **NOT**
vectorization -- they ARE loops (at the interpreter level), just done
in functionalized form. Please read background material (John
Chambers's books, MASS, or numerous others) to improve your
understanding and avoid posting erroneous comments.
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Tue, Jul 28, 2015 at 3:00 PM, John Thaden jjtha...@flash.net wrote:
Adam,The method you propose gives a different result than the prior
methods for these example vectors
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
Old method 1
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
gives
b cda
aB CD ef
Old method 2
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
sub2(patt, repl, X)
gives
[1] aB CD ef
Your method (I gave it the unique name sub3)
sub3 - function(pattern, replacement, x) { len- length(x) y -
character(length=len) patlen - length(pattern) replen -
length(replacement) if(patlen != replen) stop('Error: Pattern and
replacement length do not match') for(i in 1:replen) {
y[which(x==pattern[i])] - replacement[i] } return(y)}sub3(patt, repl, X)
gives[1]CD
Granted, whatever it does, it does it faster
#Old method 1
system.time(for(i in 1:5)
mapply(function(p,r,x) sub(p,r,x, fixed = TRUE),p=patt,r=repl,x=X))
user system elapsed
2.530.002.52
#Old method 2
system.time(for(i in 1:5)sub2(patt, repl, X)) user system elapsed
2.320.002.32
#Your proposed method
system.time(for(i in 1:5) sub3(patt, repl, X))
user system elapsed
1.020.001.01
but would it still be faster if it actually solved the same problem?
-John Thaden
On Monday, July 27, 2015 11:40 PM, Adam Erickson
adam.michael.erick...@gmail.com wrote:
I know this is an old thread, but I wrote a simple FOR loop with vectorized
pattern replacement that is much faster than either of those (it can also
accept outputs differing in length from the patterns):
sub2 - function(pattern, replacement, x) { len - length(x)y
- character(length=len)patlen - length(pattern)replen -
length(replacement)if(patlen != replen) stop('Error: Pattern and
replacement length do not match')for(i in 1:replen) {
y[which(x==pattern[i])] - replacement[i]}return(y) }
system.time(test - sub2(patt, repl, XX)) user system elapsed 0
0 0
Cheers,
Adam
On Wednesday, October 8, 2008 at 9:38:01 PM UTC-7, john wrote:
Hello Christos,
To my surprise, vectorization actually hurt processing speed!#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
system.time( for(i in 1:1) sub2(patt, repl, X) )
user system elapsed
1.180.071.26 system.time( for(i in 1:1) mapply(function(p,
r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X) )
user system elapsed
1.420.051.47
So much for avoiding loops.
John Thaden=== At 2008-10-07, 14:58:10 Christos wrote: ===John,
Try the following:
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
b cda
aB CD ef
-Christos -My Original Message-
R pattern-matching and replacement functions are
vectorized: they can operate on vectors of targets.
However, they can only use one pattern and replacement.
Here is code to apply a different pattern and replacement for
every target. My question: can it be done better?
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
sub2(patt, repl, X)
-John__
r-h...@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
Re: [R] Removing display of R row names from list.
Is this what you mean?
z - data.frame(a=1:3,b=letters[1:3],row.names=letters[1:3])
zlist - list(one=z,too =z)
for(i in 1:2){print(zlist[[i]],row.names=FALSE); cat(\n)}
a b
1 a
2 b
3 c
a b
1 a
2 b
3 c
... which could obviously be within a function.
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Wed, Jul 29, 2015 at 8:32 AM, John Kane jrkrid...@inbox.com wrote:
Beats me. You can print a single data frame with
print(dat1, rownames = FALSE) but it is not clear to me how to do it within a
function.
I am sure someone who actually know what they are doing will be along in a
moment. Sorry not to have been of more help.
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 18:11:36 +0300
To: jrkrid...@inbox.com
Subject: Re: [R] Removing display of R row names from list.
Here is a small example. it is not the real data I am working on but this can
help
#Chick weight example
names(ChickWeight) - tolower(names(ChickWeight))
chick_m - melt(ChickWeight, id=2:4, na.rm=TRUE)
class(chick_m)
test = function(data){
i = 1
table = list()
table [[i]] - dcast(chick_m, time ~ variable, mean)
i = i + 1
print(table)
}
test(chick_m )
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[https://sites.google.com/a/aims.ac.za/fredo/]
On Wed, Jul 29, 2015 at 5:12 PM, John Kane jrkrid...@inbox.com wrote:
HI Frederic,
Can you supply a small example of the problem?
John Kane
Kingston ON Canada
-Original Message-
From: ntfr...@gmail.com
Sent: Wed, 29 Jul 2015 14:15:58 +0300
To: r-help@r-project.org
Subject: [R] Removing display of R row names from list.
Dear All,
Is there a way to not include the row names when creating the list?
Thanks!
Regards,
Frederic.
Frederic Ntirenganya
Maseno University,
African Maths Initiative,
Kenya.
Mobile:(+254)718492836
Email: fr...@aims.ac.za
https://sites.google.com/a/aims.ac.za/fredo/
[https://sites.google.com/a/aims.ac.za/fredo/]
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.ethz.ch/mailman/listinfo/r-help
[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop!
Check it out at http://www.inbox.com/earth [http://www.inbox.com/earth]
FREE ONLINE PHOTOSHARING - Share your photos online with your friends and
family!
Visit http://www.inbox.com/photosharing to find out more!
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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 -- To UNSUBSCRIBE and more, see
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] Opposite color in R
Hi,
Nope. My point is that color pairs red-green, yellow-violet and blue-orange are
physically on the opposite sides of the wheel.
I'm not 100% satisfied with this circle, because three of greens and reds are
so near each other. But this is the best solution thus far :-)
The number of colors, 24, is also fixed in the application.
Atte
Hi
I plotted the 'spectrum' and it looked a little small - spectrum colours: red
orange yellow green blue indigo violet.
I suppose you could go to infinite lengths to split it up but is this an
improvement?
I have not gone into the depths of complimentary colours
library(colorspace)
ColorsRYB=rbind(colorRamp(c(purple,violet))((0:4)/4)[1:4,],
colorRamp(c(violet,blue))((0:4)/4)[1:4,],
colorRamp(c(blue,green))((0:4)/4)[1:4,],
colorRamp(c(green,yellow))((0:4)/4)[1:4,],
colorRamp(c(yellow,orange))((0:4)/4)[1:4,],
colorRamp(c(orange,red))((0:4)/4)[1:4,],
colorRamp(c(red,purple))((0:4)/4)[1:4,])
LenCol=length(ColorsRYB[,1])
ColorsRYBhex=rep(0, LenCol)
for(i in 1: LenCol)
{
ColorsRYBhex[i]=rgb(ColorsRYB[i,1]/255,ColorsRYB[i,2]/255,ColorsRYB[i,3]/255)
}
pie(rep(1, LenCol), col = ColorsRYBhex)
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home:mackay at northnet.com.au
https://stat.ethz.ch/mailman/listinfo/r-help
-Original Message-
From: R-help [mailto:r-help-bounces at r-project.org
https://stat.ethz.ch/mailman/listinfo/r-help] On Behalf Of Atte Tenkanen
Sent: Tuesday, 28 July 2015 18:22
To:r-help at r-project.org https://stat.ethz.ch/mailman/listinfo/r-help
Subject: [R] Opposite color in R
It seems that there is no implementation for the traditional artist's
color circle in R. However I'm searching for such a wheel, because my
program needs it.
