Re: [R] double-axis labels function of each other
Yes, I keep a copy in MS Word. Would Notepad be OK? I need sleep now, will work tomorrow. -- View this message in context: http://r.789695.n4.nabble.com/double-axis-labels-function-of-each-other-tp4705457p4705463.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] RInside
help Hello I'm newbee with R and RInside My question is about stderr in R. Is there a way to collect R stderr in C++ program embedding R Thanks in advance Michel --- L'absence de virus dans ce courrier électronique a été vérifiée par le logiciel antivirus Avast. __ 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] Package build system adds line break in DESCRIPTION URL
Has anybody noticed that if field URL in DESCRIPTION contains a uri with 66 or more characters, then file DESCRIPTION in the resulting package includes a line break at the beginning? So this (source DESCRIPTION): URL: http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/Pages/default.aspx becomes (again file DESCRIPTION, but inside the package) URL: http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/Pages/default.aspx This has been tested with R on Windows 8.1 (devel 01/04/2015 and 3.1.3) and Linux Mint (3.1.3). It has many sad implications including not acceptance of such packages to CRAN. __ 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] applying cumsum within groups
This small example will be applied to a problem with 1.4e6 lines of data. First, here is the dataset and a few lines of R script, followed by an explanation of what I'd like to get: dat - read.table(textConnection(ISEG IRCH val 11 265 12 260 13 234 54 39 467 54 40 468 54 41 460 54 42 489 11 265 12 276 13 217 54 39 456 54 40 507 54 41 483 54 42 457 11 265 12 287 13 224 54 39 473 54 40 502 54 41 497 54 42 447 11 230 12 251 13 199 54 39 439 54 40 474 54 41 477 54 42 413 11 230 12 262 13 217 54 39 455 54 40 493 54 41 489 54 42 431 11 1002 12 1222 13 1198 54 39 1876 54 40 1565 54 41 1455 54 42 1427 11 1002 12 1246 13 1153 54 39 1813 54 40 1490 54 41 1518 54 42 1486 11 1002 12 1229 13 1142 54 39 1797 54 40 1517 54 41 1527 54 42 1514),header=TRUE) dat$seq - ifelse(dat$ISEG==1 dat$IRCH==1, 1, 0) tmp - diff(dat[dat$seq==1,]$val)!=0 dat$idx - 0 dat[dat$seq==1,][c(TRUE,tmp),]$idx - 1 dat$ts - cumsum(dat$idx) At this point, I'd like to add one more column called iter that counts up by 1 based on seq, but within each ts. So, the result would look like this (undoubtedly this is a simple problem with something like ddply, but I've been unable to construct the R for it): dat ISEG IRCH val seq idx ts iter 11 265 1 1 11 12 260 0 0 11 13 234 0 0 11 54 39 467 0 0 11 54 40 468 0 0 11 54 41 460 0 0 11 54 42 489 0 0 11 11 265 1 0 12 12 276 0 0 12 13 217 0 0 12 54 39 456 0 0 12 54 40 507 0 0 12 54 41 483 0 0 12 54 42 457 0 0 12 11 265 1 0 13 12 287 0 0 13 13 224 0 0 13 54 39 473 0 0 13 54 40 502 0 0 13 54 41 497 0 0 13 54 42 447 0 0 13 11 230 1 1 21 12 251 0 0 21 13 199 0 0 21 54 39 439 0 0 21 54 40 474 0 0 21 54 41 477 0 0 21 54 42 413 0 0 21 11 230 1 0 22 12 262 0 0 22 13 217 0 0 22 54 39 455 0 0 22 54 40 493 0 0 22 54 41 489 0 0 22 54 42 431 0 0 22 11 1002 1 1 31 12 1222 0 0 31 13 1198 0 0 31 54 39 1876 0 0 31 54 40 1565 0 0 31 54 41 1455 0 0 31 54 42 1427 0 0 31 11 1002 1 0 32 12 1246 0 0 32 13 1153 0 0 32 54 39 1813 0 0 32 54 40 1490 0 0 32 54 41 1518 0 0 32 54 42 1486 0 0 32 11 1002 1 0 33 12 1229 0 0 33 13 1142 0 0 33 54 39 1797 0 0 33 54 40 1517 0 0 33 54 41 1527 0 0 33 54 42 1514 0 0 33 [[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] idiom for constructing data frame
On 31 Mar 2015, at 20:55 , William Dunlap wdun...@tibco.com wrote: You can use structure() to attach the names to a list that is input to data.frame. E.g., dfNames - c(First, Second Name) data.frame(lapply(structure(dfNames, names=dfNames), function(name)rep(NA_real_, 5))) Yes, I cooked up something similar: names - c(foo,bar,baz) names(names) - names # confuse 'em as.data.frame(lapply(names, function(x) rep(NA_real_,10))) but wouldn't it be more to the point to do df - as.data.frame(rep(list(rep(NA_real_, 10)),3)) names(df) - names ? The lapply() approach could be generalized to a vector of column classes, though. A general solution looks impracticable; once you start considering how to specify factor columns with each their own level set, things get a bit out of hand. -pd Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Mar 31, 2015 at 11:37 AM, Sarah Goslee sarah.gos...@gmail.com wrote: Hi, Duncan Murdoch suggested: The matrix() function has a dimnames argument, so you could do this: names - c(strat, id, pid) data.frame(matrix(NA, nrow=10, ncol=3, dimnames=list(NULL, names))) That's a definite improvement, thanks. But no way to skip matrix()? It just seems unRlike, although since it's only full of NA values there are no coercion issues with column types or anything, so it doesn't hurt. It's just inelegant. :) 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. [[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. -- 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] idiom for constructing data frame
