[R] Problems in setting up in MARSS package
I am trying to model a state space process using the MARSS package. My model has two unobservable states and 5 observable time series along with external covariates in the observation process only. None of the coefficients in either of the two processes are time varying. After running my setup and then fitting with MARSS with the fit=FALSE option turned on I get the following outputs. Model Structure is m: 2 state process(es) n: 5 observation time series Z : unconstrained A : time-varying R : unconstrained B : unconstrained U : unconstrained Q : unconstrained x0 : fixed V0 : identity Obviously I have no idea why A has been classfied as time varying. Can anyone help me as to how to do this correctly ? A is initialized like this A1 - zero The full code has been copied below #For two states B1 - matrix(c(b1,b2,b3,b4),2,2) #For 2 states U1 - matrix(c(u1,u2),2,1) #For 2 states Q1 - unconstrained Z1 - matrix(c(z1,z2,z3,z4,z5,z6,z7,z8,z9,z10),5,2) #For 2 states #Initial Parameters pi1 - matrix(c(1,2),2,1) V1 - diag(1,2) A1 - zero R1 - matrix(list(r11,r12,r13,r14,r15, r12,r22,r23,r24,r25, r13,r23,r33,r34,r35, r14,r24,r34,r44,r45, r15,r25,r35,r45,r55, r16,r26,r36,r46,r56) ,5,5) D - unconstrained d - covariate.data rownames(d) - rownames(covariate.data) model.list - list(B=B1,U=U1,Q=Q1,Z=Z1,A=A1,R=R1,D=D,d=d,x0=pi1,V0=V1,tinitx=0) fit - MARSS(yt,model=model.list,control=list(maxit=1,trace=0),fit=FALSE) Regards The woods are lovely, dark and deep But I have promises to keep And miles to go before I sleep And miles to go before I sleep - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] spacing problem in main title using car package scatterplot
On Aug 29, 2013, at 23:56 , John Fox wrote: Dear Gerard, Without your data, it's not possible to reproduce your problem exactly, but it's clear that it isn't specific to the scatterplot() function in the car package. For example, try plot(1:10) title(main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=, 100 ,))), adj=0) As far as I can tell, the issue is nothing to do with justification. It's because it is pasting the number onto a three-line block of text, so the longest line determines where the next thing goes. Using atop() might help, except that it doesn't left justify. BTW, Gerard, please don't create new threads by replying to old ones. On threading mail programs like OSX Mail, this drags in problems with abline from October 2010 and two other old threads... -Peter D. You should be able to adapt the following solution: plot(1:10) mtext(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28), side=3, line=2) mtext(paste(at Month 18 (N=, 100 ,), sep=), side=3, line=1) I hope this helps, John --- John Fox McMaster University Hamilton, Ontario, Canada -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Gerard Smits Sent: Thursday, August 29, 2013 5:00 PM To: r-help@r-project.org Subject: [R] spacing problem in main title using car package scatterplot Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ### ### ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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 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] Outliers Help
This is my a part of my data set
D[1:15,c(1,5:10)]
X. media IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010
1 1108 22.060.0 39 4.0 8.016.0 5.0
2 1479 110.0 NA NA53.0 1166.0 344.8 110.0
3 1591 86.6 247.0 8795.094.081.076.0
4 3408 807.0 302.0 322 621.0 1071.0 1301.0 1225.0
5 34239.0 NA NA NA 410.8 7.011.0
6 3872 103.25000 288.6 113 116.090.094.0 12036.6
7 5823 73.0 117.0 7080.074.069.072.0
8 6051 73.0 NA NA NA NA60.086.0
9 8099 125.16667 196.0 161 150.094.072.078.0
10 8100 70.0 NA NA NA NA48.092.0
11 10640 67.3 1256.61152 664.274.077.051.0
12 12600 2417.0 1960.02383 2453.0 2506.0 2758.0 2442.0
13 14680 38.030.0 61 373.642.019.0 220.8
14 14698 698.16667 553.0 664 847.0 800.0 679.0 646.0
15 17143 392.16667 323.0 322 434.0 383.0 459.0 432.0
I have done multiple imputation and now I have some outliers which I would
like to replace with the mean of this row or if it is possible with the mean of
the previos and the next value of this row, I mean for instance:
value 1 - Outlier- Value 2
I would like to replace the outlier with the mean of value 1 and value2, the
problem is that this values could be NA ( NA after the imputation because they
don't exist), in this case I would like to replace outlier with the mean of the
row.
An other problem I have is to detect correctly outlier values, for instance in
this example of data set for X=3872 and IE.2010, we can see an outlier, I have
thought to compare the values with the mean ( column media)
I have tried to do this code
D-datos[, c(1,16:24)]
m-as.matrix(D)
for( i in 1: nrow(D))
{
for( j in 5:(ncol(D)-1)) # I would change this in the new data set, because
I will have more years than 2010
{
if(!is.na(m[i,j]) !is.na
(m[i,j+1])!is.na(m[i,j-1])!is.na(m[i,2])((m[i,j]/m[i,2])4)){m[m[i,j]]-
(m[i,j-1]+m[i,j+1])/2 # Here I would like to find the values that are much more
bigger than the mean of this row,
#if( !is.na(m[i,j])
# and replace them by the mean of the previous and the next values of the
same row.
}
}
}
D-as.data.frame(m)
But I get a data.frame that I had previously, it changes nothing
I accept any idea.
Thanks a lot, Teresa
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
Thanks On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 01:28 AM, Shane Carey wrote: Hello all, I have decided to go ahead with gap.boxplot. I am trying to suppress the axis labels, both x and y labels. I tried using axis.labels=NULL but it would not work. Hi Shane, To suppress the axis labels, pass an empty string: gap.barplot(...,xlab=,ylab=**,...) Many default values of NULL tell the function to work out labels from the data, usually names. Jim -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] create new column to combine 2 data.frames
Hello together i have a little problem with the combine of two data.frames. I have 2 data.frames, which look like this one: first dataframe: ID Name 1 Jack 2 John 3 Jill second dataframe ID Days Type 13 Training 21 Management 34 Training At the end i want to create a data.frame like this one (the Type should be the new column and the entry of the column should be the days): ID Name TrainingManagement 1 Jack30 2 John01 3 Jill 40 maybe anyone can help me, how i can do this. Thanks a lot. Best regards. Mat -- View this message in context: http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Hello,
Suposing that your data frames are named dat1 and dat2, the following
works, but it's a bit complicated, maybe there are simpler solutions.
dat1 - read.table(text =
ID Name
1 Jack
2 John
3 Jill
, header = TRUE, stringsAsFactors = FALSE)
dat2 - read.table(text =
ID Days Type
13 Training
21 Management
34 Training
, header = TRUE, stringsAsFactors = FALSE)
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
y - integer(length(x))
y[!is.na(x)] - dat2[[Days]][!is.na(x)]
y})
result - merge(dat1, tmp)
result
Hope this helps,
Rui Barradas
Em 30-08-2013 09:37, Mat escreveu:
Hello together i have a little problem with the combine of two data.frames.
I have 2 data.frames, which look like this one:
first dataframe:
ID Name
1 Jack
2 John
3 Jill
second dataframe
ID Days Type
13 Training
21 Management
34 Training
At the end i want to create a data.frame like this one (the Type should be
the new column and the entry of the column should be the days):
ID Name TrainingManagement
1 Jack30
2 John01
3 Jill 40
maybe anyone can help me, how i can do this.
Thanks a lot.
Best regards. Mat
--
View this message in context:
http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running pre R.14 version of R with R3.0.0
On 29.08.2013 16:33, Luvalle, Michael J (Michael) wrote: I upgraded R from 2.12.1 to 3.0.0 (on windows XP(, and as soon as I saved the 3.0.0 workspace, was unable to access .Rdata from 2.12.1. The message in the R console is Error in loadNamesSpace(name): there is no package called parallel and a popup window that says Fatal error: unable to restore saved date in .Rdata Is there anything that can be done to access the old .Rdata without destroying the new? Is it not possible to read all the RData files with a recent R such as R-3.0.1? If not, what is the error message? Your is from reading new files with an old R. Uwe Ligges Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Thanks first.
that doesn't look bad. But if have a equal ID in dat2, the days are no
longer correct.
the correct data.frame has to look like this one:
ID Name Management Training
1 1 Jack 13
2 2 John 10
3 3 Jill 04
not this one:
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
dat1 - read.table(text =
+ ID Name
+ 1 Jack
+ 2 John
+ 3 Jill
+ , header = TRUE, stringsAsFactors = FALSE)
dat2 - read.table(text =
+ ID Days Type
+ 11 Management
+ 13 Training
+ 21 Management
+ 34 Training
+ , header = TRUE, stringsAsFactors = FALSE)
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
+ y - integer(length(x))
+ y[!is.na(x)] - dat2[[Days]][!is.na(x)]
+ y})
result - merge(dat1, tmp)
result
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
--
View this message in context:
http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963p4674968.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Outliers Help
Hi Ma Teresa,
Sorry, but I can't understand what you're trying to achieve.
On a statistical note, I'd tend to think more in terms of medians and would
think hard before replacing any outliers, but that's another matter.
Here I created the dataframe dd with the means column of D in its first column,
and then populated with a 1 whenever the value of D for that cell was greater
than 4 times the mean for that row -your definition of 'outlier'.
dd - rep(0,15*7)
dim(dd) - c(15,7)
dd[,1]- D[,1]
for (i in 1:15){
+ for (j in 2:7){
+ dd[i,j] - D[i,(j+1)]/D[i,2]4
+ }
+ }
dd
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1108000000
[2,] 1479 NA NA0100
[3,] 1591000000
[4,] 3408000000
[5,] 3423 NA NA NA100
[6,] 3872000001
[7,] 5823000000
[8,] 6051 NA NA NA NA00
[9,] 8099000000
[10,] 8100 NA NA NA NA00
[11,] 10640111000
[12,] 12600000000
[13,] 14680001001
[14,] 14698000000
[15,] 17143000000
So, you encounter four situations:
a) as in row 2, you have an outlier preceded and followed by values
b) as in row 5, you have an outlier preceded by an NA
c) as in row 6, there is an outlier in the last column
d) as in row 11, there are two or more consecutive outliers
The replacement rule you described would only apply to situations a) (ie
replacing the outlier by the mean of the preceding and subsequent values), and
b) (replacing it by the mean for the row).