As said, the description of complementary/opposite-function in package
colortools is misleading since, for example
opposite(green) produces violet, not red, but the description of
complementary-function says
Complementary or opposite color scheme is formed by colors that are
opposite each other on the color wheel (example: red and green).
So, there must be just a lapse in the text.
I constrained such kind of a color wheel, which is enough near of what
I need:
library(colorspace)
ColorsRYB=rbind(colorRamp(c(red,
violet))((0:4)/4)[1:4,],colorRamp(c(violet,
blue))((0:4)/4)[1:4,],colorRamp(c(blue,
green))((0:4)/4)[1:4,],colorRamp(c(green,
yellow))((0:4)/4)[1:4,],colorRamp(c(yellow,
orange))((0:4)/4)[1:4,],colorRamp(c(orange, red))((0:4)/4)[1:4,])
LenCol=length(ColorsRYB[,1])
ColorsRYBhex=rep(0, LenCol)
for(i in 1: LenCol)
{
ColorsRYBhex[i]=rgb(ColorsRYB[i,1]/255,ColorsRYB[i,2]/255,ColorsRYB[i,3]/255)
}
pie(rep(1, 24), col = ColorsRYBhex)
Atte T.
28.7.2015, 2.23, Steve Taylor kirjoitti:
/ I wonder if the hcl colour space is useful? Varying hue while keeping
chroma and luminosity constant should give varying colours of perceptually
the same colourness and brightness.
//
// ?hcl
// pie(rep(1,12),col=hcl((1:12)*30,c=70),border=NA)
//
//
// -Original Message-
// From: R-help [mailto:r-help-bounces at r-project.org
https://stat.ethz.ch/mailman/listinfo/r-help] On Behalf Of Atte Tenkanen
// Sent: Sunday, 26 July 2015 7:50a
// To:r-help at r-project.org
https://stat.ethz.ch/mailman/listinfo/r-help
// Subject: [R] Opposite color in R
//
// Hi,
//
// I have tried to find a way to find opposite or complementary colors in R.
//
// I would like to form a color circle with R like this one:
// http://nobetty.net/dandls/colorwheel/complementary_colors.jpg
//
// If you just make a basic color wheel in R, the colors do not form
// complementary color circle:
//
// palette(rainbow(24))
// Colors=palette()
// pie(rep(1, 24), col = Colors)
//
// There is a package ”colortools” where you can find function opposite(),
// but it doesn’t work as is said. I tried
//
// library(colortools)
// opposite(violet) and got green instead of yellow and
//
// opposite(blue) and got yellow instead of orange.
//
// Do you know any solutions?
//
// Atte Tenkanen
/
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Re: [R] Daily Category Revenue-Stacked Bar Chart in ggplot2
No, I'm not confused. I just posted on R help website. I don't know how to use the email client to do anything which you are speaking of. If you would like the posts to be made in an alternative way, then you will need to provide clear direction. Sending someone to a help page which doesn't reference what you are speaking about isn't actually helpful. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
Alright, I think I understand what you guys are talking about. It is still not clear the relationship between R-help and Nabble since you guys haven't actually answered that. Again, the best way to provide proper advice is to actually quote to portion of the website you are sending people to, i.e., To post a message to all the list members, send email to r-help@r-project.org. But this still doesn't address the issue of why that is necessary in the first place. Indeed, when a post is made the user sees the following message, which seems to indicate that the method of posting doesn't actually matter: This forum is an archive/gateway which will forward your post to the r-help@r-project.org mailing list. Cheers! -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710549.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Rearrange Data Frame
My question is about how to select and rearrange the data to a new data frame Here is an example: Samples counts time A 10 3 A 12 4 A 11 3 B 12 4 B 10 5 C 11 2 C 13 3 Say, if I want to make a new table that only look at “counts” as below: AB C 10 12 11 12 10 13 11 How can I do this in R? Thank you! . __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Mixed Date Formats
On Wed, Jul 29, 2015 at 2:45 PM, farnoosh sheikhi via R-help r-help@r-project.org wrote: Hi Arun, Hope all is well with you. I have a data with a column for date.The date format is mixed. There are date values with Month/Day/Year format and values with Day/Month/Year format.I don't know how to unify it.I really appreciate your help.Thanks. You sent this to the R-help list, not just to Arun, so I'm assuming this is an R question. The best way to get help is to provide a sample of your data using dput() and to clearly specify what you would like as the result - unify is a bit vague. paste(x, collapse=) could be considered unification, after all. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Mixed Date Formats
Hi Arun, Hope all is well with you. I have a data with a column for date.The date format is mixed. There are date values with Month/Day/Year format and values with Day/Month/Year format.I don't know how to unify it.I really appreciate your help.Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Mixed Date Formats
I'm assuming you actually want the date column to be character, not factor: SampleData - structure(list(id = 1:7, value = c(5813L, 8706L, 4049L, 5877L, 1375L, 2223L, 3423L), date = c(19-Dec-11, 07-Dec-11, 06/05/11, 05/12/11, 31/12/2011, 10/19/2011, 01/22/2011)), .Names = c(id, value, date), row.names = c(NA, -7L), class = data.frame) id value date 1 1 5813 19-Dec-11 2 2 8706 07-Dec-11 3 3 4049 06/05/11 4 4 5877 05/12/11 5 5 1375 31/12/2011 6 6 2223 10/19/2011 7 7 3423 01/22/2011 Given this assemblage of dates, there is a lot of ambiguity. The first two are clear, but is the third in May or June? Is the fourth May or December? I had hoped that the number of digits in the year provided a clue, but #5 is clearly dd/mm/ while #7 is mm/dd/ Or actually, #3 could be yy/mm/dd too. I don't see a way to programmatically solve your problem, because there is no clear way to parse all of these dates because of the ambiguity. You could get some of them by checking for values outside the bounds of legitimate month numbers, or just try it and discard the values that give NA results: as.Date(01/22/2011, %m/%d/%Y) [1] 2011-01-22 as.Date(01/22/2011, %d/%m/%Y) [1] NA But you need more information to figure out the rest. Sarah On Wed, Jul 29, 2015 at 5:15 PM, farnoosh sheikhi farnoosh...@yahoo.com wrote: Hi Sarah, Thanks for getting back to me. Here is an example of my data: SampleData - structure(list(id = 1:7, value = c(5813L, 8706L, 4049L, 5877L, 1375L, 2223L, 3423L), date = structure(c(4L, 3L, 2L, 1L, 7L, 6L, 5L), .Label = c(05/12/11, 06/05/11, 07-Dec-11, 19-Dec-11, 01/22/2011, 10/19/2011, 31/12/2011 ), class = factor)), .Names = c(id, value, date), row.names = c(NA, -7L), class = data.frame) SampleData Thanks for your help:). On Wednesday, July 29, 2015 1:50 PM, Sarah Goslee sarah.gos...@gmail.com wrote: On Wed, Jul 29, 2015 at 2:45 PM, farnoosh sheikhi via R-help r-help@r-project.org wrote: Hi Arun, Hope all is well with you. I have a data with a column for date.The date format is mixed. There are date values with Month/Day/Year format and values with Day/Month/Year format.I don't know how to unify it.I really appreciate your help.Thanks. You sent this to the R-help list, not just to Arun, so I'm assuming this is an R question. The best way to get help is to provide a sample of your data using dput() and to clearly specify what you would like as the result - unify is a bit vague. paste(x, collapse=) could be considered unification, after all. Sarah __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Mixed Date Formats