but wouldn't it be more to the point to do df - as.data.frame(rep(list(rep(NA_real_, 10)),3)) names(df) - names As a matter of personal style (and functional programming sensibility), I prefer not to make named objects and then modify them. Also, the names coming out of that as.data.frame call are exceedingly ugly and I'd rather not generate them at all. Also adding the names after calling data.frame means can give different results than passing them into data.frame(), which can mangle nonsyntactic names like Second Name into Second.Name. It is often preferable, but it is different. Bill Dunlap TIBCO Software wdunlap tibco.com On Fri, Apr 3, 2015 at 5:51 AM, peter dalgaard pda...@gmail.com wrote: On 31 Mar 2015, at 20:55 , William Dunlap wdun...@tibco.com wrote: You can use structure() to attach the names to a list that is input to data.frame. E.g., dfNames - c(First, Second Name) data.frame(lapply(structure(dfNames, names=dfNames), function(name)rep(NA_real_, 5))) Yes, I cooked up something similar: names - c(foo,bar,baz) names(names) - names # confuse 'em as.data.frame(lapply(names, function(x) rep(NA_real_,10))) but wouldn't it be more to the point to do df - as.data.frame(rep(list(rep(NA_real_, 10)),3)) names(df) - names ? The lapply() approach could be generalized to a vector of column classes, though. A general solution looks impracticable; once you start considering how to specify factor columns with each their own level set, things get a bit out of hand. -pd Bill Dunlap TIBCO Software wdunlap tibco.com On Tue, Mar 31, 2015 at 11:37 AM, Sarah Goslee sarah.gos...@gmail.com wrote: Hi, Duncan Murdoch suggested: The matrix() function has a dimnames argument, so you could do this: names - c(strat, id, pid) data.frame(matrix(NA, nrow=10, ncol=3, dimnames=list(NULL, names))) That's a definite improvement, thanks. But no way to skip matrix()? It just seems unRlike, although since it's only full of NA values there are no coercion issues with column types or anything, so it doesn't hurt. It's just inelegant. :) 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. [[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. -- 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 [[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] applying cumsum within groups
ave() is your friend (unfortunately named as it may be): ave(dat$seq, dat$ts, FUN=cumsum) [1] 1 1 1 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 1 1 1 1 1 1 1 2 2 2 2 2 2 2 1 1 1 [39] 1 1 1 1 2 2 2 2 2 2 2 3 3 3 3 3 3 3 On 03 Apr 2015, at 14:17 , Morway, Eric emor...@usgs.gov wrote: This small example will be applied to a problem with 1.4e6 lines of data. First, here is the dataset and a few lines of R script, followed by an explanation of what I'd like to get: dat - read.table(textConnection(ISEG IRCH val 11 265 12 260 13 234 54 39 467 54 40 468 54 41 460 54 42 489 11 265 12 276 13 217 54 39 456 54 40 507 54 41 483 54 42 457 11 265 12 287 13 224 54 39 473 54 40 502 54 41 497 54 42 447 11 230 12 251 13 199 54 39 439 54 40 474 54 41 477 54 42 413 11 230 12 262 13 217 54 39 455 54 40 493 54 41 489 54 42 431 11 1002 12 1222 13 1198 54 39 1876 54 40 1565 54 41 1455 54 42 1427 11 1002 12 1246 13 1153 54 39 1813 54 40 1490 54 41 1518 54 42 1486 11 1002 12 1229 13 1142 54 39 1797 54 40 1517 54 41 1527 54 42 1514),header=TRUE) dat$seq - ifelse(dat$ISEG==1 dat$IRCH==1, 1, 0) tmp - diff(dat[dat$seq==1,]$val)!=0 dat$idx - 0 dat[dat$seq==1,][c(TRUE,tmp),]$idx - 1 dat$ts - cumsum(dat$idx) At this point, I'd like to add one more column called iter that counts up by 1 based on seq, but within each ts. So, the result would look like this (undoubtedly this is a simple problem with something like ddply, but I've been unable to construct the R for it): dat ISEG IRCH val seq idx ts iter 11 265 1 1 11 12 260 0 0 11 13 234 0 0 11 54 39 467 0 0 11 54 40 468 0 0 11 54 41 460 0 0 11 54 42 489 0 0 11 11 265 1 0 12 12 276 0 0 12 13 217 0 0 12 54 39 456 0 0 12 54 40 507 0 0 12 54 41 483 0 0 12 54 42 457 0 0 12 11 265 1 0 13 12 287 0 0 13 13 224 0 0 13 54 39 473 0 0 13 54 40 502 0 0 13 54 41 497 0 0 13 54 42 447 0 0 13 11 230 1 1 21 12 251 0 0 21 13 199 0 0 21 54 39 439 0 0 21 54 40 474 0 0 21 54 41 477 0 0 21 54 42 413 0 0 21 11 230 1 0 22 12 262 0 0 22 13 217 0 0 22 54 39 455 0 0 22 54 40 493 0 0 22 54 41 489 0 0 22 54 42 431 0 0 22 11 1002 1 1 31 12 1222 0 0 31 13 1198 0 0 31 54 39 1876 0 0 31 54 40 1565 0 0 31 54 41 1455 0 0 31 54 42 1427 0 0 31 11 1002 1 0 32 12 1246 0 0 32 13 1153 0 0 32 54 39 1813 0 0 32 54 40 1490 0 0 32 54 41 1518 0 0 32 54 42 1486 0 0 32 11 1002 1 0 33 12 1229 0 0 33 13 1142 0 0 33 54 39 1797 0 0 33 54 40 1517 0 0 33 54 41 1527 0 0 33 54 42 1514 0 0 33 [[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. -- 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] Color US counties on US map using a numeric variable for color intensity
Dimitri,
To answer your questions:
The colorRamp() function creates a new function, newpal().
The value returned by newpal() is a numeric matrix of RGB color values.
The rgb() function is then used to convert this numeric matrix to colors,
with the argument maxColorValue
giving the maximum of the color values range. Typically either 255 or 1.
See the help files for more information
?colorRamp
?rgb
I think Jim Lemon's suggestion to use color.scale() function is a handier
solution.
Jean
On Thu, Apr 2, 2015 at 6:05 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This is really cool, Jim - thanks a lot!
On Thu, Apr 2, 2015 at 6:18 PM, Jim Lemon drjimle...@gmail.com wrote:
Hi Dimitri,
You can also try the color.scale function in plotrix, which allows you to
specify the NA color in the call.
newcol-color.scale(mydata.final$Mean.Wait,extremes=c(yellow,red),na.color=white)
Jim
On Fri, Apr 3, 2015 at 8:08 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Jean, I think I fixed it:
newpal - colorRamp(c(yellow, red))
missing - is.na(mydata.final$Mean.Wait)
newcol - ifelse(missing, white,
rgb(newpal(mydata.final$Mean.Wait[!is.na(mydata.final$Mean.Wait)]/
max(mydata.final$Mean.Wait,
na.rm=T)), maxColorValue=255))
map('county', fill=TRUE, col=newcol,
resolution=0, lty=0, bg=transparent)
map('state', lwd=1, add=TRUE)
One understanding question: what exactly does this rgb line do and why
do we have to say maxColorValue=255?