But what of situations c) and d)?
And, because this is just a chunk of a bigger dataset, you can also get an
outlier in the first column, followed by a number. Again, your rule has not
accounted for this situation either.
Hope this helps,
José
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mª Teresa Martinez Soriano
Sent: 30 August 2013 09:13
To: r-help@r-project.org
Subject: [R] Outliers Help
This is my a part of my data set
D[1:15,c(1,5:10)]
X. media IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010
1 1108 22.060.0 39 4.0 8.016.0 5.0
2 1479 110.0 NA NA53.0 1166.0 344.8 110.0
3 1591 86.6 247.0 8795.094.081.076.0
4 3408 807.0 302.0 322 621.0 1071.0 1301.0 1225.0
5 34239.0 NA NA NA 410.8 7.011.0
6 3872 103.25000 288.6 113 116.090.094.0 12036.6
7 5823 73.0 117.0 7080.074.069.072.0
8 6051 73.0 NA NA NA NA60.086.0
9 8099 125.16667 196.0 161 150.094.072.078.0
10 8100 70.0 NA NA NA NA48.092.0
11 10640 67.3 1256.61152 664.274.077.051.0
12 12600 2417.0 1960.02383 2453.0 2506.0 2758.0 2442.0
13 14680 38.030.0 61 373.642.019.0 220.8
14 14698 698.16667 553.0 664 847.0 800.0 679.0 646.0
15 17143 392.16667 323.0 322 434.0 383.0 459.0 432.0
I have done multiple imputation and now I have some outliers which I would
like to replace with the mean of this row or if it is possible with the mean of
the previos and the next value of this row, I mean for instance:
value 1 - Outlier- Value 2
I would like to replace the outlier with the mean of value 1 and value2, the
problem is that this values could be NA ( NA after the imputation because they
don't exist), in this case I would like to replace outlier with the mean of the
row.
An other problem I have is to detect correctly outlier values, for instance in
this example of data set for X=3872 and IE.2010, we can see an outlier, I have
thought to compare the values with the mean ( column media)
I have tried to do this code
D-datos[, c(1,16:24)]
m-as.matrix(D)
for( i in 1: nrow(D))
{
for( j in 5:(ncol(D)-1)) # I would change this in the new data set, because
I will have more years than 2010
{
if(!is.na(m[i,j]) !is.na
(m[i,j+1])!is.na(m[i,j-1])!is.na(m[i,2])((m[i,j]/m[i,2])4)){m[m[i,j]]-
(m[i,j-1]+m[i,j+1])/2 # Here I would like to find the values that are much more
bigger than the mean of this row,
#if( !is.na(m[i,j])
# and replace them by the mean of the previous and the next values of the
same row.
}
}
}
D-as.data.frame(m)
But I get a data.frame that I had previously, it changes nothing
I accept any idea.
Thanks a lot, Teresa
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Re: [R] scale breaks
Hi, xlab=, ylab, Would not work. Thanks On Fri, Aug 30, 2013 at 9:37 AM, Shane Carey careys...@gmail.com wrote: Thanks On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 01:28 AM, Shane Carey wrote: Hello all, I have decided to go ahead with gap.boxplot. I am trying to suppress the axis labels, both x and y labels. I tried using axis.labels=NULL but it would not work. Hi Shane, To suppress the axis labels, pass an empty string: gap.barplot(...,xlab=,ylab=**,...) Many default values of NULL tell the function to work out labels from the data, usually names. Jim -- Shane -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
On 30-08-2013, at 11:49, Shane Carey careys...@gmail.com wrote: Hi, xlab=, ylab, You were told to use xlab=, ylab= You seem to have omitted the = after ylab Berend Would not work. Thanks On Fri, Aug 30, 2013 at 9:37 AM, Shane Carey careys...@gmail.com wrote: Thanks On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 01:28 AM, Shane Carey wrote: Hello all, I have decided to go ahead with gap.boxplot. I am trying to suppress the axis labels, both x and y labels. I tried using axis.labels=NULL but it would not work. Hi Shane, To suppress the axis labels, pass an empty string: gap.barplot(...,xlab=,ylab=**,...) Many default values of NULL tell the function to work out labels from the data, usually names. Jim -- Shane -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
This is what I put in: gap.boxplot(DATA$Conductivity~factor(DATA$UnitName_1),ylim=c(LOWER_Y_Conductivity,UPPER_Y_Conductivity_int),gap=gap_Conductivity, col=colours,outwex=one,whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize, range=1.5,xlab=,ylab=) My aim is to not show the labels at the tick marks as I will add them to the plot afterwards On Fri, Aug 30, 2013 at 10:55 AM, Berend Hasselman b...@xs4all.nl wrote: On 30-08-2013, at 11:49, Shane Carey careys...@gmail.com wrote: Hi, xlab=, ylab, You were told to use xlab=, ylab= You seem to have omitted the = after ylab Berend Would not work. Thanks On Fri, Aug 30, 2013 at 9:37 AM, Shane Carey careys...@gmail.com wrote: Thanks On Fri, Aug 30, 2013 at 1:49 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 01:28 AM, Shane Carey wrote: Hello all, I have decided to go ahead with gap.boxplot. I am trying to suppress the axis labels, both x and y labels. I tried using axis.labels=NULL but it would not work. Hi Shane, To suppress the axis labels, pass an empty string: gap.barplot(...,xlab=,ylab=**,...) Many default values of NULL tell the function to work out labels from the data, usually names. Jim -- Shane -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Changing string to date
Hello again, I have a string which I need to put in some legitimate date format. My string is: MAY-14 And output format would be 05/01/2014, this should be of Date class, so that I can make some sensible calculation with it. I have tried this without any success: as.Date(MAY-14, format = %b-%y) [1] NA Can somebody help me how I can achieve that? Thanks and regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
On 08/30/2013 07:57 PM, Shane Carey wrote: This is what I put in: gap.boxplot(DATA$Conductivity~factor(DATA$UnitName_1),ylim=c(LOWER_Y_Conductivity,UPPER_Y_Conductivity_int),gap=gap_Conductivity, col=colours,outwex=one,whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize, range=1.5,xlab=,ylab=) My aim is to not show the labels at the tick marks as I will add them to the plot afterwards Ah, not the axis labels but the tick labels. This is something I had not put into the function. You can edit the gap.boxplot function to do this by commenting out line 53: # axis(1,labels=bxpt$names,at=1:nboxes) saving the function (call it gap.boxplot2.R) and sourcing the edited function: library(plotrix) source(gap.boxplot2.R) gap.boxplot(...) Remember that you have to load plotrix, _then_ source the new function. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Hello,
Ok, try instead
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
y - integer(length(x))
s - as.character(x[!is.na(x)])[1]
idx - which(as.character(dat2[[Type]]) == s)
y[!is.na(x)] - dat2[[Days]][idx]
y})
result - merge(dat1, tmp)
result
Rui Barradas
Em 30-08-2013 10:27, Mat escreveu:
Thanks first.
that doesn't look bad. But if have a equal ID in dat2, the days are no
longer correct.