Hi Sarah, Thanks for getting back to me.Here is an example of my data:SampleData - structure(list(id = 1:7, value = c(5813L, 8706L, 4049L, 5877L, 1375L, 2223L, 3423L), date = structure(c(4L, 3L, 2L, 1L, 7L, 6L, 5L), .Label = c(05/12/11, 06/05/11, 07-Dec-11, 19-Dec-11, 01/22/2011, 10/19/2011, 31/12/2011 ), class = factor)), .Names = c(id, value, date), row.names = c(NA, -7L), class = data.frame)SampleData Thanks for your help:). On Wednesday, July 29, 2015 1:50 PM, Sarah Goslee sarah.gos...@gmail.com wrote: On Wed, Jul 29, 2015 at 2:45 PM, farnoosh sheikhi via R-help r-help@r-project.org wrote: Hi Arun, Hope all is well with you. I have a data with a column for date.The date format is mixed. There are date values with Month/Day/Year format and values with Day/Month/Year format.I don't know how to unify it.I really appreciate your help.Thanks. You sent this to the R-help list, not just to Arun, so I'm assuming this is an R question. The best way to get help is to provide a sample of your data using dput() and to clearly specify what you would like as the result - unify is a bit vague. paste(x, collapse=) could be considered unification, after all. Sarah -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Rearrange Data Frame
Hi Stella, On Wed, Jul 29, 2015 at 1:14 PM, Stella Xu stella...@hli.ubc.ca wrote: My question is about how to select and rearrange the data to a new data frame Here is an example: Samples counts time A 10 3 A 12 4 A 11 3 B 12 4 B 10 5 C 11 2 C 13 3 Say, if I want to make a new table that only look at “counts” as below: AB C 10 12 11 12 10 13 11 How can I do this in R? Thank you! Your example data doesn't use time at all, and contains a duplicate pair of A,?,3 - what do you want to have happen there? How should duplicates be handled? How should the ordering of values work? If instead that should be a 5, here's something that is almost what you want (but I find more useful): x - structure(list(Samples = c(A, A, A, B, B, C, C), counts = c(10L, 12L, 11L, 12L, 10L, 11L, 13L), time = c(3L, 4L, 5L, 4L, 5L, 2L, 3L)), .Names = c(Samples, counts, time), class = data.frame, row.names = c(NA, -7L)) library(reshape2) dcast(x, time ~ Samples, value.var=counts, sum) time A B C 12 0 0 11 23 10 0 13 34 12 12 0 45 11 10 0 If you want the results scooted up, I think there was recently a discussion on this list on doing so. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
On Jul 29, 2015, at 1:02 PM, Hidden Markov Model wrote: Alright, I think I understand what you guys are talking about. It is still not clear the relationship between R-help and Nabble since you guys haven't actually answered that. Nabble's relationship to Rhelp? Nabble is an ongoing annoyance to the regular users and to the list moderators. It's web display format deceives naive users (who usually fail to read the Posting Guide as they were directed) into thinking everyone sees what they see, and so they often fail to maintain the context of the discussion. Many regular contributors simply ignore the content from Nabble-originated postings. Again, the best way to provide proper advice is to actually quote to portion of the website you are sending people to, i.e., To post a message to all the list members, send email to r-help@r-project.org. But this still doesn't address the issue of why that is necessary in the first place. Indeed, when a post is made the user sees the following message, which seems to indicate that the method of posting doesn't actually matter: This forum is an archive/gateway which will forward your post to the r-help@r-project.org mailing list. Nabble lies. It is not an archive. I periodically get notices telling me that my old posts are going to be dropped. The real archive is: https://stat.ethz.ch/pipermail/r-help/ Nabble does forward messages to Rhelp and if a message passes the spam filters and the human moderators, it gets posted. Nabble then strips off the footers from the returned messages. If you want to complain about Nabble not living up to your expectations, you should contact them. We did not ask them to mirror Rhelp. You are expected to know how to control your own mail-client. Expecting us to educate you on the basics of computer use is unreasonable. View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710549.html R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] vectorized sub, gsub, grep, etc.
Further refining the vectorized (within a loop) exact string match
function, I get times below 0.9 seconds while maintaining error checking.
This is accomplished by removing which() and replacing 1:length() with
seq_along().
sub2 - function(pattern, replacement, x) {
len- length(x)
y - character(length=len)
patlen - length(pattern)
replen - length(replacement)
if(patlen != replen) stop('Error: Pattern and replacement length do not
match')
for(i in seq_along(pattern)) {
y[x==pattern[i]] - replacement[i]
}
return(y)
}
system.time(for(i in 1:5) sub2(patt, repl, X))
user system elapsed
0.860.000.86
Since the ordered vectors are perfectly aligned, might as well do an exact
string match. Hence, I think this is not off-topic.
Cheers,
Adam
On Wednesday, July 29, 2015 at 8:15:52 AM UTC-7, Bert Gunter wrote:
There is confusion here. apply() family functions are **NOT**
vectorization -- they ARE loops (at the interpreter level), just done
in functionalized form. Please read background material (John
Chambers's books, MASS, or numerous others) to improve your
understanding and avoid posting erroneous comments.
Cheers,
Bert
Bert Gunter
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
-- Clifford Stoll
On Tue, Jul 28, 2015 at 3:00 PM, John Thaden jjth...@flash.net
javascript: wrote:
Adam,The method you propose gives a different result than the prior
methods for these example vectors
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)
Old method 1
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl,
x=X)
gives
b cda
aB CD ef
Old method 2
sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
sub2(patt, repl, X)
gives
[1] aB CD ef
Your method (I gave it the unique name sub3)
sub3 - function(pattern, replacement, x) { len- length(x) y
- character(length=len) patlen - length(pattern) replen -
length(replacement) if(patlen != replen) stop('Error: Pattern and
replacement length do not match') for(i in 1:replen) {
y[which(x==pattern[i])] - replacement[i] } return(y)}sub3(patt, repl,
X)
gives[1]CD
Granted, whatever it does, it does it faster
#Old method 1
system.time(for(i in 1:5)
mapply(function(p,r,x) sub(p,r,x, fixed = TRUE),p=patt,r=repl,x=X))
user system elapsed
2.530.002.52
#Old method 2
system.time(for(i in 1:5)sub2(patt, repl, X)) user system elapsed
2.320.002.32
#Your proposed method
system.time(for(i in 1:5) sub3(patt, repl, X))
user system elapsed
1.020.001.01
but would it still be faster if it actually solved the same problem?