Thank you!
On Thu, Apr 2, 2015 at 5:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you, Jean, but I think this newcol line is not working. I am
running:
newcol - ifelse(missing, white,
rgb(newpal(mydata.final$Mean.Wait/max(mydata.final$Mean.Wait,
na.rm=T)),
maxColorValue=255))
# And I am getting:
Error in rgb(newpal(mydata.final$Mean.Wait/max(mydata.final$Mean.Wait,
:
color intensity NA, not in 0:255
I think it's not liking the NAs - despite the ifelse...
On Thu, Apr 2, 2015 at 4:26 PM, Adams, Jean jvad...@usgs.gov wrote:
Dimitri,
You could use colorRamp() and rgb() to get more continuous colors.
For example
newpal - colorRamp(c(yellow, red))
missing - is.na(mydata.final$Mean.Wait)
newcol - ifelse(missing, white,
rgb(newpal(mydat$Mean.Wait/max(mydat$Mean.Wait)),
maxColorValue=255))
map('county', fill=TRUE, col=newcol,
resolution=0, lty=0, bg=transparent)
map('state', lwd=1, add=TRUE)
Jean
On Thu, Apr 2, 2015 at 12:03 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
I have a data frame 'mydata.final' (see below) that contains US
counties and a continuous numeric variable 'Mean.Wait' that ranges
from zero to 10 or so. I also created variable 'wait' that is based
on
the 'Mean.Wait' and takes on discrete values from 1 (lowest values
on
'Mean.Wait') to 5 (highest values on 'Mean.Wait').
I can create a map of the US with the counties colored based on the
values of 'wait' using R package 'maps':
#
### Generating an artificial data file:
#
library(maps)
mydata.final - data.frame(county = (map('county', plot =
FALSE)$names),
stringsAsFactors = F)
### My numeric variable:
set.seed(123)
mydata.final$Mean.Wait - runif(nrow(mydata.final)) * 10
### Introducing NAs to mimic my real data set:
set.seed(1234)
mydata.final$Mean.Wait[sample(1:nrow(mydata.final), 1500)] - NA
### Cutting the original numeric variable into categories
### because I don't know how to color based on 'Mean.Wait':
mydata.final$wait - cut(mydata.final$Mean.Wait, breaks = 5)
levels(mydata.final$wait) - 1:5
mydata.final$wait - as.numeric(as.character(mydata.final$wait))
Building a US map based on 'wait' (5 categories)
#
### Creating my 5 colors:
pal - colorRampPalette(c(yellow, red))
allcolors - pal(5)
### Looking at my 5 colors:
barplot(1:5, rep(1,5), col = allcolors, horiz = T)
### Builiding the US map using 5 categories in 'wait':
map('county', fill = TRUE, col = allcolors[mydata.final$wait],
resolution = 0, lty = 0, bg = transparent)
map('state', lwd=1, add=TRUE)
My goal is: instead of splitting 'Mean.Wait' into 5 ordered
categories
('wait'), I'd like to color the counties on the map based on the
intensity of my (continuous) 'Mean.Wait'. What would be the way to
do
it and maybe even to add a legend?
Thanks a lot!
--
Dimitri Liakhovitski
[R] question on waveletcomp plot image
Dear R experts, I have used waveletcomp package of R and was trying to get the dates formatted as month year but get below error while trying it with the example provided in http://www.hs-stat.com/projects/WaveletComp/WaveletComp_guided_tour.pdf Kindly help me with the dateformat so that I can get %Y-%m as xaxis label data(FXtrade.transactions) my.data.a = FXtrade.transactions[FXtrade.transactions$active == T, ] my.w.a = analyze.wavelet( + my.data.a, transactions, + loess.span = 0.0, # no detrending required + dt = 1/(12*24), + # one day has 12*24 5-minute time slots + dj = 1/250, + # resolution along period axis + lowerPeriod = 1/8, # lowest period of interest: 3 hours + make.pval = T, + # draws white lines indicating significance + n.sim = 10) wt.image(my.w.a, n.levels = 250, periodlab = periods (days), + legend.params = list(lab = wavelet power levels), + show.date = T, date.format = %Y %m, timelab = ) Error in plot.window(...) : need finite 'xlim' values In addition: Warning messages: 1: In min(x) : no non-missing arguments to min; returning Inf 2: In max(x) : no non-missing arguments to max; returning -In with best regards Sudheer ** Dr. Sudheer Joseph Scientist, INCOIS, MoES, Govt. of India. OCEAN VALLEY , Pragathi Nagar (BO), Nizampet SO, Telangana, India. PIN- 500 090. Tel:+91-9440832534(Mobile) Tel:+91-40-23886047(O),Fax:+91-40-23892910(O) E-mail: sjo.in...@gmail.com; s...@incois.gov.in. ---* The ultimate measure of a man is not where he stands in moments of comfort and convenience, but where he stands at times of challenge and controversy. Martin Luther King, Jr. *** [[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] Cut breaks in descending order
Hi,
Is it a requirement to provide the break intervals of the cut function in
ascending order? The help documentation for cut didn't specify this but the
labels returned are reversed if I indicate the break intervals in a descending
order. Here is an example
tbl-data.frame(x=c(0:10))
tbl$ascending-cut(tbl$x, breaks=c(0,3,5,9,99),
labels=c('3','4-5','6-9','9'), include.lowest=T)
tbl$descending-cut(tbl$x, breaks=c(99,9,5,3,0),
labels=c('9','6-8','4-5','3'), include.lowest=T)
tbl
x ascending descending
1 0 3 9
2 1 3 9
3 2 3 9
4 3 3 9
5 4 4-5 6-8
6 5 4-5 6-8
7 6 6-9 4-5
8 7 6-9 4-5
9 8 6-9 4-5
10 9 6-9 4-5
11 10 9 3
Appreciate any guidance on this.