the correct data.frame has to look like this one:
ID Name Management Training
1 1 Jack 13
2 2 John 10
3 3 Jill 04
not this one:
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
dat1 - read.table(text =
+ ID Name
+ 1 Jack
+ 2 John
+ 3 Jill
+ , header = TRUE, stringsAsFactors = FALSE)
dat2 - read.table(text =
+ ID Days Type
+ 11 Management
+ 13 Training
+ 21 Management
+ 34 Training
+ , header = TRUE, stringsAsFactors = FALSE)
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
+ y - integer(length(x))
+ y[!is.na(x)] - dat2[[Days]][!is.na(x)]
+ y})
result - merge(dat1, tmp)
result
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
--
View this message in context:
http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963p4674968.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing string to date
On 30.08.2013 11:59, Christofer Bogaso wrote: Hello again, I have a string which I need to put in some legitimate date format. My string is: MAY-14 And output format would be 05/01/2014, this should be of Date class, so that I can make some sensible calculation with it. I have tried this without any success: as.Date(MAY-14, format = %b-%y) [1] NA You must have set an English locale (to recognize the word May as a month) and you have to add a day. Best, Uwe Ligges Can somebody help me how I can achieve that? Thanks and regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
Oooff, Right, I will give it a go and see how I get on. Thanks On Fri, Aug 30, 2013 at 11:17 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 07:57 PM, Shane Carey wrote: This is what I put in: gap.boxplot(DATA$Conductivity~**factor(DATA$UnitName_1),ylim=** c(LOWER_Y_Conductivity,UPPER_**Y_Conductivity_int),gap=gap_** Conductivity, col=colours,outwex=one,**whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize, range=1.5,xlab=,ylab=) My aim is to not show the labels at the tick marks as I will add them to the plot afterwards Ah, not the axis labels but the tick labels. This is something I had not put into the function. You can edit the gap.boxplot function to do this by commenting out line 53: # axis(1,labels=bxpt$names,at=1:**nboxes) saving the function (call it gap.boxplot2.R) and sourcing the edited function: library(plotrix) source(gap.boxplot2.R) gap.boxplot(...) Remember that you have to load plotrix, _then_ source the new function. Jim -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] scale breaks
It worked perfectly, your a star!!! Thanks On Fri, Aug 30, 2013 at 11:40 AM, Shane Carey careys...@gmail.com wrote: Oooff, Right, I will give it a go and see how I get on. Thanks On Fri, Aug 30, 2013 at 11:17 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 07:57 PM, Shane Carey wrote: This is what I put in: gap.boxplot(DATA$Conductivity~**factor(DATA$UnitName_1),ylim=** c(LOWER_Y_Conductivity,UPPER_**Y_Conductivity_int),gap=gap_** Conductivity, col=colours,outwex=one,**whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize, range=1.5,xlab=,ylab=) My aim is to not show the labels at the tick marks as I will add them to the plot afterwards Ah, not the axis labels but the tick labels. This is something I had not put into the function. You can edit the gap.boxplot function to do this by commenting out line 53: # axis(1,labels=bxpt$names,at=1:**nboxes) saving the function (call it gap.boxplot2.R) and sourcing the edited function: library(plotrix) source(gap.boxplot2.R) gap.boxplot(...) Remember that you have to load plotrix, _then_ source the new function. Jim -- Shane -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Calculation with Times Series
Thanks a lot! It works like I want to! Am 29.08.2013 15:44 schrieb arun kirshna [via R] ml-node+s789695n4674877...@n4.nabble.com: HI, May be this helps: ts1- ts(1:20) ts2- ts(1:25) ts1[-(1:3)]- ts1[-(1:3)]+ts2[1:17] as.numeric(ts1) # [1] 1 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 A.K. Hey everyone, I`m an absolut beginner in R and need some help for an exercise: I want to do ordinary calculations with 2 time series. The issue with this, that I want to use different elements of time series. Let me give you an example: I want to sum let`s say the 10th element of time series 1 with the 7th element of time series 2. And 9th element of TS 1 with 6th element of TS 2 and 8th element of TS 1 with 5th element of TS 2 ... This pattern of the summation should go all over the time series. Is there a function, which allows me to do this, if possible a function in which I can change the difference of the position with a variable. Thanks a lot for your support. I´m for any advice thankful! __ [hidden email] http://user/SendEmail.jtp?type=nodenode=4674877i=0mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Calculation-with-Times-Series-tp4674852p4674877.html To unsubscribe from Calculation with Times Series, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4674852code=Yy5icmVua2VuQGdtYWlsLmNvbXw0Njc0ODUyfDk2MDYzMzE= . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml -- View this message in context: http://r.789695.n4.nabble.com/Calculation-with-Times-Series-tp4674852p4674956.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Running pre R.14 version of R with R3.0.0
Hi Do you have 2 versions of .RData? If yes you probably need to install packages which were installed and used before saving old .RData file to be able to access this file with new R version. If this does not help, you could return to previous R version, save necessary objects to separate .rdata file and load them to R 3.0.0 environment. see ?save, ?load Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Luvalle, Michael J (Michael) Sent: Thursday, August 29, 2013 4:33 PM To: r-help@r-project.org Subject: [R] Running pre R.14 version of R with R3.0.0 I upgraded R from 2.12.1 to 3.0.0 (on windows XP(, and as soon as I saved the 3.0.0 workspace, was unable to access .Rdata from 2.12.1. The message in the R console is Error in loadNamesSpace(name): there is no package called parallel and a popup window that says Fatal error: unable to restore saved date in .Rdata Is there anything that can be done to access the old .Rdata without destroying the new? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Omitted/blank variables in R function
Hi It would be good if you provided some more specific inforamation. What function do you use for transfering data to R?. E.g. read.table has option to specify NA values during data transfer. If you want to set some values to NA in already read objects you can do it object[object==some value] - NA Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Greg Snow Sent: Friday, August 30, 2013 5:23 AM To: newruser12345 Cc: r-help Subject: Re: [R] Omitted/blank variables in R function Look at the missing function. Or set the default value of the arguments to NA. On Thu, Aug 29, 2013 at 3:23 PM, newruser12345 smetc...@gelbergroup.comwrote: Hi All, I'm very green user and have little programming background, but appreciate any and all help/direction. I have a spreadsheet that successfully sends values from Excel cells to R as variables for a function, which then runs and generates a plot. I cannot figure out how to make R recognize those variables as NA if one of the cells in Excel is blank - or, for that matter, I don't know how to get an R function to recognize variables as NA if no value is assigned to that variable. I have unsuccessfully tried using : if(is.na(four)) return(NA) My function is very simple: mtmatches - c(one,two,three,four) Everything runs smoothly if the four variables have values assigned to them. Any advice on how to get it to run when one of the variables has no value? No worries about the Excel element...figure I can decipher that puzzle later! Thanks! -- View this message in context: http://r.789695.n4.nabble.com/Omitted-blank-variables-in-R-function- tp 4674931.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Packaging
Hi, I have a problem when I try to generate the Documentation pdf (from .rda files)in Spanish during the package creation. Could you tell me the way I can do it?. Thanks in advance. Regards. Eva [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing string to date
Hi as.Date(paste(MAY-14,-01, sep=), format = %b-%y-%d) Shall be OK. Change output format by ?format. Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Christofer Bogaso Sent: Friday, August 30, 2013 12:00 PM To: r-help Subject: [R] Changing string to date Hello again, I have a string which I need to put in some legitimate date format. My string is: MAY-14 And output format would be 05/01/2014, this should be of Date class, so that I can make some sensible calculation with it. I have tried this without any success: as.Date(MAY-14, format = %b-%y) [1] NA Can somebody help me how I can achieve that? Thanks and regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem in while loop?
I have three datasets that I want to compute the errors between them using
linear regression.for this, I want to iterate to reach certain criteria for
the calibration. if changes become smaller than eps the iteration is
successful, hence stop and write parameters into cal:eps=0.1 if number
of iterations is itermax the iteration failed, hence stop and fill cal
with missing value itermax=400
So I tried this code:
x= c(5,2,4,2,1)
y= c(5,3,4,6,9)
z= c(5,8,4,7,3)
itermax=400
get initial calibration parameters, here we assume that:x is the reference
dataset offset x_a=0, slope x_b=1 the other two datasets y, z are
calibrated to x using a simple linear regression
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))#calculate error of x
iter - 0
while(((x_e-x) 0.1) (iter itermax)) {
iter - 0 ##start iteration
x = x_e
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))
iter - iter + 1 # increase iteration counter
}
But I got the same result for X_e before and after the loop:
x_e
[1] 6.454089
--
View this message in context:
http://r.789695.n4.nabble.com/problem-in-while-loop-tp4674962.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Hi see ?merge something like merge(df1, df2) Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Mat Sent: Friday, August 30, 2013 10:37 AM To: r-help@r-project.org Subject: [R] create new column to combine 2 data.frames Hello together i have a little problem with the combine of two data.frames. I have 2 data.frames, which look like this one: first dataframe: ID Name 1 Jack 2 John 3 Jill second dataframe ID Days Type 13 Training 21 Management 34 Training At the end i want to create a data.frame like this one (the Type should be the new column and the entry of the column should be the days): ID Name TrainingManagement 1 Jack30 2 John01 3 Jill 40 maybe anyone can help me, how i can do this. Thanks a lot. Best regards. Mat -- View this message in context: http://r.789695.n4.nabble.com/create-new- column-to-combine-2-data-frames-tp4674963.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Hi
Sorry, I did not read your question to the end
library(reshape)
merge(dat1,cast(melt(dat2, c(ID, Type)), ID~Type))
ID Name Management Training
1 1 Jack 13
2 2 John 1 NA
3 3 Jill NA4
Is close enough, you can easily change NA to 0 if you want.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Mat
Sent: Friday, August 30, 2013 11:27 AM
To: r-help@r-project.org
Subject: Re: [R] create new column to combine 2 data.frames
Thanks first.
that doesn't look bad. But if have a equal ID in dat2, the days are no
longer correct.
the correct data.frame has to look like this one:
ID Name Management Training
1 1 Jack 13
2 2 John 10
3 3 Jill 04
not this one:
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
dat1 - read.table(text =
+ ID Name
+ 1 Jack
+ 2 John
+ 3 Jill
+ , header = TRUE, stringsAsFactors = FALSE)
dat2 - read.table(text =
+ ID Days Type
+ 11 Management
+ 13 Training
+ 21 Management
+ 34 Training
+ , header = TRUE, stringsAsFactors = FALSE)
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
+ y - integer(length(x))
+ y[!is.na(x)] - dat2[[Days]][!is.na(x)]
+ y})
result - merge(dat1, tmp)
result
ID Name Management Training
1 1 Jack 11
2 2 John 30
3 3 Jill 01
--
View this message in context: http://r.789695.n4.nabble.com/create-new-
column-to-combine-2-data-frames-tp4674963p4674968.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem in while loop?
On 30-08-2013, at 09:44, Jonsson amen.alya...@bordeaux.inra.fr wrote:
I have three datasets that I want to compute the errors between them using
linear regression.for this, I want to iterate to reach certain criteria for
the calibration. if changes become smaller than eps the iteration is
successful, hence stop and write parameters into cal:eps=0.1 if number
of iterations is itermax the iteration failed, hence stop and fill cal
with missing value itermax=400
So I tried this code:
x= c(5,2,4,2,1)
y= c(5,3,4,6,9)
z= c(5,8,4,7,3)
itermax=400
get initial calibration parameters, here we assume that:x is the reference
dataset offset x_a=0, slope x_b=1 the other two datasets y, z are
calibrated to x using a simple linear regression
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))#calculate error of x
iter - 0
while(((x_e-x) 0.1) (iter itermax)) {
iter - 0 ##start iteration
x = x_e
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))
iter - iter + 1 # increase iteration counter
}
But I got the same result for X_e before and after the loop:
x_e
[1] 6.454089
I tried your code and got this error message:
Error in model.frame.default(formula = x ~ y, drop.unused.levels = TRUE) :
variable lengths differ (found for 'y')
Calls: lm - eval - eval - Anonymous - model.frame.default
And looking at your code: x is a vector, x_e is a scalar and in the while loop
you are assigning x_e to x so x is then a scalar.