-John Thaden
On Monday, July 27, 2015 11:40 PM, Adam Erickson
adam.micha...@gmail.com javascript: wrote:
I know this is an old thread, but I wrote a simple FOR loop with
vectorized pattern replacement that is much faster than either of those (it
can also accept outputs differing in length from the patterns):
sub2 - function(pattern, replacement, x) { len - length(x)
y - character(length=len)patlen - length(pattern)replen -
length(replacement)if(patlen != replen) stop('Error: Pattern and
replacement length do not match')for(i in 1:replen) {
y[which(x==pattern[i])] - replacement[i]}return(y) }
system.time(test - sub2(patt, repl, XX)) user system elapsed 0
0 0
Cheers,
Adam
On Wednesday, October 8, 2008 at 9:38:01 PM UTC-7, john wrote:
Hello Christos,
To my surprise, vectorization actually hurt processing speed!#Example
X - c(ab, cd, ef)
patt - c(b, cd, a)
repl - c(B, CD, A)sub2 - function(pattern, replacement, x) {
len - length(x)
if (length(pattern) == 1)
pattern - rep(pattern, len)
if (length(replacement) == 1)
replacement - rep(replacement, len)
FUN - function(i, ...) {
sub(pattern[i], replacement[i], x[i], fixed = TRUE)
}
idx - 1:length(x)
sapply(idx, FUN)
}
system.time( for(i in 1:1) sub2(patt, repl, X) )
user system elapsed
1.180.071.26 system.time( for(i in 1:1)
mapply(function(p, r, x) sub(p, r, x, fixed = TRUE), p=patt, r=repl, x=X)
)
user system elapsed
1.420.051.47
So much for avoiding loops.
John Thaden=== At 2008-10-07, 14:58:10 Christos wrote: ===John,
Try the
Re: [R] Daily Category Revenue-Stacked Bar Chart in ggplot2
Since there IS NO R-help website that you can post on, you are definitely being confused by the Nabble website. This is a MAILING list that you interact with by sending emails to r-help@r-project.org, not a website forum. The trouble with Nabble is that it does confuse users, and breaks the continuity of emails that people who use it as it was intended to be used depend on. For example, as a Nabble user you may not even be seeing the following footer that the rest of us live by: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. A more accurate web rendition of R-help than Nabble presents may be seen here: https://stat.ethz.ch/pipermail/r-help/ which is mentioned at the link that you complained did not help you. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On July 29, 2015 2:37:56 PM EDT, Hidden Markov Model huss...@touchofmodern.com wrote: No, I'm not confused. I just posted on R help website. I don't know how to use the email client to do anything which you are speaking of. If you would like the posts to be made in an alternative way, then you will need to provide clear direction. Sending someone to a help page which doesn't reference what you are speaking about isn't actually helpful. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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-es] Duda ' raster package '
Hola, Tengo una matriz ' elev.rad.mx ' que tiene tres columnas: 1� columna: coordenada x del punto geogr�fico. 2� columna: coordenada y del punto geogr�fico. 3� columna: valores de un radar en el punto geogr�fico (x,y). Quiero convertir esta matriz a un objeto raster. Para ello, pongo la siguiente instrucci�n: elev.rad.raster - raster(elev.rad.mx, xmn=355349, xmx=366549, ymn=4468331, ymx=4476651) Lo que me sale por consola al poner ' elev.rad.raster ' es: elev.rad.raster class : RasterLayer dimensions : 13306, 3, 39918 (nrow, ncol, ncell) resolution : 3733.333, 0.6252818 (x, y) extent : 355349, 366549, 4468331, 4476651 (xmin, xmax, ymin, ymax) coord. ref. : NA data source : in memory names :layer values : 723.13, 4476611 (min, max) Entonces, mi duda es: �C�mo puedo convertir o transformar este raster a otro raster (con los mismos datos y la misma extensi�n que tiene ' elev.rad.raster ') para que tenga de dimensiones 140 140 y de resoluci�n (x,y)=(80,80)? Lo necesito porque tengo otro raster de ese tipo y quiero juntar los dos en un Grid. Para que todo encaje, tienen que tener la misma resoluci�n. Muchas gracias de antemano. Un saludo, [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] Daily Category Revenue-Stacked Bar Chart in ggplot2
Hi John, Great thanks for the examples! I am not sure what you are referring to when you say a post from Nabble - I posted this directly on R-help. I actually never heard of Nabble. Stacked Bar charts are great for when you have a lot of moving parts and need to be able to zero in on one of them (e.g. Advertising Placements). Unfortunately, I can't post the data since it is confidential. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710540.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
Then why do your messages contain the Nabble footer: View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710540.html Sent from the R help mailing list archive at Nabble.com. I think you are confused. You should be looking at https://stat.ethz.ch/mailman/listinfo/r-help and using your email client to send emails rather than posting on a web page. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On July 29, 2015 2:23:54 PM EDT, Hidden Markov Model huss...@touchofmodern.com wrote: Hi John, Great thanks for the examples! I am not sure what you are referring to when you say a post from Nabble - I posted this directly on R-help. I actually never heard of Nabble. Stacked Bar charts are great for when you have a lot of moving parts and need to be able to zero in on one of them (e.g. Advertising Placements). Unfortunately, I can't post the data since it is confidential. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710540.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
On 29-07-2015, at 20:37, Hidden Markov Model huss...@touchofmodern.com wrote: No, I'm not confused. I just posted on R help website. The R help website is NOT on Nabble. You posted from Nabble as can be seen from your message. It can also be seen in what the internet browser displays. It is what we see in our email application. Jeff gave you the correct advice. Use your email client. Don’t use Nabble and your internet browser. See the section Using R help on https://stat.ethz.ch/mailman/listinfo/r-help Berend I don't know how to use the email client to do anything which you are speaking of. If you would like the posts to be made in an alternative way, then you will need to provide clear direction. Sending someone to a help page which doesn't reference what you are speaking about isn't actually helpful. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710543.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
Alright thanks for clarifying. That's all a bit esoteric. Quite different from basics off computer use. None of the documentation anyone mentioned actually contains any of this. If Nabble is so bad, then why does anyone use it? It doesn't make sense. You should probably write a wiki going forward so you won't get upset every-time somebody asks a reasonable question. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710560.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