Regards
Wing Keong
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] applying cumsum within groups
On Apr 3, 2015, at 5:17 AM, Morway, Eric wrote: This small example will be applied to a problem with 1.4e6 lines of data. First, here is the dataset and a few lines of R script, followed by an explanation of what I'd like to get: dat - read.table(textConnection(ISEG IRCH val 11 265 12 260 13 234 54 39 467 54 40 468 54 41 460 54 42 489 11 265 12 276 13 217 54 39 456 54 40 507 54 41 483 54 42 457 11 265 12 287 13 224 54 39 473 54 40 502 54 41 497 54 42 447 11 230 12 251 13 199 54 39 439 54 40 474 54 41 477 54 42 413 11 230 12 262 13 217 54 39 455 54 40 493 54 41 489 54 42 431 11 1002 12 1222 13 1198 54 39 1876 54 40 1565 54 41 1455 54 42 1427 11 1002 12 1246 13 1153 54 39 1813 54 40 1490 54 41 1518 54 42 1486 11 1002 12 1229 13 1142 54 39 1797 54 40 1517 54 41 1527 54 42 1514),header=TRUE) dat$seq - ifelse(dat$ISEG==1 dat$IRCH==1, 1, 0) tmp - diff(dat[dat$seq==1,]$val)!=0 dat$idx - 0 dat[dat$seq==1,][c(TRUE,tmp),]$idx - 1 dat$ts - cumsum(dat$idx) At this point, I'd like to add one more column called iter that counts up by 1 based on seq, but within each ts. So, the result would look like this (undoubtedly this is a simple problem with something like ddply, but I've been unable to construct the R for it): dat$iter2 - ave(dat$seq, dat$ts,FUN=cumsum) dat ISEG IRCH val seq idx ts iter iter2 1 11 265 1 1 1 1_1 1 2 12 260 0 0 1 1_1 1 3 13 234 0 0 1 1_1 1 454 39 467 0 0 1 1_1 1 554 40 468 0 0 1 1_1 1 654 41 460 0 0 1 1_1 1 754 42 489 0 0 1 1_1 1 8 11 265 1 0 1 1_2 2 9 12 276 0 0 1 1_2 2 1013 217 0 0 1 1_2 2 11 54 39 456 0 0 1 1_2 2 12 54 40 507 0 0 1 1_2 2 13 54 41 483 0 0 1 1_2 2 14 54 42 457 0 0 1 1_2 2 1511 265 1 0 1 1_3 3 1612 287 0 0 1 1_3 3 1713 224 0 0 1 1_3 3 18 54 39 473 0 0 1 1_3 3 19 54 40 502 0 0 1 1_3 3 20 54 41 497 0 0 1 1_3 3 21 54 42 447 0 0 1 1_3 3 2211 230 1 1 2 2_4 1 2312 251 0 0 2 2_4 1 snipped- -- David dat ISEG IRCH val seq idx ts iter 11 265 1 1 11 12 260 0 0 11 13 234 0 0 11 54 39 467 0 0 11 54 40 468 0 0 11 54 41 460 0 0 11 54 42 489 0 0 11 11 265 1 0 12 12 276 0 0 12 13 217 0 0 12 54 39 456 0 0 12 54 40 507 0 0 12 54 41 483 0 0 12 54 42 457 0 0 12 11 265 1 0 13 12 287 0 0 13 13 224 0 0 13 54 39 473 0 0 13 54 40 502 0 0 13 54 41 497 0 0 13 54 42 447 0 0 13 11 230 1 1 21 12 251 0 0 21 13 199 0 0 21 54 39 439 0 0 21 54 40 474 0 0 21 54 41 477 0 0 21 54 42 413 0 0 21 11 230 1 0 22 12 262 0 0 22 13 217 0 0 22 54 39 455 0 0 22 54 40 493 0 0 22 54 41 489 0 0 22 54 42 431 0 0 22 11 1002 1 1 31 12 1222 0 0 31 13 1198 0 0 31 54 39 1876 0 0 31 54 40 1565 0 0 31 54 41 1455 0 0 31 54 42 1427 0 0 31 11 1002 1 0 32 12 1246 0 0 32 13 1153 0 0 32 54 39 1813 0 0 32 54 40 1490 0 0 32 54 41 1518 0 0 32 54 42 1486 0 0 32 11 1002 1 0 33 12 1229 0 0 33 13 1142 0 0 33 54 39 1797 0 0 33 54 40 1517 0 0 33 54 41 1527 0 0 33 54 42 1514 0 0 33 [[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. 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
[R] open xlsx file using read.xls function of gdata package
Dear all, I am trying to open excel files using the gdata package. I can do that using a .xls file, but the same file, containing the same data, formatted in .xlsx gives error (R does not recognize the pattern from where to start reading the data). Doen anybody knows whether it is possible to read .xlslx with this package? Am I missing another package to implement the reading of the .xlsx? Thank you Luigi PS: this is the error I get: my.file - array.xlsx my.data-read.xls( + my.file, + sheet=sheet x, + verbose=FALSE, + pattern=row name, + na.strings=c(NA,#DIV/0!), + method=tab, + perl=perl + ) Warning message: In read.xls(my.file, sheet = sheet x, verbose = FALSE, : pattern not found The verbose version runs like this: “array.xlsx” to tab file “/tmp/Rtmp2tAjzz/filef06102dd018.tab” ... Executing ' '/usr/bin/perl' '/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl' 'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '... Loading 'array.xlsx'... Done. Orignal Filename: array.xlsx Number of Sheets: 2 Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab' Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4 (Ignored 0 blank lines.) 0 Done. Searching for lines tfntaining pattern row name ... Warning message: In read.xls(my.file, sheet = sheet x, verbose = TRUE, : pattern not found __ 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] Cut breaks in descending order
On Apr 3, 2015, at 5:09 AM, Wing Keong Lew wrote:
Hi,
Is it a requirement to provide the break intervals of the cut function in
ascending order?
Apparently not. I get teh sam splits even with random permutations. It is
apparently a requirement to make sure you labels match the sorted order of
the breaks.
The findInterval function does require that its `vec` argument be
non-decreasing, but I do not see a discussion of break order in the help page.
Looking at cut.default the first think it does to the breaks is sort them.
#... snipped code that deals with length(breaks)==1
else nb - length(breaks - sort.int(as.double(breaks)))
--
David.