Berend
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
thanks. Works perfectly you made my day :-) -- View this message in context: http://r.789695.n4.nabble.com/create-new-column-to-combine-2-data-frames-tp4674963p4674994.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Memory usage bar plot
Hi, I haven't tried the code yet. Is there a way to parse this data using R and create bar plots so that each program's 'RAM used' figures are grouped together. So 'uuidd' bars will be together. The data will have about 50 sets. So if there are 100 processes each will have about 50 bars. What is the recommended way to graph these big barplots ? I am looking for only 'RAM used' figures. Thanks, Mohan Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check - 453.9 MiB = Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check -- 453.9 MiB = This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] create new column to combine 2 data.frames
Hi,
You can use 'fill=0' in ?cast()
merge(dat1,cast(melt(dat2, c(ID, Type)), ID~Type,fill=0))
# ID Name Management Training
#1 1 Jack 1 3
#2 2 John 1 0
#3 3 Jill 0 4
A.K.
- Original Message -
From: PIKAL Petr petr.pi...@precheza.cz
To: Mat matthias.we...@fnt.de; r-help@r-project.org r-help@r-project.org
Cc:
Sent: Friday, August 30, 2013 5:57 AM
Subject: Re: [R] create new column to combine 2 data.frames
Hi
Sorry, I did not read your question to the end
library(reshape)
merge(dat1,cast(melt(dat2, c(ID, Type)), ID~Type))
ID Name Management Training
1 1 Jack 1 3
2 2 John 1 NA
3 3 Jill NA 4
Is close enough, you can easily change NA to 0 if you want.
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Mat
Sent: Friday, August 30, 2013 11:27 AM
To: r-help@r-project.org
Subject: Re: [R] create new column to combine 2 data.frames
Thanks first.
that doesn't look bad. But if have a equal ID in dat2, the days are no
longer correct.
the correct data.frame has to look like this one:
ID Name Management Training
1 1 Jack 1 3
2 2 John 1 0
3 3 Jill 0 4
not this one:
ID Name Management Training
1 1 Jack 1 1
2 2 John 3 0
3 3 Jill 0 1
dat1 - read.table(text =
+ ID Name
+ 1 Jack
+ 2 John
+ 3 Jill
+ , header = TRUE, stringsAsFactors = FALSE)
dat2 - read.table(text =
+ ID Days Type
+ 1 1 Management
+ 1 3 Training
+ 2 1 Management
+ 3 4 Training
+ , header = TRUE, stringsAsFactors = FALSE)
library(reshape2)
tmp - dcast(data = dat2, ID ~ Type, value.var = Type)
tmp[-1] - lapply(tmp[-1], function(x){
+ y - integer(length(x))
+ y[!is.na(x)] - dat2[[Days]][!is.na(x)]
+ y})
result - merge(dat1, tmp)
result
ID Name Management Training
1 1 Jack 1 1
2 2 John 3 0
3 3 Jill 0 1
--
View this message in context: http://r.789695.n4.nabble.com/create-new-
column-to-combine-2-data-frames-tp4674963p4674968.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Outliers Help
HI,
Also,
dd1-matrix(cbind(D[,1],(D[-c(1:2)]/D[,2]4)*1),dimnames=NULL,ncol=7)
identical(dd,dd1)
#[1] TRUE
A.K.
- Original Message -
From: Jose Iparraguirre jose.iparragui...@ageuk.org.uk
To: Mª Teresa Martinez Soriano teresama...@hotmail.com;
r-help@r-project.org r-help@r-project.org
Cc:
Sent: Friday, August 30, 2013 5:39 AM
Subject: Re: [R] Outliers Help
Hi Ma Teresa,
Sorry, but I can't understand what you're trying to achieve.
On a statistical note, I'd tend to think more in terms of medians and would
think hard before replacing any outliers, but that's another matter.
Here I created the dataframe dd with the means column of D in its first column,
and then populated with a 1 whenever the value of D for that cell was greater
than 4 times the mean for that row -your definition of 'outlier'.
dd - rep(0,15*7)
dim(dd) - c(15,7)
dd[,1]- D[,1]
for (i in 1:15){
+ for (j in 2:7){
+ dd[i,j] - D[i,(j+1)]/D[i,2]4
+ }
+ }
dd
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1108 0 0 0 0 0 0
[2,] 1479 NA NA 0 1 0 0
[3,] 1591 0 0 0 0 0 0
[4,] 3408 0 0 0 0 0 0
[5,] 3423 NA NA NA 1 0 0
[6,] 3872 0 0 0 0 0 1
[7,] 5823 0 0 0 0 0 0
[8,] 6051 NA NA NA NA 0 0
[9,] 8099 0 0 0 0 0 0
[10,] 8100 NA NA NA NA 0 0
[11,] 10640 1 1 1 0 0 0
[12,] 12600 0 0 0 0 0 0
[13,] 14680 0 0 1 0 0 1
[14,] 14698 0 0 0 0 0 0
[15,] 17143 0 0 0 0 0 0
So, you encounter four situations:
a) as in row 2, you have an outlier preceded and followed by values
b) as in row 5, you have an outlier preceded by an NA
c) as in row 6, there is an outlier in the last column
d) as in row 11, there are two or more consecutive outliers
The replacement rule you described would only apply to situations a) (ie
replacing the outlier by the mean of the preceding and subsequent values), and
b) (replacing it by the mean for the row).
But what of situations c) and d)?
And, because this is just a chunk of a bigger dataset, you can also get an
outlier in the first column, followed by a number. Again, your rule has not
accounted for this situation either.
Hope this helps,
José
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Mª Teresa Martinez Soriano
Sent: 30 August 2013 09:13
To: r-help@r-project.org
Subject: [R] Outliers Help
This is my a part of my data set
D[1:15,c(1,5:10)]
X. media IE.2005 IE.2006 IE.2007 IE.2008 IE.2009 IE.2010
1 1108 22.0 60.0 39 4.0 8.0 16.0 5.0
2 1479 110.0 NA NA 53.0 1166.0 344.8 110.0
3 1591 86.6 247.0 87 95.0 94.0 81.0 76.0
4 3408 807.0 302.0 322 621.0 1071.0 1301.0 1225.0
5 3423 9.0 NA NA NA 410.8 7.0 11.0
6 3872 103.25000 288.6 113 116.0 90.0 94.0 12036.6
7 5823 73.0 117.0 70 80.0 74.0 69.0 72.0
8 6051 73.0 NA NA NA NA 60.0 86.0
9 8099 125.16667 196.0 161 150.0 94.0 72.0 78.0
10 8100 70.0 NA NA NA NA 48.0 92.0
11 10640 67.3 1256.6 1152 664.2 74.0 77.0 51.0
12 12600 2417.0 1960.0 2383 2453.0 2506.0 2758.0 2442.0
13 14680 38.0 30.0 61 373.6 42.0 19.0 220.8
14 14698 698.16667 553.0 664 847.0 800.0 679.0 646.0
15 17143 392.16667 323.0 322 434.0 383.0 459.0 432.0
I have done multiple imputation and now I have some outliers which I would like
to replace with the mean of this row or if it is possible with the mean of the
previos and the next value of this row, I mean for instance:
value 1 - Outlier- Value 2
I would like to replace the outlier with the mean of value 1 and value2, the
problem is that this values could be NA ( NA after the imputation because they
don't exist), in this case I would like to replace outlier with the mean of the
row.
An other problem I have is to detect correctly outlier values, for instance in
this example of data set for X=3872 and IE.2010, we can see an outlier, I have
thought to compare the values with the mean ( column media)
I have tried to do this code
D-datos[, c(1,16:24)]
m-as.matrix(D)
for( i in 1: nrow(D))
{
for( j in 5:(ncol(D)-1)) # I would change this in the new data set, because
I will have more years than 2010
{
if(!is.na(m[i,j]) !is.na
(m[i,j+1])!is.na(m[i,j-1])!is.na(m[i,2])((m[i,j]/m[i,2])4)){m[m[i,j]]-
(m[i,j-1]+m[i,j+1])/2 # Here I would like to find the values that are much more
bigger than the mean of this row,
#if( !is.na(m[i,j])
# and replace them by the mean of the previous and the next values of the
same row.
}
}
}
Re: [R] mean
Hi, Better would be to show a reproducible example using ?dput() and the codes you used. Assuming that you tried something like this: lst1- list(1:10,c(5,4,3),4:15) mean(lst1) #[1] NA #Warning message: #In mean.default(lst1) : argument is not numeric or logical: returning NA sapply(lst1,mean) #also depends upon the list elements. In the example I used, it is a vector. #[1] 5.5 4.0 9.5 #or lapply(lst1,mean) #[[1]] #[1] 5.5 # #[[2]] #[1] 4 # #[[3]] #[1] 9.5 #Suppose, your list elements are: set.seed(24) lst2- list(as.data.frame(matrix(sample(1:10,5*4,replace=TRUE),ncol=5)),as.data.frame(matrix(sample(1:40,8*5,replace=TRUE),ncol=8))) #and you wanted to find the column means lapply(lst2,colMeans) #[[1]] # V1 V2 V3 V4 V5 #5.00 7.00 5.75 7.00 2.75 # #[[2]] # V1 V2 V3 V4 V5 V6 V7 V8 #14.4 16.8 16.4 26.4 11.4 12.0 24.8 16.8 Hope it helps. A.K. When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
Hello, This memory usage should be graphed with time. Are there examples of scatterplots that can clearly show usage vs time ? This is for memory leak detection. Thanks, Mohan From: PIKAL Petr petr.pi...@precheza.cz To: mohan.radhakrish...@polarisft.com mohan.radhakrish...@polarisft.com, r-help@r-project.org r-help@r-project.org Date: 08/30/2013 05:33 PM Subject:RE: [R] Memory usage bar plot Hi For reading data into R you shall look to read.table and similar. For plotting ggplot could handle it. However I wonder if 100 times 50 bars is the way how to present your data. You shall think over what do you want to show to yourself or your audience. Maybe boxplots or scatterplots could be better. Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of mohan.radhakrish...@polarisft.com Sent: Friday, August 30, 2013 1:25 PM To: r-help@r-project.org Subject: [R] Memory usage bar plot Hi, I haven't tried the code yet. Is there a way to parse this data using R and create bar plots so that each program's 'RAM used' figures are grouped together. So 'uuidd' bars will be together. The data will have about 50 sets. So if there are 100 processes each will have about 50 bars. What is the recommended way to graph these big barplots ? I am looking for only 'RAM used' figures. Thanks, Mohan Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check - 453.9 MiB = Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check -- 453.9 MiB = This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e- mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: ddply for comparing simulation results
forgot to send it back to the list.