Below. Bert Gunter Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. -- Clifford Stoll On Wed, Jul 29, 2015 at 3:43 PM, Hidden Markov Model huss...@touchofmodern.com wrote: Alright thanks for clarifying. That's all a bit esoteric. Quite different from basics off computer use. None of the documentation anyone mentioned actually contains any of this. If Nabble is so bad, then why does anyone use it? It doesn't make sense. **You should probably write a wiki going forward so you won't get upset every-time somebody asks a reasonable question.** Fortune nomination! (I don't actually agree, but I thought it was amusing) -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710560.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
Alright kids, I think we should close this topic since we appear to be getting further and further off-topic. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431p4710563.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] dplyr help
Hello, I've recently discovered the helpful dplyr package. I'm using the 'aggregate' function as such: bevs - data.frame(cbind(name = c(Bill, Mary), drink = c(coffee, tea, cocoa, water), cost = seq(1:8), sex = c(male,female))); bevs$cost - seq(1:8) bevs name drink costsex 1 Bill coffee1 male 2 Marytea2 female 3 Bill cocoa3 male 4 Mary water4 female 5 Bill coffee5 male 6 Marytea6 female 7 Bill cocoa7 male 8 Mary water8 female aggregate(cost ~ name + drink, data = bevs, sum) name drink cost 1 Bill cocoa 10 2 Bill coffee6 3 Marytea8 4 Mary water 12 My issue is that I would like to keep a column for 'sex', for which there is a 1:1 mapping with 'name', such that every time 'Bill' appears, it is always 'male'. Does anyone know of a way to accomplish this, with or without dplyr? The ideal command(s) would produce this: name drink cost sex 1 Bill cocoa 10 male 2 Bill coffee6 male 3 Marytea8 female 4 Mary water 12 female I would be thankful for any suggestion! Thanks, Jonathan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] dplyr help
Hi Brian, Thanks for the suggestion, although the command is throwing an error as such: bevs %% group_by(name, sex, drink) %% summarise( cost = sum(cost)) %% select(name, drink, cost, sex) Error: unexpected input in bevs %% group_by(name, sex, drink) %% summarise( Your syntax is new to me so I'm not immediately clear on how to fix it; any idea how? Thanks again, Jonathan On Wed, Jul 29, 2015 at 11:07 PM, Brian Kreeger brian.kree...@gmail.com wrote: dplyr solution: bevs %% group_by(name, sex, drink) %% summarise(cost = sum(cost)) %% select(name, drink, cost, sex) The last select statement puts the output in the column order you wanted in your result. I hope this helps. Brian On Wed, Jul 29, 2015 at 9:37 PM, Jon BR jonsle...@gmail.com wrote: Hello, I've recently discovered the helpful dplyr package. I'm using the 'aggregate' function as such: bevs - data.frame(cbind(name = c(Bill, Mary), drink = c(coffee, tea, cocoa, water), cost = seq(1:8), sex = c(male,female))); bevs$cost - seq(1:8) bevs name drink costsex 1 Bill coffee1 male 2 Marytea2 female 3 Bill cocoa3 male 4 Mary water4 female 5 Bill coffee5 male 6 Marytea6 female 7 Bill cocoa7 male 8 Mary water8 female aggregate(cost ~ name + drink, data = bevs, sum) name drink cost 1 Bill cocoa 10 2 Bill coffee6 3 Marytea8 4 Mary water 12 My issue is that I would like to keep a column for 'sex', for which there is a 1:1 mapping with 'name', such that every time 'Bill' appears, it is always 'male'. Does anyone know of a way to accomplish this, with or without dplyr? The ideal command(s) would produce this: name drink cost sex 1 Bill cocoa 10 male 2 Bill coffee6 male 3 Marytea8 female 4 Mary water 12 female I would be thankful for any suggestion! Thanks, Jonathan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Vignette using knitr, devtools, and R markdown
Hi All, I am writing a Vignette with Knitr using devtools ... actually quite awesome. However, I would like to create a table of contents. I have read the knitr and r markdown instructions on TOC and I can create a TOC in a markdown document but I cannot translate the command to a vignette - maybe doing something wrong. Does anyone know if I can create a TOC in a vignette using r-markdown and devtools? Glenn __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] dplyr help
David, I do appreciate your help, if not the dose of contempt. I hope you feel OK. Thanks for the tips, -Jonathan On Wed, Jul 29, 2015 at 11:14 PM, David Winsemius dwinsem...@comcast.net wrote: On Jul 29, 2015, at 7:37 PM, Jon BR wrote: Hello, I've recently discovered the helpful dplyr package. I'm using the 'aggregate' function as such: The `aggregate` function is part of base-R: bevs - data.frame(cbind(name = c(Bill, Mary), drink = c(coffee, tea, cocoa, water), cost = seq(1:8), sex = c(male,female))); bevs$cost - seq(1:8) bevs name drink costsex 1 Bill coffee1 male 2 Marytea2 female 3 Bill cocoa3 male 4 Mary water4 female 5 Bill coffee5 male 6 Marytea6 female 7 Bill cocoa7 male 8 Mary water8 female aggregate(cost ~ name + drink, data = bevs, sum) name drink cost 1 Bill cocoa 10 2 Bill coffee6 3 Marytea8 4 Mary water 12 My issue is that I would like to keep a column for 'sex', for which there is a 1:1 mapping with 'name', such that every time 'Bill' appears, it is always 'male'. Does anyone know of a way to accomplish this, with or without dplyr? As pointed out you have not yet demonstrated any dplyr functions. The ideal command(s) would produce this: name drink cost sex 1 Bill cocoa 10 male 2 Bill coffee6 male 3 Marytea8 female 4 Mary water 12 female Doesn't this (glaringly obvious?) approach succeed? aggregate(cost ~ name + drink+sex, data = bevs, sum) name drinksex cost 1 Marytea female8 2 Mary water female 12 3 Bill cocoa male 10 4 Bill coffee male6 I would be thankful for any suggestion! Thanks, Jonathan [[alternative HTML version deleted]] Please learn to post in plain text. -- David Winsemius Alameda, CA, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] dplyr help
dplyr solution: bevs %% group_by(name, sex, drink) %% summarise(cost = sum(cost)) %% select(name, drink, cost, sex) The last select statement puts the output in the column order you wanted in your result. I hope this helps. Brian On Wed, Jul 29, 2015 at 9:37 PM, Jon BR jonsle...@gmail.com wrote: Hello, I've recently discovered the helpful dplyr package. I'm using the 'aggregate' function as such: bevs - data.frame(cbind(name = c(Bill, Mary), drink = c(coffee, tea, cocoa, water), cost = seq(1:8), sex = c(male,female))); bevs$cost - seq(1:8) bevs name drink costsex 1 Bill coffee1 male 2 Marytea2 female 3 Bill cocoa3 male 4 Mary water4 female 5 Bill coffee5 male 6 Marytea6 female 7 Bill cocoa7 male 8 Mary water8 female aggregate(cost ~ name + drink, data = bevs, sum) name drink cost 1 Bill cocoa 10 2 Bill coffee6 3 Marytea8 4 Mary water 12 My issue is that I would like to keep a column for 'sex', for which there is a 1:1 mapping with 'name', such that every time 'Bill' appears, it is always 'male'. Does anyone know of a way to accomplish this, with or without dplyr? The ideal command(s) would produce this: name drink cost sex 1 Bill cocoa 10 male 2 Bill coffee6 male 3 Marytea8 female 4 Mary water 12 female I would be thankful for any suggestion! Thanks, Jonathan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Mixed Date Formats