The help documentation for cut didn't specify this but the labels returned
are reversed if I indicate the break intervals in a descending order. Here is
an example
tbl-data.frame(x=c(0:10))
tbl$ascending-cut(tbl$x, breaks=c(0,3,5,9,99),
labels=c('3','4-5','6-9','9'), include.lowest=T)
tbl$descending-cut(tbl$x, breaks=c(99,9,5,3,0),
labels=c('9','6-8','4-5','3'), include.lowest=T)
tbl
x ascending descending
1 039
2 139
3 239
4 339
5 4 4-5 6-8
6 5 4-5 6-8
7 6 6-9 4-5
8 7 6-9 4-5
9 8 6-9 4-5
10 9 6-9 4-5
11 1093
Appreciate any guidance on this.
Regards
Wing Keong
__
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David Winsemius
Alameda, CA, USA
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Re: [R] open xlsx file using read.xls function of gdata package
I had poor luck with gdata. I have had better luck with XLConnect. There is no single best package for this, since each seems to leverage efforts made in other languages (so there are non-R configuration requirements to keep working) and Excel is a proprietary moving target. In general YMMV when it comes to Excel data. In most cases I just export the data to CSV and avoid the issue. --- 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 April 3, 2015 11:24:29 AM PDT, Luigi Marongiu marongiu.lu...@gmail.com wrote: Dear all, I am trying to open excel files using the gdata package. I can do that using a .xls file, but the same file, containing the same data, formatted in .xlsx gives error (R does not recognize the pattern from where to start reading the data). Doen anybody knows whether it is possible to read .xlslx with this package? Am I missing another package to implement the reading of the .xlsx? Thank you Luigi PS: this is the error I get: my.file - array.xlsx my.data-read.xls( + my.file, + sheet=sheet x, + verbose=FALSE, + pattern=row name, + na.strings=c(NA,#DIV/0!), + method=tab, + perl=perl + ) Warning message: In read.xls(my.file, sheet = sheet x, verbose = FALSE, : pattern not found The verbose version runs like this: “array.xlsx” to tab file “/tmp/Rtmp2tAjzz/filef06102dd018.tab” ... Executing ' '/usr/bin/perl' '/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl' 'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '... Loading 'array.xlsx'... Done. Orignal Filename: array.xlsx Number of Sheets: 2 Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab' Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4 (Ignored 0 blank lines.) 0 Done. Searching for lines tfntaining pattern row name ... Warning message: In read.xls(my.file, sheet = sheet x, verbose = TRUE, : pattern not found __ 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] Kruskal-Wallace power calculations.
Thank you very much Greg, I will give that a try.
Best,
Collin.
On Fri, Apr 3, 2015 at 1:43 PM, Greg Snow 538...@gmail.com wrote:
Here is some sample code:
## Simulation function to create data, analyze it using
## kruskal.test, and return the p-value
## change rexp to change the simulation distribution
simfun - function(means, k=length(means), n=rep(50,k)) {
mydata - lapply( seq_len(k), function(i) {
rexp(n[i], 1) - 1 + means[i]
})
kruskal.test(mydata)$p.value
}
# simulate under the null to check proper sizing
B - 1
out1 - replicate(B, simfun(rep(3,4)))
hist(out1)
mean( out1 = 0.05 )
binom.test( sum(out1 = 0.05), B, p=0.05)
### Now simulate for power
B - 1
out2 - replicate(B, simfun( c(3,3,3.2,3.3)))
hist(out2)
mean( out2 = 0.05 )
binom.test( sum(out2 = 0.05), B, p=0.05 )
This simulates from a continuous exponential (skewed) and shifts to
get the means (shifted location is a common assumption, though not
required for the actual test).
On Thu, Apr 2, 2015 at 8:19 PM, Collin Lynch cfly...@ncsu.edu wrote:
Thank you Jim, I did see those (though not my typo :) and am still
pondering the warning about post-hoc analyses.
The situation that I am in is that I have a set of individuals who
have been assigned a course grade. We have then clustered these
individuals into about 50 communities using standard community
detection algorithms with the goal of determining whether community
membership affects one of their grades. We are using the KW test as
the grade data is strongly non-normal and my coauthors preferred KW as
an alternative.
The two issues that I am struggling with are: 1) whether the post-hoc
power analysis would be useful; and 2) how to code the simulation
studies that are described in:
http://onlinelibrary.wiley.com/doi/10.1002/bimj.4710380510/abstract
Problem #1 is of course beyond the scope of this e-mail list though I
would welcome anyone's suggestions on that point. I am not sure that
I buy the arguments against it offered here:
http://graphpad.com/support/faq/why-it-is-not-helpful-to-compute-the-power-of-an-experiment-to-detect-the-difference-actually-observed-why-is-post-hoc-power-analysis-futile/
It seems that the rationale boils down to you didn't find it so you
couldn't find it but that does not tell me how far off I was from the
goal. I am still perusing the articles the author cites however.
With respect to question #2 I am trying to lay my hands on the article
and did find this old r-help discussion:
http://r.789695.n4.nabble.com/Power-of-Kruskal-Wallis-Test-td4671188.html
however I am not sure how to adapt the simulation studies that it
links to to my current problem. The links it leads to focus on
mixed-effects models. This may be more of a pure stats question and
not suited for this list but I thought I'd ask in the hopes that
anyone had any more specific KW code or knew of a good tutorial for
the right kinds of simulation studies.
Thank you,
Collin.
On Thu, Apr 2, 2015 at 6:35 PM, Jim Lemon drjimle...@gmail.com wrote:
Hi Collin,
Have a look at this:
http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
Although, thinking about it, this might have constituted your perusal of
the literature.
Plus it always looks better when you spell the names properly
Jim
On Fri, Apr 3, 2015 at 2:23 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
Please stop... you are acting like a broken record, and are also posting
in HTML format. Please read the Posting Guide and demonstrate that you have
used a search engine on this topic before posting again.
---
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 April 2, 2015 7:25:20 AM PDT, Collin Lynch cfly...@ncsu.edu wrote:
Greetings, I am working on a project where we are applying the
Kruskal-Wallace test to some factor data to evaluate their correlation
with
existing grade data. I know that the grade data is nonnormal therefore
we
cannot rely on ANOVA or a similar parametric test. What I would like
to
find is a mechanism for making power calculations for the KW test given
the
nonparametric assumptions. My perusal of the literature has suggested
that
a simulation would be the best method.
Can anyone point me to good examples of such simulations for KW in R?