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
-- Forwarded message --
From: jim holtman jholt...@gmail.com
Date: Fri, Aug 30, 2013 at 8:10 AM
Subject: Re: ddply for comparing simulation results
To: john doe anon.r.u...@gmail.com
try the 'data.table package. It gives the answer in less than a second.
# 1 million leads, half of which were simulated, half of which were not
id=1:100
isSimulated = c(rep(0,50), rep(1, 50))
userId=sample(1:10, 100, replace=T)
df_leads=data.frame(id, isSimulated, userId)
require(data.table)
Loading required package: data.table
data.table 1.8.8 For help type: help(data.table)
system.time({
+ df_leads - data.table(df_leads)
+ df_leads_sum - df_leads[
+ , list(count = .N)
+ , keyby = c('isSimulated', 'userId')
+ ]
+ })
user system elapsed
0.750.010.76
head(df_leads_sum)
isSimulated userId count
1: 0 1 5
2: 0 2 9
3: 0 3 5
4: 0 4 4
5: 0 5 3
6: 0 6 7
you can use 'setdiff' to find userIDs that are missing from one group
or the other:
#see which userIDs are missing between the groups
not_in - setdiff(df_leads_sum$userId[df_leads_sum$isSimulated == 0]
+ , df_leads_sum$userId[df_leads_sum$isSimulated == 1]
+ )
str(not_in)
int [1:697] 59 100 204 584 656 828 840 999 1012 1046 ...
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Aug 29, 2013 at 11:33 PM, john doe anon.r.u...@gmail.com wrote:
I am trying to use R and plyr to compare the effectiveness of various
algorithms for online advertising. At the core, I am simply counting when a
user receives a lead: this is measured with the userId column. Leads that
were sent in production have a 0 in the isSimulated column, and leads that
were sent in our simulation have isSimulated=1. I have two questions: one
about performance and one about how to use plyr to get the data in a form
that I want.
Here is an example of my code:
# 1 million leads, half of which were simulated, half of which were not
id=1:100
isSimulated = c(rep(0,50), rep(1, 50))
userId=sample(1:10, 100, replace=T)
df_leads=data.frame(id, isSimulated, userId)
# split by simulated and userid, and then sum
system.time(df_leads_sum - ddply(df_leads, .(isSimulated, userId), nrow))
user system elapsed
38.167 0.212 38.386
The above call to ddply is great because it allows me to create histograms
of how many people receive just a few leads, or a lot of leads, both in
production and in the simulator.
Question 1: The above ddply call takes a while to execute. With production
data it takes several minutes in R, but only a few seconds in MySQL. Is
there a way to improve the performance of the above call?
Question 2: What I would really like to do is create a histogram which
measures the distribution of change in leads between non-simulated and
simulated data. A complicating fact is that some users might only appear in
simulated or non-simulated data, so I need to correclty handle the absense
of a userId. (In production, users are actually guaranteed to appear in
production - but the crux of the problem is the same: userIds might be
missing in one of the splits). Can someone help me with this? I've read
the documentation a few times, and think that the summarize function might
be able to help, but I'm not quite sure how to do this.
Thanks.
--
You received this message because you are subscribed to the Google Groups
manipulatr group.
To unsubscribe from this group and stop receiving emails from it, send an
email to manipulatr+unsubscr...@googlegroups.com.
To post to this group, send email to manipul...@googlegroups.com.
Visit this group at http://groups.google.com/group/manipulatr.
For more options, visit https://groups.google.com/groups/opt_out.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Ordering a matrix (and not losing the rownames)
Hi Ramón, May be this helps: tags_totals-matrix(c(15,11,23,7,5),ncol=1,dimnames=list(c(Wikis,Glosarios,Grupos,Bases de datos,Taller),NULL)) tags_totals[order(tags_totals[,1],decreasing=TRUE),,drop=FALSE] # [,1] #Grupos 23 #Wikis 15 #Glosarios 11 #Bases de datos 7 #Taller 5 A.K. Hello, I have a matrix like this: tags_totals [,1] Wikis 15 Glosarios 11 Grupos 23 Bases de datos 7 Taller 5 And I want to order by the value of the first column. I do this: ordered_matrix - as.matrix(tags_totals[order(tags_totals[,1],decreasing=TRUE)]) It orders alright, but I lose the rownames, that I need for the graphics ordered_matrix [,1] [1,] 23 [2,] 15 [3,] 11 [4,] 7 [5,] 5 rownames(ordered_matrix) NULL If I try to do it after converting to a dataframe I get an error that I don't understand: tags_totals_frame - as.data.frame(tags_totals) tags_totals_frame[,1] [1] 15 11 23 7 5 ordered_frame - tags_totals_frame[order(tags_totals_frame[,1],decreasing=TRUE)] Error en `[.data.frame`(tags_totals_frame, order(tags_totals_frame[, 1], : undefined columns selected Thanks on any help, -- == Ramón Ovelar Campus Virtual Birtuala UPV/EHU Tel: (34) 94 601 3407 http://campusvirtual.ehu.es __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
Here is how to parse the data and put it into groups. Not sure what
the 'timing' of each group is since not time information was given.
Also not sure is there is an 'MiB' qualifier on the data, but you have
the matrix of data which is easy to do with as you want.
input - readLines(textConnection(
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ -
+ 453.9 MiB
+
+ =
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ --
+ 453.9 MiB
+ =))
# keep only the data
input - input[grepl('=', input)]
# separate into groups
grps - split(input, cumsum(grepl(= RAM, input)))
# parse the data (not sure if there is also 'MiB')
parsed - lapply(grps, function(.grp){
+ # parse ignoring first and last lines
+ .data - sub(.*= ([^ ]+) ([^ ]+)\\s+(.*), \\1 \\2 \\3
+ , .grp[2:(length(.grp) - 1L)]
+ )
+ # return matrix
+ do.call(rbind, strsplit(.data, ' '))
+ })
parsed
$`1`
[,1][,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB hald-addon-storage
[6,] 159.0 KiB gpm
[7,] 162.5 KiB pam_timestamp_check
$`2`
[,1][,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB hald-addon-storage
[6,] 159.0 KiB gpm
[7,] 162.5 KiB pam_timestamp_check
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Aug 30, 2013 at 7:24 AM, mohan.radhakrish...@polarisft.com wrote:
Hi,
I haven't tried the code yet. Is there a way to parse this data
using R and create bar plots so that each program's 'RAM used' figures are
grouped together.
So 'uuidd' bars will be together. The data will have about 50 sets. So if
there are 100 processes each will have about 50 bars.
What is the recommended way to graph these big barplots ? I am looking for
only 'RAM used' figures.
Thanks,
Mohan
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
-
453.9 MiB
=
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
--
453.9 MiB
=
This e-Mail may contain proprietary and confidential information and is sent
for the intended recipient(s) only. If by an addressing or transmission
error this mail has been misdirected to you, you are requested to delete this
mail immediately. You are also hereby notified that any use, any form of
reproduction, dissemination, copying, disclosure, modification, distribution
and/or publication of this e-mail message, contents or its attachment other
than by its intended recipient/s is strictly prohibited.