Hi I wonder if it is easier to convert the dates to character format and then reformat using gsub or the like str(SampleData) SampleData$date - as.character(SampleData$date) str(SampleData) as.Date( ifelse(nchar(SampleData[,date]) == 9, as.Date(SampleData[,date], format = %d-%b-%y), ifelse(nchar(SampleData[,date]) == 8, as.Date(SampleData[,date], format = %d/%m/%y), ifelse(as.numeric(substr(SampleData[,date],1,2)) 12, as.Date(SampleData[,date], format = %d/%m/%Y), as.Date(SampleData[,date], format = %m/%d/%Y)) )), origin = as.Date(1970-01-01)) Beware of the American format in months jan feb mar oct nov -- will need more conditions to be imposed Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of farnoosh sheikhi via R-help Sent: Thursday, 30 July 2015 07:16 To: Sarah Goslee Cc: R. Help Subject: Re: [R] Mixed Date Formats Hi Sarah, Thanks for getting back to me.Here is an example of my data:SampleData - structure(list(id = 1:7, value = c(5813L, 8706L, 4049L, 5877L, 1375L, 2223L, 3423L), date = structure(c(4L, 3L, 2L, 1L, 7L, 6L, 5L), .Label = c(05/12/11, 06/05/11, 07-Dec-11, 19-Dec-11, 01/22/2011, 10/19/2011, 31/12/2011 ), class = factor)), .Names = c(id, value, date), row.names = c(NA, -7L), class = data.frame)SampleData Thanks for your help:). On Wednesday, July 29, 2015 1:50 PM, Sarah Goslee sarah.gos...@gmail.com wrote: On Wed, Jul 29, 2015 at 2:45 PM, farnoosh sheikhi via R-help r-help@r-project.org wrote: Hi Arun, Hope all is well with you. I have a data with a column for date.The date format is mixed. There are date values with Month/Day/Year format and values with Day/Month/Year format.I don't know how to unify it.I really appreciate your help.Thanks. You sent this to the R-help list, not just to Arun, so I'm assuming this is an R question. The best way to get help is to provide a sample of your data using dput() and to clearly specify what you would like as the result - unify is a bit vague. paste(x, collapse=) could be considered unification, after all. Sarah -- Sarah Goslee http://www.functionaldiversity.org [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] dplyr help
On Jul 29, 2015, at 7:37 PM, Jon BR wrote: Hello, I've recently discovered the helpful dplyr package. I'm using the 'aggregate' function as such: The `aggregate` function is part of base-R: bevs - data.frame(cbind(name = c(Bill, Mary), drink = c(coffee, tea, cocoa, water), cost = seq(1:8), sex = c(male,female))); bevs$cost - seq(1:8) bevs name drink costsex 1 Bill coffee1 male 2 Marytea2 female 3 Bill cocoa3 male 4 Mary water4 female 5 Bill coffee5 male 6 Marytea6 female 7 Bill cocoa7 male 8 Mary water8 female aggregate(cost ~ name + drink, data = bevs, sum) name drink cost 1 Bill cocoa 10 2 Bill coffee6 3 Marytea8 4 Mary water 12 My issue is that I would like to keep a column for 'sex', for which there is a 1:1 mapping with 'name', such that every time 'Bill' appears, it is always 'male'. Does anyone know of a way to accomplish this, with or without dplyr? As pointed out you have not yet demonstrated any dplyr functions. The ideal command(s) would produce this: name drink cost sex 1 Bill cocoa 10 male 2 Bill coffee6 male 3 Marytea8 female 4 Mary water 12 female Doesn't this (glaringly obvious?) approach succeed? aggregate(cost ~ name + drink+sex, data = bevs, sum) name drinksex cost 1 Marytea female8 2 Mary water female 12 3 Bill cocoa male 10 4 Bill coffee male6 I would be thankful for any suggestion! Thanks, Jonathan [[alternative HTML version deleted]] Please learn to post in plain text. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Element-by-element division
On 28 Jul 2015, at 15:53 , Sarah Goslee sarah.gos...@gmail.com wrote: Sure, there are lots of ways to do everything in R. But mixing in apply muddles the issue, since apply() and sweep() use different logic to determine MARGIN. Actually, apply() and sweep() were designed together and use exactly the SAME logic to determine the margin. E.g., to sweep out means according to the last two dimensions of an array, you do m - array(1:24, c(4,3,2)) (mm - apply(m, c(2,3), mean)) [,1] [,2] [1,] 2.5 14.5 [2,] 6.5 18.5 [3,] 10.5 22.5 sweep(m, c(2,3), mm, -) , , 1 [,1] [,2] [,3] [1,] -1.5 -1.5 -1.5 [2,] -0.5 -0.5 -0.5 [3,] 0.5 0.5 0.5 [4,] 1.5 1.5 1.5 , , 2 [,1] [,2] [,3] [1,] -1.5 -1.5 -1.5 [2,] -0.5 -0.5 -0.5 [3,] 0.5 0.5 0.5 [4,] 1.5 1.5 1.5 The trouble comes when people miss the point and start using apply() in ways it wasn't designed for... -pd -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] indices of mismatch element in two vector with missing values
Hervé Pagès hpa...@fredhutch.org on Tue, 28 Jul 2015 23:07:14 -0700 writes: On 07/28/2015 09:58 PM, Peter Alspach wrote: One way seq(test1)[-which(test1==test2)] One question is whether 2 NAs should be considered to match or not. The OP doesn't tell but I guess he wants them to match: test1 - c(1, 2, NA, 4, NA, 6) test2 - c(1, 2, 3, NA, NA, 66) seq(test1)[-which(test1==test2)] # [1] 3 4 5 6 5 should probably not be there! but I imagine there are better ones . Yes, indeed, as logical indexing is considerably safer than negative indexing with which(.) : PLEASE, everyone, do remember: %%===%% %% There is one ___big__ disadvantage to [ - which(logical) ] : %% %% It entirely __fails__ when the logical vector is all FALSE %% %%===%% ## Example (remember, we want the indices of *differing* entries): ## --- Here, all entries differ, so we want to get 1:8 : x - c(NA, 1:7) y - c(11:17, NA) ## now this seq(x)[ - which(x == y) ] ## gives not what you expect ## But logical indexing always works: x == y is.na(x) == is.na(y) seq(x)[!(x == y is.na(x) == is.na(y))] With output : x - c(NA, 1:7) y - c(11:17, NA) ## now this seq(x)[ - which(x == y) ] ## gives not what you expect integer(0) ## But logical indexing always works: x == y is.na(x) == is.na(y) [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE seq(x)[!(x == y is.na(x) == is.na(y))] [1] 1 2 3 4 5 6 7 8 Martin Maechler ETH Zurich -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of baccts Sent: Wednesday, 29 July 2015 8:26 a.m. To: r-help@r-project.org Subject: [R] indices of mismatch element in two vector with missing values How would you return the index where two vectors differs if they may contain missing (NA) values? For example: test1 - c(1,2,NA); test2 - c(1,2,3); which(test1!=test2) does not return 3! Thanks in advance. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] indices of mismatch element in two vector with missing values
On 07/28/2015 09:58 PM, Peter Alspach wrote:
One way
seq(test1)[-which(test1==test2)]
One question is whether 2 NAs should be considered to match or not.
The OP doesn't tell but I guess he wants them to match:
test1 - c(1, 2, NA, 4, NA, 6)
test2 - c(1, 2, 3, NA, NA, 66)
seq(test1)[-which(test1==test2)]
# [1] 3 4 5 6
5 should probably not be there!