And
does anyone have a favourite package for generating simulated
Re: [R] Repeated failures to install caret package (of Max Kuhn)
Hi Ron, Without knowing more it's really hard to help. What error messages are you getting? failed and no success are utterly uninformative - there are many things that could be wrong. The first thing, though, is always to check whether you have the necessary dependencies installed, eg devel libraries. Sarah On Fri, Apr 3, 2015 at 5:07 PM, Ronald Wyllys wyl...@ischool.utexas.edu wrote: For an edx course, MIT's The Analtics Edge, I need to install the caret package that was originated and is maintained by Dr. Max Kuhn of Pfizer. So far, every effort I've made to try to install.packages(caret) has failed. (I'm using R v. 3.1.3 and RStudio v. 0.98.1103 in LinuxMint 17.1) Here are some of the things I've tried unsuccessfully: install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;)) install.packages(caret, dependencies=TRUE) install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;), dependencies=TRUE) install.packages(caret, dependencies = c(Depends, Suggests)) install.packages(caret, repos=http://cran.rstudio.com/;) I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas, and tried the above installs again, unsuccessfully. I've succeeded in individually installing a number of packages on which caret appears to be dependent. Specifically, I've been able to install nloptr, minqa, Rcpp, reshape2, stringr, and scales. But I've had no success with trying to do individual installs of BradleyTerry2, car, lme4, quantreg, and RcppEigen. Any suggestions will be very gratefully received (and tried out quickly). Thanks in advance. Ron Wyllys -- 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] WEIBULL or EXPONENTIAL?
Hi Xavier, I use the fitdistrplus and logspline packages to know which distribution fits better my data. Here is an example : install.packages(fitdistrplus) library(fitdistrplus) instal.packages(logspline) library(logspline) x=c(44986,18288,56147,44488,41018,40631,27301,39025,45688,47172,12300,21558,16103,48874,67245,36119,10398,42630,12879,34058,84443,30639) descdist(x,discrete=FALSE) Cheers, S. De : CHIRIBOGA Xavier xavier.chirib...@unine.ch À : r-help@r-project.org r-help@r-project.org Envoyé le : Vendredi 3 avril 2015 16h33 Objet : [R] WEIBULL or EXPONENTIAL? Dear members, I am doing a survival analysis wiith the function coxph...however I am wondering how can I know if my data follows a EXPONENTIAL or WEIBULL distribution? I have 3 censored datum. Using R studio. Thanks for the suggestions, Xavier __ 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] Repeated failures to install caret package (of Max Kuhn)
For an edx course, MIT's The Analtics Edge, I need to install the caret package that was originated and is maintained by Dr. Max Kuhn of Pfizer. So far, every effort I've made to try to install.packages(caret) has failed. (I'm using R v. 3.1.3 and RStudio v. 0.98.1103 in LinuxMint 17.1) Here are some of the things I've tried unsuccessfully: install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;)) install.packages(caret, dependencies=TRUE) install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;), dependencies=TRUE) install.packages(caret, dependencies = c(Depends, Suggests)) install.packages(caret, repos=http://cran.rstudio.com/;) I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas, and tried the above installs again, unsuccessfully. I've succeeded in individually installing a number of packages on which caret appears to be dependent. Specifically, I've been able to install nloptr, minqa, Rcpp, reshape2, stringr, and scales. But I've had no success with trying to do individual installs of BradleyTerry2, car, lme4, quantreg, and RcppEigen. Any suggestions will be very gratefully received (and tried out quickly). Thanks in advance. Ron Wyllys __ 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] Repeated failures to install caret package (of Max Kuhn)
What is the error message? Best, Uwe Ligges On 03.04.2015 23:07, Ronald Wyllys wrote: For an edx course, MIT's The Analtics Edge, I need to install the caret package that was originated and is maintained by Dr. Max Kuhn of Pfizer. So far, every effort I've made to try to install.packages(caret) has failed. (I'm using R v. 3.1.3 and RStudio v. 0.98.1103 in LinuxMint 17.1) Here are some of the things I've tried unsuccessfully: install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;)) install.packages(caret, dependencies=TRUE) install.packages(caret, repos=c(http://rstudio.org/_packages;, http://cran.rstudio.com;), dependencies=TRUE) install.packages(caret, dependencies = c(Depends, Suggests)) install.packages(caret, repos=http://cran.rstudio.com/;) I've changed my CRAN mirror from UCLA to Revolution Analytics in Dallas, and tried the above installs again, unsuccessfully. I've succeeded in individually installing a number of packages on which caret appears to be dependent. Specifically, I've been able to install nloptr, minqa, Rcpp, reshape2, stringr, and scales. But I've had no success with trying to do individual installs of BradleyTerry2, car, lme4, quantreg, and RcppEigen. Any suggestions will be very gratefully received (and tried out quickly). Thanks in advance. Ron Wyllys __ 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] species names on a RDA plot
Dear R users I'm trying to do a RDA analysis based on Borcard et al. 2011 Numerical Ecology with R examples. What I cannot understand is why when I run the script (RDA.R) using the dataset from book (dataset1.rar) RDA triplot shows species names and when I use my dataset (dataset2.rar) species names are not shown. Data and script can be downloaded at https://app.box.com/s/oayq7tglbmdsu72fj05h83dlzclsiglg Does anyone know why this happens? Thanks for any clue. Best regards Antônio Olinto Ávila da Silva Fisheries Institute São Paulo, Brasil [[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] command help
Dear colleagues, I wanted to run a command to do iteration for the following equation: Pj+1=TT11+FtransposeTT22F-FtransposeTT21-TT12 F+y[(Atranspose-FtransposeBtranspose)P(A-BF)] where TT11, TT22,TT12, TT21, F transpose, A, B and F are matrices and y is a scalar we want to find the iteration for P using loop. Can you send me the right command for the loop for the above equation? Thank you in advance, Ali [[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] open xlsx file using read.xls function of gdata package