Visit us at http://www.polarisFT.com
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
Re: [R] scale breaks
Hi Jim et al, I want to remove the upper bounding box, I did this by #box() in the gap.plot function. It still leaves me with two horizontal lines. I would like to remove them also, where are the created within the function? Thanks, this is great, exactly what I need. Cheers On Fri, Aug 30, 2013 at 11:44 AM, Shane Carey careys...@gmail.com wrote: It worked perfectly, your a star!!! Thanks On Fri, Aug 30, 2013 at 11:40 AM, Shane Carey careys...@gmail.com wrote: Oooff, Right, I will give it a go and see how I get on. Thanks On Fri, Aug 30, 2013 at 11:17 AM, Jim Lemon j...@bitwrit.com.au wrote: On 08/30/2013 07:57 PM, Shane Carey wrote: This is what I put in: gap.boxplot(DATA$Conductivity~**factor(DATA$UnitName_1),ylim=** c(LOWER_Y_Conductivity,UPPER_**Y_Conductivity_int),gap=gap_** Conductivity, col=colours,outwex=one,**whisklty = solid,whisklwd=lwth,outcol= black, outpch=dtsym, outcex=dtsize, range=1.5,xlab=,ylab=) My aim is to not show the labels at the tick marks as I will add them to the plot afterwards Ah, not the axis labels but the tick labels. This is something I had not put into the function. You can edit the gap.boxplot function to do this by commenting out line 53: # axis(1,labels=bxpt$names,at=1:**nboxes) saving the function (call it gap.boxplot2.R) and sourcing the edited function: library(plotrix) source(gap.boxplot2.R) gap.boxplot(...) Remember that you have to load plotrix, _then_ source the new function. Jim -- Shane -- Shane -- Shane [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] XLSX package + Excel creation question
You can also look at the XLConnect package. Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Thu, Aug 29, 2013 at 9:40 AM, Zsurzsa Laszlo zsurzsalas...@gmail.com wrote: I understand you response but it does not solve the problem. I'am aware that one can simply color every cell in an excel file by using his own algorithm. The question was if I can write my data to a *single* cells and use different formatting for every piece of data. - - László-András Zsurzsa,- - Msc. Infromatics, Technical University Munich, Germany - - Scientific Employee, TUM - - On Thu, Aug 29, 2013 at 3:36 PM, Rainer Hurling rhur...@gwdg.de wrote: Am 29.08.2013 15:03 (UTC+1) schrieb Zsurzsa Laszlo: First of all thank you for the quick resposen. I know I can color and set up every cell. I will take a look again * CellStyle* but is it possbile for example to write an array to a single cell that has different colors for some data. Basically the color depends on the data. As far as I know there is no ready to use functionality to mask groups of selected cells. You have to write your own function, which selects the right cells and changes their style with setCellStyle(cell, cellStyle). Some hints are given in the examples section of ?CellStyle. - - László-András Zsurzsa,- - Msc. Infromatics, Technical University Munich, Germany - - Scientific Employee, TUM - - On Thu, Aug 29, 2013 at 2:55 PM, Rainer Hurling rhur...@gwdg.de wrote: Am 29.08.2013 12:08 (UTC+1) schrieb Zsurzsa Laszlo: Dear R users, I have a question about the xlsx package. It's possible to create excel files and color cells and etc. yes, with package xlsx you can colourize you data sheets, even the fonts. See for example ?CellStyle . A good demonstration of the capabilities is on http://tradeblotter.wordpress.com/2013/05/02/writing-from-r-to-excel-with-xlsx/ My question would be that is it possible to color only some part of the data hold in a cell. Let's assume I've got the following data : 167,153,120,100 and I want to color to red everything that is bigger then 120. How can I achive this using R. Example file setup with a few lines in attachment. (SEL_MASS column can be used for example) Attachment missing ... HTH, Rainer Thank you in advance, - - László-András Zsurzsa, - - Msc. Infromatics, Technical University Munich, Germany - - Scientific Employee, TUM - - [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
HI,
You could also parse the data by:
input1- input
library(stringr)
input2-str_trim(gsub([=+],,input1))
dat1-read.table(text=word(input2[!grepl(---,input2) input2!=
!grepl(RAM|MiB,input2)],8,15),sep=,header=FALSE,stringsAsFactors=FALSE)
lst1-split(dat1,cumsum(dat1$V3==uuidd))
lst1
#$`1`
# V1 V2 V3
#1 107.5 KiB uuidd
#2 120.5 KiB klogd
#3 141.0 KiB hidd
#4 146.0 KiB acpid
#5 153.5 KiB hald-addon-storage
#6 159.0 KiB gpm
#7 162.5 KiB pam_timestamp_check
#
#$`2`
# V1 V2 V3
#8 107.5 KiB uuidd
#9 120.5 KiB klogd
#10 141.0 KiB hidd
#11 146.0 KiB acpid
#12 153.5 KiB hald-addon-storage
#13 159.0 KiB gpm
#14 162.5 KiB pam_timestamp_check
A.K.
- Original Message -
From: jim holtman jholt...@gmail.com
To: mohan.radhakrish...@polarisft.com
Cc: R mailing list r-help@r-project.org
Sent: Friday, August 30, 2013 9:44 AM
Subject: Re: [R] Memory usage bar plot
Here is how to parse the data and put it into groups. Not sure what
the 'timing' of each group is since not time information was given.
Also not sure is there is an 'MiB' qualifier on the data, but you have
the matrix of data which is easy to do with as you want.
input - readLines(textConnection(
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ -
+ 453.9 MiB
+
+ =
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ --
+ 453.9 MiB
+ =))
# keep only the data
input - input[grepl('=', input)]
# separate into groups
grps - split(input, cumsum(grepl(= RAM, input)))
# parse the data (not sure if there is also 'MiB')
parsed - lapply(grps, function(.grp){
+ # parse ignoring first and last lines
+ .data - sub(.*= ([^ ]+) ([^ ]+)\\s+(.*), \\1 \\2 \\3
+ , .grp[2:(length(.grp) - 1L)]
+ )
+ # return matrix
+ do.call(rbind, strsplit(.data, ' '))
+ })
parsed
$`1`
[,1] [,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB hald-addon-storage
[6,] 159.0 KiB gpm
[7,] 162.5 KiB pam_timestamp_check
$`2`
[,1] [,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB hald-addon-storage
[6,] 159.0 KiB gpm
[7,] 162.5 KiB pam_timestamp_check
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Aug 30, 2013 at 7:24 AM, mohan.radhakrish...@polarisft.com wrote:
Hi,
I haven't tried the code yet. Is there a way to parse this data
using R and create bar plots so that each program's 'RAM used' figures are
grouped together.
So 'uuidd' bars will be together. The data will have about 50 sets. So if
there are 100 processes each will have about 50 bars.
What is the recommended way to graph these big barplots ? I am looking for
only 'RAM used' figures.
Thanks,
Mohan
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
-
453.9 MiB
=
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
Re: [R] (no subject)
On 30.08.2013 15:27, Joana Costa wrote: good afternoon, I´m writting because I´m having some problems with R installation. I would like to install R essentials with spss Version 20 but I can´t find wher in Cran site I can do the dowload speciffically of Essentials. Can you help me? I do not understand musch of R software and I just need to run it in SPSS menus Hope forward to hearning from you soon Please ask your SPSS support. The R project does not offer anything called R essentials. Best, Uwe Ligges Best regards Joana Joana Costa, MSc, Ph.D.Postdoctoral Researcher(joanasco...@hotmail.com) CINEICC, University of Coimbra Rua do Colégio Novo, Apartado 6153 3001-802 Coimbra, Portugal Telefone: (+351) 239 851450 Fax: (+351) 239851462 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] strange conversion char to date
Hi R-friends, Can anyone explain the following strange behavior to me? as.Date( 4/25/71,%m/%d/%y) [1] 1971-04-25 as.Date( 4/25/62,%m/%d/%y) [1] 2062-04-25 so 71 is converted to 1971, while 62 is converted to 2062? Does anyone know why? And is there a simple way to specify the date? (does works the same way in R 2.01 as well as in 1.9) --- dr F.H.G. (Frans) Marcelissen DigiPsy (www.DigiPsy.nl http://www.digipsy.nl/) Pomperschans 26 5595 AV Leende tel: 040 2065030/06 2325 06 53 skype adres: frans.marcelissen email: frans.marcelis...@digipsy.nl [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean
It would be easier to diagnose the problem if you included an example illustrating exactly what you did. I'll guess: a - list(3,4,5) mean(a) [1] NA Warning message: In mean.default(a) : argument is not numeric or logical: returning NA mean(as.numeric(a)) [1] 4 But that's just a guess, as I don't know the actual contents of your list! albyn On Fri, Aug 30, 2013 at 5:18 AM, agnes69 fes...@gredeg.cnrs.fr wrote: When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installing R for use in SPSS (was: no subject)
On Aug 30, 2013, at 8:27 AM, Joana Costa joanasco...@hotmail.com wrote: good afternoon, I´m writting because I´m having some problems with R installation. I would like to install R essentials with spss Version 20 but I can´t find wher in Cran site I can do the dowload speciffically of Essentials. Can you help me? I do not understand musch of R software and I just need to run it in SPSS menus Hope forward to hearning from you soon Best regards Joana Joana Costa, MSc, Ph.D.Postdoctoral Researcher(joanasco...@hotmail.com) CINEICC, University of Coimbra First and foremost: Do not reply to a prior post with a new query, which is known as hijacking a thread. In the future, start a new thread by using a new e-mail and a proper subject line. If you try to find your post in the list archives, it is buried within another thread, which makes it problematic. There is no Essentials installation for R. The basic R installation installs Base R plus Recommended Packages and are a part of the standard R distribution. You can download the proper binary installation from CRAN for your OS (presuming Windows, since you are on hotmail). Just download it and run the installer. If there is any other configuration that SPSS requires, you will need to contact them for support. To being your journey with R, you should start with the manuals: http://cran.r-project.org/manuals.html and the FAQs: http://cran.r-project.org/faqs.html and consider reading the Posting Guide: http://www.r-project.org/posting-guide.html Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strange conversion char to date
Take a look at the documentation as to what %y means: ‘%y’ Year without century (00-99). On input, values 00 to 68 are prefixed by 20 and 69 to 99 by 19 - that is the behaviour specified by the 2004 and 2008 POSIX standards, but they do also say ‘it is expected that in a future version the default century inferred from a 2-digit year will change’. This is due to the Y2K problem that maybe you never had to deal with. Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. On Fri, Aug 30, 2013 at 9:12 AM, Frans Marcelissen frans.marcelis...@digipsy.nl wrote: Hi R-friends, Can anyone explain the following strange behavior to me? as.Date( 4/25/71,%m/%d/%y) [1] 1971-04-25 as.Date( 4/25/62,%m/%d/%y) [1] 2062-04-25 so 71 is converted to 1971, while 62 is converted to 2062? Does anyone know why? And is there a simple way to specify the date? (does works the same way in R 2.01 as well as in 1.9) --- dr F.H.G. (Frans) Marcelissen DigiPsy (www.DigiPsy.nl http://www.digipsy.nl/) Pomperschans 26 5595 AV Leende tel: 040 2065030/06 2325 06 53 skype adres: frans.marcelissen email: frans.marcelis...@digipsy.nl [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help with Fisher test for a more than two group of genes
From: gabriel.wajnb...@hotmail.com To: r-help@r-project.org Subject: Fisher test for a more than two group of genesþ Date: Wed, 28 Aug 2013 15:53:59 -0300 Good Afternoon, My name is Gabriel, I'm doing an analysis if there is increase or decrease in dependence on the mutated genes, using 3 or more genes using the fisher exact test.I performed with success an analysis for two genes using fisher.test( ). example of the 2x2 contigency table: Gene A mutated | Gene A normalGene B mutated| 26| 12Gene B normal | 10 | 50 Now I'm wondering how can I perform the analysis for 3 genes (and construct the contigency table), as follows: Gene A mutated, Gene A normal, Gene B mutated, Gene B normal, Gene C mutated and Gene C normal. How do I perform a fisher test using fisher.test( ) function using this data (3x3 contigency table)?Can someone help me ? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean
Use is.numeric(list) on the list to see if it is a list of numbers or if it is list of characters. If it is a list of characters (which could be the case if you read it in using read.csv or the like) and it makes sense to convert to numeric, then do something like: list-as.numeric(list) -Roy On Aug 30, 2013, at 5:18 AM, agnes69 fes...@gredeg.cnrs.fr wrote: When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ** The contents of this message do not reflect any position of the U.S. Government or NOAA. ** Roy Mendelssohn Supervisory Operations Research Analyst NOAA/NMFS Environmental Research Division Southwest Fisheries Science Center 1352 Lighthouse Avenue Pacific Grove, CA 93950-2097 e-mail: roy.mendelss...@noaa.gov (Note new e-mail address) voice: (831)-648-9029 fax: (831)-648-8440 www: http://www.pfeg.noaa.gov/ Old age and treachery will overcome youth and skill. From those who have been given much, much will be expected the arc of the moral universe is long, but it bends toward justice -MLK Jr. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sensitivy / Specificity and nulls
Maybe you want to check out the prevalence package (http://cran.r-project.org/web/packages/prevalence/index.html), and its development version on GitHub (https://github.com/brechtdv/prevalence/). When you have two tests, and neither can be considered to be a 'gold standard' (ie, SE SP = 100%), Bayesian latent class approaches can help you to estimate true prevalence, sensitivity and specificity. Brecht Good Day All, I am working with a diagnostic test and comparing the new test to an old test. Normally I would be able to calculate sensitivity and specificity quite easily. However, the 'gold standard' that I am comparing my new diagnostic with is really 'gold-plated' in that sometimes the 'gold standard' fails completely and I have no data from the 'gold standard' but I might have data from the diagnostic test. Of course sometimes my new diagnostic fails but I have data from my 'gold standard' To me this really starts moving towards classification but I cannot seem to find the appropriate calculations. Can someone point me to some web resources to determine the appropriate method to be able to deal with the NULLs ? Resources within the medical realm would be better (because the rest of the folks would understand them better) but not required. -- Devleesschauwer Brecht Doctoral Researcher, MVSc DVM Department of Virology, Parasitology and Immunology Faculty of Veterinary Medicine Ghent University Salisburylaan 133 9820 Merelbeke Belgium Telephone: +32 9 264 7328 Mobile (Belgium): +32 476 365743 Mobile (Nepal): +977 9842 223323 E-mail: brecht.devleesschau...@ugent.be http://users.ugent.be/~bdvleess/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] (no subject)
Hi Joana, As far as I can see, R Essentials has nothing to do with R in terms of being an official R package so you will not find it on CRAN. It seems to be an IBM add-in for SPSS that let's you run R code within SPSS. It should be available from http://sourceforge.net/projects/ibmspssstat/ Good luck. Oh and R is much nicer than SPSS. Come join us. John Kane Kingston ON Canada -Original Message- From: joanasco...@hotmail.com Sent: Fri, 30 Aug 2013 14:27:37 +0100 To: r-help@r-project.org Subject: [R] (no subject) good afternoon, I4m writting because I4m having some problems with R installation. I would like to install R essentials with spss Version 20 but I can4t find wher in Cran site I can do the dowload speciffically of Essentials. Can you help me? I do not understand musch of R software and I just need to run it in SPSS menus Hope forward to hearning from you soon Best regards Joana Joana Costa, MSc, Ph.D.Postdoctoral Researcher(joanasco...@hotmail.com) CINEICC, University of Coimbra Rua do Coligio Novo, Apartado 6153 3001-802 Coimbra, Portugal Telefone: (+351) 239 851450 Fax: (+351) 239851462 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. 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 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] calculate with different columns from different datasets
Hi, Using the same datasets: dat1- read.table(text= V1 V2 V3 2 6 8 4 3 4 1 9 8 ,sep=,header=TRUE) dat2- read.table(text= V1 V2 V3 6 8 4 2 0 7 8 1 3 ,sep=,header=TRUE) sapply(seq_len(ncol(dat1)),function(i) cor(dat1[,i],dat2[,i],method=spearman)) #[1] -1.000 0.500 -0.8660254 #or diag(cor(dat1,dat2,method=spearman)) # V1 V2 V3 #-1.000 0.500 -0.8660254 A.K. - Original Message - From: laro l_roh...@gmx.ch To: r-help@r-project.org Cc: Sent: Friday, August 30, 2013 12:52 PM Subject: Re: [R] calculate with different columns from different datasets Thank you a lot A.K.! One more question: I'd like to compute the Spearman's rank correlation coefficients for V1 (from dat1) and V1 (from dat2) and so on... Do you know how to do that? I managed to write the Pearson's correlation product moment coefficient with your sapply-approach, but I have no idea how to rank it. Kind regards -- View this message in context: http://r.789695.n4.nabble.com/calculate-with-different-columns-from-different-datasets-tp4674918p4675034.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mean
When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
good afternoon, I´m writting because I´m having some problems with R installation. I would like to install R essentials with spss Version 20 but I can´t find wher in Cran site I can do the dowload speciffically of Essentials. Can you help me? I do not understand musch of R software and I just need to run it in SPSS menus Hope forward to hearning from you soon Best regards Joana Joana Costa, MSc, Ph.D.Postdoctoral Researcher(joanasco...@hotmail.com) CINEICC, University of Coimbra Rua do Colégio Novo, Apartado 6153 3001-802 Coimbra, Portugal Telefone: (+351) 239 851450 Fax: (+351) 239851462 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] https://stat.ethz.ch/pipermail/r-help/20
https://stat.ethz.ch/pipermail/r-help/20。。。！！。！。。。！。。！。。@@@。@。。une/243667.html__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] https://stat.ethz.ch/pipermail/r-help/20
https://stat.ethz.ch/pipermail/r-help/20。。。！！。！。。。！。。！。。@@@。@。。une/243667.html__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with newline using bquote(paste())
Hi All, This is a variant of a problem I posted yesterday (see below) where I found I had a large gap between my N= and he number I had evaluated using .(x). I seem to have trouble with newlines in a main title. I find now that all works as expected (no unsightly gap between my N= and the value, if all of the title is put on the same line. Whenever I try the newline, I run into problems. Below, I have one example that gives me no syntax errors, but simply does not print the information after the \n (the N= xxx) part. Any help appreciated. Thanks, Gerard PS using R 3.0.0 ss-n(m18_das28*b_dascore) par(oma=c(2,2,2,2)) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=as.expression(bquote(paste(Baseline xyz with Month 18 DAS28\n)), bquote(paste((N=,.(ss),, xlab=Baseline xyz, ylab=Month 18 DAS28, legend.plot=F) Prior, related post: On Aug 29, 2013, at 2:00 PM, Gerard Smits g_sm...@verizon.net wrote: Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ## ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Changing string to date
On 8/30/2013 3:36 AM, Uwe Ligges wrote: On 30.08.2013 11:59, Christofer Bogaso wrote: Hello again, I have a string which I need to put in some legitimate date format. My string is: MAY-14 And output format would be 05/01/2014, this should be of Date class, so that I can make some sensible calculation with it. I have tried this without any success: as.Date(MAY-14, format = %b-%y) [1] NA You must have set an English locale (to recognize the word May as a month) and you have to add a day. You can use the yearmon class from the zoo package as an intermediary to get around having to explicitly add a day library(zoo) as.Date(as.yearmon(MAY-14, format=%b-%y)) gives as.Date(as.yearmon(MAY-14, format=%b-%y)) [1] 2014-05-01 Best, Uwe Ligges Can somebody help me how I can achieve that? Thanks and regards, __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian S. Diggs, PhD Senior Research Associate, Department of Surgery Oregon Health Science University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strange conversion char to date
Maybe there is some built-in assumption about dating and when to change to the new century. It seems to kick-in at 1968 as.Date(25/04/69, %d/%m/%y) as.Date(25/04/68, %d/%m/%y) as.Date(25/04/60, %d/%m/%y) John Kane Kingston ON Canada -Original Message- From: frans.marcelis...@digipsy.nl Sent: Fri, 30 Aug 2013 15:12:40 +0200 To: r-help@r-project.org Subject: [R] strange conversion char to date Hi R-friends, Can anyone explain the following strange behavior to me? as.Date( 4/25/71,%m/%d/%y) [1] 1971-04-25 as.Date( 4/25/62,%m/%d/%y) [1] 2062-04-25 so 71 is converted to 1971, while 62 is converted to 2062? Does anyone know why? And is there a simple way to specify the date? (does works the same way in R 2.01 as well as in 1.9) --- dr F.H.G. (Frans) Marcelissen DigiPsy (www.DigiPsy.nl http://www.digipsy.nl/) Pomperschans 26 5595 AV Leende tel: 040 2065030/06 2325 06 53 skype adres: frans.marcelissen email: frans.marcelis...@digipsy.nl [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. GET FREE SMILEYS FOR YOUR IM EMAIL - Learn more at http://www.inbox.com/smileys Works with AIM®, MSN® Messenger, Yahoo!® Messenger, ICQ®, Google Talk™ and most webmails __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
Hi From: mohan.radhakrish...@polarisft.com [mailto:mohan.radhakrish...@polarisft.com] Sent: Friday, August 30, 2013 3:16 PM To: PIKAL Petr Cc: r-help@r-project.org Subject: RE: [R] Memory usage bar plot Hello, This memory usage should be graphed with time. Are there examples of scatterplots that can clearly show usage vs time ? This is for memory leak detection. Hm, Actually I do not understand what do you want. No data, no code just some vague description. If you have data frame with variables usage and time you can plot plot(time, usage) Regards Petr Thanks, Mohan From:PIKAL Petr petr.pi...@precheza.czmailto:petr.pi...@precheza.cz To: mohan.radhakrish...@polarisft.commailto:mohan.radhakrish...@polarisft.com mohan.radhakrish...@polarisft.commailto:mohan.radhakrish...@polarisft.com, r-help@r-project.orgmailto:r-help@r-project.org r-help@r-project.orgmailto:r-help@r-project.org Date:08/30/2013 05:33 PM Subject:RE: [R] Memory usage bar plot Hi For reading data into R you shall look to read.table and similar. For plotting ggplot could handle it. However I wonder if 100 times 50 bars is the way how to present your data. You shall think over what do you want to show to yourself or your audience. Maybe boxplots or scatterplots could be better. Petr -Original Message- From: r-help-boun...@r-project.orgmailto:r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of mohan.radhakrish...@polarisft.commailto:mohan.radhakrish...@polarisft.com Sent: Friday, August 30, 2013 1:25 PM To: r-help@r-project.orgmailto:r-help@r-project.org Subject: [R] Memory usage bar plot Hi, I haven't tried the code yet. Is there a way to parse this data using R and create bar plots so that each program's 'RAM used' figures are grouped together. So 'uuidd' bars will be together. The data will have about 50 sets. So if there are 100 processes each will have about 50 bars. What is the recommended way to graph these big barplots ? I am looking for only 'RAM used' figures. Thanks, Mohan Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check - 453.9 MiB = Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check -- 453.9 MiB = This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e- mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.comhttp://www.polarisft.com/ [[alternative HTML version deleted]] __ R-help@r-project.orgmailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-http://www.r-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e-mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list
Re: [R] strange conversion char to date
Hi,
You may try:
x1- c(4/25/71,4/20/64)
fun1- function(x, year){
m1-as.numeric(format(as.Date(x1,%m/%d/%y),%y))
m1- ifelse(m1year%%100,1900+m1,2000+m1)
m2- paste0(gsub((.*\\/).*$,\\1,x),m1)
as.Date(m2,%m/%d/%Y)
}
fun1(x1,1950)
#[1] 1971-04-25 1964-04-20
str(fun1(x1,1950))
# Date[1:2], format: 1971-04-25 1964-04-20
A.K.