H.
but I imagine there are better ones .
Peter Alspach
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of baccts
Sent: Wednesday, 29 July 2015 8:26 a.m.
To: r-help@r-project.org
Subject: [R] indices of mismatch element in two vector with missing values
How would you return the index where two vectors differs if they may contain
missing (NA) values?
For example:
test1 - c(1,2,NA);
test2 - c(1,2,3);
which(test1!=test2) does not return 3!
Thanks in advance.
--
View this message in context:
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--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
P.O. Box 19024
Seattle, WA 98109-1024
E-mail: hpa...@fredhutch.org
Phone: (206) 667-5791
Fax:(206) 667-1319
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Re: [R] Reading data with two rows of variable names using read.zoo
Thanks, Dan. Your codes work fine. But I have tens of countries UK, JP, BR, US..., each of which has ten columns a1, a2, ..., a10 of data. So a little more automation is needed. I have been trying to make a list of each country's data and use sapply thing to get UK JP 2009 Q2 65 2009 Q3 75 2009 Q4 87 2010 Q1 67 2010 Q2 63 But for me, it was not easy as it looks... Thank you in advance! -- View this message in context: http://r.789695.n4.nabble.com/Reading-data-with-two-rows-of-variable-names-using-read-zoo-tp4710496p4710510.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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-es] Método S3 paquete
Hola Rubén,
Muchas gracias.
Un saludo,
Guillermo
Hola a ambos,
Otra referencia que puede ser de interés es:
http://r-pkgs.had.co.nz y también http://adv-r.had.co.nz (las dos de
Hadley Wickham...)
Un saludo, Rubén.
El 27/07/2015 a las 10:46, guillermo.vi...@uv.es escribió:
Hola Carlos,
Muchas gracias por el enlace, me ha sido de gran ayuda. Ya he entendido
cómo funciona el sistema S3.
Un saludo,
Guillermo
Hola, ¿qué tal?
Sigue
http://www.datanalytics.com/2011/08/04/desarrollo-de-paquetes-con-r-iv-funciones-genericas/
a rajatabla y lo tendrás.
Un saludo,
Carlos J. Gil Bellosta
http://www.datanalytics.com
P.D.: Si te fijas bien, no estás siguiendo esa guía a rajatabla.
El día 23 de julio de 2015, 16:26, guillermo.vi...@uv.es escribió:
Hola,
Estoy tratando de crear un método S3 llamado anthr dentro del
paquete
que estoy desarrollando, cuyo argumento principal es res que
básicamente es una lista con un solo componente. Pero si el segundo
argumento llamado oneSize es FALSE, res es una lista de listas.
Lo que he escrito hasta el momento es lo siguiente:
anthr - function(res, oneSize, nsizes){
UseMethod(anthr)
}
anthr.tri - function(res, oneSize, nsizes){
if(oneSize){
cases - c()
cases - res$meds
}else{
cases - list()
for (i in 1 : (nsizes - 1)){
cases[[i]] - res[[i]]$meds
}
}
return(cases)
}
El problema cuando instalo el paquete y utilizo este método, es
que R no
me reconoce que res sea una lista. En concreto, me aparece este
error:
Error in UseMethod(anthr) :
no applicable method for 'anthr' applied to an object of class
list
He tratado de añadir esto:
tri - function(x){
value - list(meds = x$meds)
attr(value, class) - tri
value
}
pero sigue sin funcionarme. ¿Alguien puede ofrecerme alguna ayuda? .
Muchas gracias de antemano.
Un saludo,
Guillermo
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Re: [R] Stop tkbind
Not any ideas? Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Stop-tkbind-tp4710476p4710514.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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-es] Método S3 paquete
Hola a ambos,
Otra referencia que puede ser de interés es:
http://r-pkgs.had.co.nz y también http://adv-r.had.co.nz (las dos de
Hadley Wickham...)
Un saludo, Rubén.
El 27/07/2015 a las 10:46, guillermo.vi...@uv.es escribió:
Hola Carlos,
Muchas gracias por el enlace, me ha sido de gran ayuda. Ya he entendido
cómo funciona el sistema S3.
Un saludo,
Guillermo
Hola, ¿qué tal?
Sigue
http://www.datanalytics.com/2011/08/04/desarrollo-de-paquetes-con-r-iv-funciones-genericas/
a rajatabla y lo tendrás.
Un saludo,
Carlos J. Gil Bellosta
http://www.datanalytics.com
P.D.: Si te fijas bien, no estás siguiendo esa guía a rajatabla.
El día 23 de julio de 2015, 16:26, guillermo.vi...@uv.es escribió:
Hola,
Estoy tratando de crear un método S3 llamado anthr dentro del paquete
que estoy desarrollando, cuyo argumento principal es res que
básicamente es una lista con un solo componente. Pero si el segundo
argumento llamado oneSize es FALSE, res es una lista de listas.
Lo que he escrito hasta el momento es lo siguiente:
anthr - function(res, oneSize, nsizes){
UseMethod(anthr)
}
anthr.tri - function(res, oneSize, nsizes){
if(oneSize){
cases - c()
cases - res$meds
}else{
cases - list()
for (i in 1 : (nsizes - 1)){
cases[[i]] - res[[i]]$meds
}
}
return(cases)
}
El problema cuando instalo el paquete y utilizo este método, es que R no
me reconoce que res sea una lista. En concreto, me aparece este error:
Error in UseMethod(anthr) :
no applicable method for 'anthr' applied to an object of class list
He tratado de añadir esto:
tri - function(x){
value - list(meds = x$meds)
attr(value, class) - tri
value
}
pero sigue sin funcionarme. ¿Alguien puede ofrecerme alguna ayuda? .
Muchas gracias de antemano.
Un saludo,
Guillermo
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[R] Removing display of R row names from list.
Dear All, Is there a way to not include the row names when creating the list? Thanks! Regards, Frederic. Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] How to read CSV from web?
data-read.csv(https://raw.githubusercontent.com/sjkiss/Survey/master/mlogit.out.csv,header=T,sep=,;) -- View this message in context: http://r.789695.n4.nabble.com/How-to-read-CSV-from-web-tp4710502p4710513.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Stop tkbind
The solution is to create a new tkbind function that has not argument tkbind(img,button-1,) -- View this message in context: http://r.789695.n4.nabble.com/Stop-tkbind-tp4710476p4710517.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] help_ReverseGeocoding
Thank you for your suggestion. Is there any open source api that won't
rate-limit it?
On Tue, Jul 28, 2015 at 9:38 PM, boB Rudis b...@rudis.net wrote:
You should use ggmap::revgeocode (it calls google's api) and google
will rate-limit you. There are also packages to use HERE maps
geo/revgeo lookups
http://blog.corynissen.com/2014/10/making-r-package-to-use-here-geocode-api.html
and the geocode package has GNfindNearestAddress, so tons of options
to choose from.
On Tue, Jul 28, 2015 at 11:30 AM, MacQueen, Don macque...@llnl.gov
wrote:
My first guess, after a quick glance, is that Google only lets you do a
limited number of lookups within some period of time.