You might try the readxl package - it's only available on github but it reads both xlsx and xls. All going well, it should be on its way to CRAN next week. Hadley On Friday, April 3, 2015, Luigi Marongiu marongiu.lu...@gmail.com wrote: Dear all, I am trying to open excel files using the gdata package. I can do that using a .xls file, but the same file, containing the same data, formatted in .xlsx gives error (R does not recognize the pattern from where to start reading the data). Doen anybody knows whether it is possible to read .xlslx with this package? Am I missing another package to implement the reading of the .xlsx? Thank you Luigi PS: this is the error I get: my.file - array.xlsx my.data-read.xls( + my.file, + sheet=sheet x, + verbose=FALSE, + pattern=row name, + na.strings=c(NA,#DIV/0!), + method=tab, + perl=perl + ) Warning message: In read.xls(my.file, sheet = sheet x, verbose = FALSE, : pattern not found The verbose version runs like this: “array.xlsx” to tab file “/tmp/Rtmp2tAjzz/filef06102dd018.tab” ... Executing ' '/usr/bin/perl' '/home/gigiux/R/x86_64-pc-linux-gnu-library/3.0/gdata/perl/xls2tab.pl' 'array.xlsx' '/tmp/Rtmp2tAjzz/filef06102dd018.tab' 'sheet x' '... Loading 'array.xlsx'... Done. Orignal Filename: array.xlsx Number of Sheets: 2 Writing sheet 'sheet x' to file '/tmp/Rtmp2tAjzz/filef06102dd018.tab' Minrow=31 Maxrow=17310 Mincol=0 Maxcol=4 (Ignored 0 blank lines.) 0 Done. Searching for lines tfntaining pattern row name ... Warning message: In read.xls(my.file, sheet = sheet x, verbose = TRUE, : pattern not found __ R-help@r-project.org javascript:; 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. -- http://had.co.nz/ [[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] glmnet: converting coefficients back to original scale
This is interesting, can you post your lm.ridge solution as well? I suspect in glmnet, you need to use model.matrix with intercept, that could be the reason. -m __ 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] command help
I cannot make sense of your email. This is partly due to your use of HTML, which the Posting Guidelines warn you not to do. The symbolic language used on this mailing list is R. The linear algebra operations available in R are rather straightforward. Please read the documentation and convey your request using R do we can understand 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 April 3, 2015 11:16:11 AM PDT, ali alRubaiee ali.alrubaiee1...@gmail.com wrote: Dear colleagues, I wanted to run a command to do iteration for the following equation: Pj+1=TT11+FtransposeTT22F-FtransposeTT21-TT12 F+y[(Atranspose-FtransposeBtranspose)P(A-BF)] where TT11, TT22,TT12, TT21, F transpose, A, B and F are matrices and y is a scalar we want to find the iteration for P using loop. Can you send me the right command for the loop for the above equation? Thank you in advance, Ali [[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] Kruskal-Wallace power calculations.
Here is some sample code:
## Simulation function to create data, analyze it using
## kruskal.test, and return the p-value
## change rexp to change the simulation distribution
simfun - function(means, k=length(means), n=rep(50,k)) {
mydata - lapply( seq_len(k), function(i) {
rexp(n[i], 1) - 1 + means[i]
})
kruskal.test(mydata)$p.value
}
# simulate under the null to check proper sizing
B - 1
out1 - replicate(B, simfun(rep(3,4)))
hist(out1)
mean( out1 = 0.05 )
binom.test( sum(out1 = 0.05), B, p=0.05)
### Now simulate for power
B - 1
out2 - replicate(B, simfun( c(3,3,3.2,3.3)))
hist(out2)
mean( out2 = 0.05 )
binom.test( sum(out2 = 0.05), B, p=0.05 )
This simulates from a continuous exponential (skewed) and shifts to
get the means (shifted location is a common assumption, though not
required for the actual test).
On Thu, Apr 2, 2015 at 8:19 PM, Collin Lynch cfly...@ncsu.edu wrote:
Thank you Jim, I did see those (though not my typo :) and am still
pondering the warning about post-hoc analyses.
The situation that I am in is that I have a set of individuals who
have been assigned a course grade. We have then clustered these
individuals into about 50 communities using standard community
detection algorithms with the goal of determining whether community
membership affects one of their grades. We are using the KW test as
the grade data is strongly non-normal and my coauthors preferred KW as
an alternative.
The two issues that I am struggling with are: 1) whether the post-hoc
power analysis would be useful; and 2) how to code the simulation
studies that are described in:
http://onlinelibrary.wiley.com/doi/10.1002/bimj.4710380510/abstract
Problem #1 is of course beyond the scope of this e-mail list though I
would welcome anyone's suggestions on that point. I am not sure that
I buy the arguments against it offered here:
http://graphpad.com/support/faq/why-it-is-not-helpful-to-compute-the-power-of-an-experiment-to-detect-the-difference-actually-observed-why-is-post-hoc-power-analysis-futile/
It seems that the rationale boils down to you didn't find it so you
couldn't find it but that does not tell me how far off I was from the
goal. I am still perusing the articles the author cites however.
With respect to question #2 I am trying to lay my hands on the article
and did find this old r-help discussion:
http://r.789695.n4.nabble.com/Power-of-Kruskal-Wallis-Test-td4671188.html
however I am not sure how to adapt the simulation studies that it
links to to my current problem. The links it leads to focus on
mixed-effects models. This may be more of a pure stats question and
not suited for this list but I thought I'd ask in the hopes that
anyone had any more specific KW code or knew of a good tutorial for
the right kinds of simulation studies.
Thank you,
Collin.
On Thu, Apr 2, 2015 at 6:35 PM, Jim Lemon drjimle...@gmail.com wrote:
Hi Collin,
Have a look at this:
http://stats.stackexchange.com/questions/70643/power-analysis-for-kruskal-wallis-or-mann-whitney-u-test-using-r
Although, thinking about it, this might have constituted your perusal of
the literature.
Plus it always looks better when you spell the names properly
Jim
On Fri, Apr 3, 2015 at 2:23 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
Please stop... you are acting like a broken record, and are also posting
in HTML format. Please read the Posting Guide and demonstrate that you have
used a search engine on this topic before posting again.
---
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 April 2, 2015 7:25:20 AM PDT, Collin Lynch cfly...@ncsu.edu wrote:
Greetings, I am working on a project where we are applying the
Kruskal-Wallace test to some factor data to evaluate their correlation
with
existing grade data. I know that the grade data is nonnormal therefore
we
cannot rely on ANOVA or a similar parametric test. What I would like
to
find is a mechanism for making power calculations for the KW test given
the
nonparametric assumptions. My perusal of the literature has suggested
that
a simulation would be the best method.
Can anyone point me to good examples of such simulations for KW in R?
And
does anyone have a favourite package for generating simulated data or
conducting such tests?
Thank you,
Collin.