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 7/27/15, 10:14 PM, R-help on behalf of shreya ghosh
r-help-boun...@r-project.org on behalf of shreya@gmail.com wrote:
Hi,
I'm trying to do reversegeocoding on a large dataset. I'm using RJSONIO
library and using Google map API to get the location of the given lat-lon
in the dataset. After 100 or 150 successful displaying location
information
it is showing
Warning message - In readLines(con) : cannot open: HTTP status was '0
(null)'
and Error : Error in fromJSON(paste(readLines(con), collapse = )) :
error in evaluating the argument 'content' in selecting a method for
function 'fromJSON': Error in readLines(con) : cannot open the
connection
Please help me to solve the issue.
location function is as follows :
location-function(latlng){
latlngStr - gsub(' ','%20', paste(latlng, collapse=,))
library(RJSONIO) #Load Library
#Open Connection
connectStr - paste('
http://maps.google.com/maps/api/geocode/json?sensor=falselatlng=
',latlngS
tr,
sep=)
con - url(connectStr)
data.json - fromJSON(paste(readLines(con), collapse=))
close(con)
data.json - unlist(data.json)
if(data.json[status]==OK)
address - data.json[results.formatted_address]
print (address)
}
I'm using R version 3.2.1 and Ubuntu 14.10 OS.
Thank you.
--
Shreya Ghosh
*9007448845*
-- The mind is not a vessel to be filled, but a fire to be kindled
[[alternative HTML version deleted]]
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PLEASE do read the posting guide
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--
Shreya Ghosh
*9007448845*
-- The mind is not a vessel to be filled, but a fire to be kindled
[[alternative HTML version deleted]]
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[R] Imbalanced random forest
¿How can i set up a study with random forest where the response is highly imbalanced? - Guided Tours Basque Country Guided tours in the three capitals of the Basque Country: Bilbao, Vitoria-Gasteiz and San Sebastian, as well as in their provinces. Available languages. Travel planners for groups and design of tourist routes across the Basque Country. -- View this message in context: http://r.789695.n4.nabble.com/Imbalanced-random-forest-tp4710524.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Daily Category Revenue-Stacked Bar Chart in ggplot2
First http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and http://adv-r.had.co.nz/Reproducibility.html Second. It is very annoying to have posts come in from Nabble. Very few R-help readers use it and the total context of some post to R-help is usually lost. If possible could you post directly to R-help. We have no idea of what your problem is (see above) but here is a very simple example of a stacked barchart using ggplot2. And since I am already complaining I have included a side-by-side version of the barchart as well. I think that stacked barcharts are not a good idea unless obfuscation is the desired outcome. It is too hard to compare quantities with no common baseline. Good luck. dat1 - structure(list(dates = structure(c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L), .Label = c(1, 2, 3, 4, 5), class = factor), revs = c(40, 7, 40, 20, 35, 20, 15, 20, 15, 20, 15, 15, 35, 20, 20, 7, 7, 20, 7, 35), typrep.LETTERS.1.2...10. = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L), .Label = c(A, B), class = factor)), .Names = c(dates, revs, typrep.LETTERS.1.2...10.), row.names = c(NA, -20L ), class = data.frame) #Stacked barchart ggplot(dat1, aes(dates, revs, fill = typ)) + geom_bar(stat = identity) #Grouped or dodged barchart (I don't think these are the real names) ggplot(dat1, aes(dates, revs, fill = typ)) + geom_bar(stat = identity, position=dodge) John Kane Kingston ON Canada -Original Message- From: huss...@touchofmodern.com Sent: Mon, 27 Jul 2015 13:53:30 -0700 (PDT) To: r-help@r-project.org Subject: [R] Daily Category Revenue-Stacked Bar Chart in ggplot2 I am trying to use the ggplot2 to build a stacked bar chart for daily Revenue by category. The chart would look have date on the x-axis, and revenue on the y axis. The fill would be the categories themselves. I have searched a great deal and have been unable to find exactly how to do this. -- View this message in context: http://r.789695.n4.nabble.com/Daily-Category-Revenue-Stacked-Bar-Chart-in-ggplot2-tp4710431.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords protects your account. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Removing display of R row names from list.
HI Frederic, Can you supply a small example of the problem? John Kane Kingston ON Canada -Original Message- From: ntfr...@gmail.com Sent: Wed, 29 Jul 2015 14:15:58 +0300 To: r-help@r-project.org Subject: [R] Removing display of R row names from list. Dear All, Is there a way to not include the row names when creating the list? Thanks! Regards, Frederic. Frederic Ntirenganya Maseno University, African Maths Initiative, Kenya. Mobile:(+254)718492836 Email: fr...@aims.ac.za https://sites.google.com/a/aims.ac.za/fredo/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] R load error
Try restarting R-Studio. I have found that every once in a while it seems to do something squirrelly but I have never isolated the problem enough to do a report. Otherwise,perhaps run R in a terminal and see if it will load the data from there to check if the file is actually okay. John Kane Kingston ON Canada -Original Message- From: ypetsc...@fcrr.org Sent: Mon, 27 Jul 2015 15:03:52 -0400 To: r-help@r-project.org Subject: [R] R load error Greetings - I'm using RStudio and recently updated both it and R. When loadings up, I'm now receiving the following error: Error: ReadItem: unknown type 63, perhaps written by later version of R I've tried using rm(list=ls()) rm(list=ls(all.names=TRUE)) and detach() but nothing works in R. Within RStudio it just continues to try and load and I'm unable to interrupt or restart. Any ideas would be greatly appreciated. Yaacov Petscher, Ph.D. Associate Director, Florida Center for Reading Research Senior Research Associate, Regional Educational Laboratory-Southeast (REL-SE) at Florida State University 2010 Levy Ave Suite 100 Tallahassee, Florida, 32310 850-645-8963 (p) 850-644-9085 (f) http://www.fcrr.org/for-researchers/petscher2.asp [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing to find out more! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] VIF threshold implying multicollinearity
Quite, but apparently not a boisterous one? John Kane Kingston ON Canada -Original Message- From: cfly...@ncsu.edu Sent: Mon, 27 Jul 2015 14:35:06 -0400 To: jrkrid...@inbox.com Subject: Re: [R] VIF threshold implying multicollinearity No actually it is a quiet good paper! :) On Mon, Jul 27, 2015 at 8:14 AM, John Kane jrkrid...@inbox.com wrote: +1 I, originally, read it as a stringent criticism of the first paper. John Kane Kingston ON Canada -Original Message- From: r.tur...@auckland.ac.nz Sent: Mon, 27 Jul 2015 15:12:43 +1200 To: cfly...@ncsu.edu Subject: Re: [R] VIF threshold implying multicollinearity On 27/07/15 13:36, Collin Lynch wrote: The following sources discuss the issues generally and may be a goof pointer to the literature ... SNIP I think that the foregoing merits fortune status! :-) cheers, Rolf -- Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help [https://stat.ethz.ch/mailman/listinfo/r-help] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html [http://www.R-project.org/posting-guide.html] and provide commented, minimal, self-contained, reproducible code. FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing [http://www.inbox.com/photosharing] to find out more! FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