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__
Re: [R] Package build system adds line break in DESCRIPTION URL
Daniel Lewandowski daniel at nextpagesoft.net writes: Has anybody noticed that if field URL in DESCRIPTION contains a uri with 66 or more characters, then file DESCRIPTION in the resulting package includes a line break at the beginning? So this (source DESCRIPTION): URL: http://ecdc.europa.eu/en/data-tools/ seroincidence-calculator-tool/Pages/default.aspx becomes (again file DESCRIPTION, but inside the package) URL: http://ecdc.europa.eu/en/data-tools/seroincidence-calculator-tool/ Pages/default.aspx This has been tested with R on Windows 8.1 (devel 01/04/2015 and 3.1.3) and Linux Mint (3.1.3). It has many sad implications including not acceptance of such packages to CRAN. (links line-broken above to make gmane happy). Haven't tested, but seems it would not be *too* hard to trace through the code to see what's happening in the package-building process. Two thoughts: (1) as a workaround, could you use a URL-shortener such as tinyurl? tinyurl allows you to specify a somewhat meaningful name for the shortened URL, e.g. http://tinyurl.com/reproducible-000 (2) this feels like it is more suitable for r-de...@r-project.org Ben Bolker __ 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] WEIBULL or EXPONENTIAL?
Dear members, I am doing a survival analysis wiith the function coxph...however I am wondering how can I know if my data follows a EXPONENTIAL or WEIBULL distribution? I have 3 censored datum. Using R studio. Thanks for the suggestions, Xavier __ 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] Mapas con spplot
Hola,
Yo lo resuelvo usando grid.rect. Tienes una posible solución aquí:
https://github.com/oscarperpinan/spacetime-vis/blob/master/choropleth.R#L216
(que es una versión actualizada de este artículo
https://procomun.wordpress.com/2012/02/18/maps_with_r_1/)
Saludos.
Oscar.
-
Oscar Perpiñán Lamigueiro
Dpto. Ing. Eléctrica, Electrónica, Automática y Física Aplicada (ETSIDI-UPM)
Grupo de Sistemas Fotovoltaicos (IES-UPM)
URL: http://oscarperpinan.github.io
2015-04-02 23:33 GMT+02:00 Ivan Aguilar iagui...@gmail.com:
Hola
He hecho el típico mapa con gadm y spplot:
library(sp)
ES -
local(get(load(url(http://biogeo.ucdavis.edu/data/gadm2/R/ESP_adm2.RData
acerco las canarias:
A - which(ES@data$NAME_1 == Islas Canarias)
L - length(ES@polygons[[A]]@Polygons)
L - length(ES@polygons[[A]]@Polygons)
for (i in 1:L){
ES@polygons[[A]]@Polygons[[i]]@coords - cbind(ES@polygons
[[A]]@Polygons[[i]]@coords[,1]+6,ES@polygons
[[A]]@Polygons[[i]]@coords[,2]+5)
}
El tema es q quiero añadirle un recuadro a las islas usando spplot (o
lattice) ya que es lo q voy a usar.
Ayuda?
--
Ivan
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Re: [R] idiom for constructing data frame
On Tue, Mar 31, 2015 at 6:42 PM, Sarah Goslee sarah.gos...@gmail.com wrote: On Tue, Mar 31, 2015 at 6:35 PM, Richard M. Heiberger r...@temple.edu wrote: I got rid of the extra column. data.frame(r=seq(8), foo=NA, bar=NA, row.names=r) Brilliant! After much fussing, including a disturbing detour into nested lapply statements from which I barely emerged with my sanity (arguable, I suppose), here is a one-liner that creates a data frame of arbitrary number of rows given an existing data frame as template for column number and name: n - 8 df1 - data.frame(A=runif(9), B=runif(9)) do.call(data.frame, setNames(c(list(seq(n), r), as.list(rep(NA, ncol(df1, c(r, row.names, colnames(df1 It's not elegant, but it is fairly R-ish. I should probably stop hunting for an elegant solution now. Given a template df, you can create a new df with subsetting: df2 - df1[rep(NA_real_, 8), ] rownames(df2) - NULL df2 This has the added benefit of preserving the types. Hadley -- http://had.co.nz/ __ 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] WEIBULL or EXPONENTIAL?
As a start you can use an exploratory approach. Standard survival analysis texts show you how to use a log-log plot to assess whether a distribution is Weibull. Of course, the exponential is a special case of the Weibull. CHIRIBOGA Xavier xavier.chirib...@unine.ch Sent by: R-help r-help-boun...@r-project.org 04/03/2015 10:33 AM To r-help@r-project.org r-help@r-project.org, cc Subject [R] WEIBULL or EXPONENTIAL? Dear members, I am doing a survival analysis wiith the function coxph...however I am wondering how can I know if my data follows a EXPONENTIAL or WEIBULL distribution? I have 3 censored datum. Using R studio. Thanks for the suggestions, Xavier __ 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] idiom for constructing data frame
On 03 Apr 2015, at 16:46 , William Dunlap wdun...@tibco.com wrote: df - as.data.frame(rep(list(rep(NA_real_, 10)),3)) names(df) - names As a matter of personal style (and functional programming sensibility), I prefer not to make named objects and then modify them. Also, the names coming out of that as.data.frame call are exceedingly ugly and I'd rather not generate them at all. Ah, yes, I missed the generation of intermediate names. You can name the list before as.data.frame, though: l - rep(list(rep(NA_real_, 10)),3) names(l) - names as.data.frame(l) or as a one-liner: as.data.frame(structure(rep(list(rep(NA_real_, 10)), 3) , .Names=names)) -- 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] double-axis labels function of each other
Notepad should be fine but you will, I think, still have to correct the apostrophes if you are copying from Word. Any code written in Notepad should be fine. There are several dedicated editors or IDE's for use with R that you might want to look into. RStudio, Tinn-R and EMACs with ESS are some that come to mind. Among other things, they usually offer code highlighting which can be very useful for finding where there are typos, missing commas, and so on. I run Linux (Ubuntu) and find gedit with an R plug-in to be very good also. John Kane Kingston ON Canada -Original Message- From: hill0...@umn.edu Sent: Thu, 2 Apr 2015 23:59:17 -0700 (PDT) To: r-help@r-project.org Subject: Re: [R] double-axis labels function of each other Yes, I keep a copy in MS Word. Would Notepad be OK? I need sleep now, will work tomorrow. -- View this message in context: http://r.789695.n4.nabble.com/double-axis-labels-function-of-each-other-tp4705457p4705463.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.