- Original Message -
From: Frans Marcelissen frans.marcelis...@digipsy.nl
To: R-help@r-project.org R-help@r-project.org
Cc:
Sent: Friday, August 30, 2013 9:12 AM
Subject: [R] strange conversion char to date
Hi R-friends,
Can anyone explain the following strange behavior to me?
as.Date( 4/25/71,%m/%d/%y)
[1] 1971-04-25
as.Date( 4/25/62,%m/%d/%y)
[1] 2062-04-25
so 71 is converted to 1971, while 62 is converted to 2062? Does anyone know
why? And is there a simple way to specify the date?
(does works the same way in R 2.01 as well as in 1.9)
---
dr F.H.G. (Frans) Marcelissen
DigiPsy (www.DigiPsy.nl http://www.digipsy.nl/)
Pomperschans 26
5595 AV Leende
tel: 040 2065030/06 2325 06 53
skype adres: frans.marcelissen
email: frans.marcelis...@digipsy.nl
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
Hi For reading data into R you shall look to read.table and similar. For plotting ggplot could handle it. However I wonder if 100 times 50 bars is the way how to present your data. You shall think over what do you want to show to yourself or your audience. Maybe boxplots or scatterplots could be better. Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of mohan.radhakrish...@polarisft.com Sent: Friday, August 30, 2013 1:25 PM To: r-help@r-project.org Subject: [R] Memory usage bar plot Hi, I haven't tried the code yet. Is there a way to parse this data using R and create bar plots so that each program's 'RAM used' figures are grouped together. So 'uuidd' bars will be together. The data will have about 50 sets. So if there are 100 processes each will have about 50 bars. What is the recommended way to graph these big barplots ? I am looking for only 'RAM used' figures. Thanks, Mohan Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check - 453.9 MiB = Private + Shared = RAM used Program 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd 108.0 KiB + 12.5 KiB = 120.5 KiB klogd 124.0 KiB + 17.0 KiB = 141.0 KiB hidd 116.0 KiB + 30.0 KiB = 146.0 KiB acpid 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage 144.0 KiB + 15.0 KiB = 159.0 KiB gpm 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check -- 453.9 MiB = This e-Mail may contain proprietary and confidential information and is sent for the intended recipient(s) only. If by an addressing or transmission error this mail has been misdirected to you, you are requested to delete this mail immediately. You are also hereby notified that any use, any form of reproduction, dissemination, copying, disclosure, modification, distribution and/or publication of this e- mail message, contents or its attachment other than by its intended recipient/s is strictly prohibited. Visit us at http://www.polarisFT.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ddply for comparing simulation results
This might do it:
lhs=c('a','a','a','b')
rhs=c('a','b','b','b')
# function to determine differences
f_diff - function(l, r){
+ t_l - table(l)
+ t_r - table(r)
+ # compare 'l' to 'r'
+ sapply(names(t_l), function(x){
+ if (is.na(t_r[x])) return(t_l[x])
+ t_l[x] - t_r[x]
+ })
+ }
f_diff(lhs, rhs)
a.a b.b
2 -2
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Fri, Aug 30, 2013 at 1:28 PM, john doe anon.r.u...@gmail.com wrote:
Hi Jim,
Thanks for the quick reply! Data.table sounds like it will help me with my
performance problem. However, I think that setdiff does not do quite what I
need. Consider this example:
lhs=c('a','a','a','b')
rhs=c('a','b','b','b')
setdiff(lhs,rhs)
character(0)
I need to do an operation between lhs and rhs which gives this result:
a: 2
b: -2
It looks like base set operations call unique on their vectors before
performing the intersection, which does not allow me to measure the
magnitude of the difference between the sets.
Ari
On Fri, Aug 30, 2013 at 5:10 AM, jim holtman jholt...@gmail.com wrote:
try the 'data.table package. It gives the answer in less than a second.
# 1 million leads, half of which were simulated, half of which were not
id=1:100
isSimulated = c(rep(0,50), rep(1, 50))
userId=sample(1:10, 100, replace=T)
df_leads=data.frame(id, isSimulated, userId)
require(data.table)
Loading required package: data.table
data.table 1.8.8 For help type: help(data.table)
system.time({
+ df_leads - data.table(df_leads)
+ df_leads_sum - df_leads[
+ , list(count = .N)
+ , keyby = c('isSimulated', 'userId')
+ ]
+ })
user system elapsed
0.750.010.76
head(df_leads_sum)
isSimulated userId count
1: 0 1 5
2: 0 2 9
3: 0 3 5
4: 0 4 4
5: 0 5 3
6: 0 6 7
you can use 'setdiff' to find userIDs that are missing from one group
or the other:
#see which userIDs are missing between the groups
not_in - setdiff(df_leads_sum$userId[df_leads_sum$isSimulated == 0]
+ , df_leads_sum$userId[df_leads_sum$isSimulated == 1]
+ )
str(not_in)
int [1:697] 59 100 204 584 656 828 840 999 1012 1046 ...
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Aug 29, 2013 at 11:33 PM, john doe anon.r.u...@gmail.com wrote:
I am trying to use R and plyr to compare the effectiveness of various
algorithms for online advertising. At the core, I am simply counting
when a
user receives a lead: this is measured with the userId column. Leads
that
were sent in production have a 0 in the isSimulated column, and leads
that
were sent in our simulation have isSimulated=1. I have two questions:
one
about performance and one about how to use plyr to get the data in a
form
that I want.
Here is an example of my code:
# 1 million leads, half of which were simulated, half of which were not
id=1:100
isSimulated = c(rep(0,50), rep(1, 50))
userId=sample(1:10, 100, replace=T)
df_leads=data.frame(id, isSimulated, userId)
# split by simulated and userid, and then sum
system.time(df_leads_sum - ddply(df_leads, .(isSimulated, userId),
nrow))
user system elapsed
38.167 0.212 38.386
The above call to ddply is great because it allows me to create
histograms
of how many people receive just a few leads, or a lot of leads, both in
production and in the simulator.
Question 1: The above ddply call takes a while to execute. With
production
data it takes several minutes in R, but only a few seconds in MySQL. Is
there a way to improve the performance of the above call?
Question 2: What I would really like to do is create a histogram which
measures the distribution of change in leads between non-simulated and
simulated data. A complicating fact is that some users might only
appear in
simulated or non-simulated data, so I need to correclty handle the
absense
of a userId. (In production, users are actually guaranteed to appear in
production - but the crux of the problem is the same: userIds might be
missing in one of the splits). Can someone help me with this? I've
read
the documentation a few times, and think that the summarize function
might
be able to help, but I'm not quite sure how to do this.
Thanks.
--
You received this message because you are subscribed to the Google
Groups
manipulatr group.
To unsubscribe from this group and stop receiving emails from it, send
an
email to manipulatr+unsubscr...@googlegroups.com.
To post to this group, send email to manipul...@googlegroups.com.
Visit this
Re: [R] problem with newline using bquote(paste())
That explains it. Thanks for the info. Gerard On Aug 30, 2013, at 8:42 AM, Marc Schwartz marc_schwa...@me.com wrote: On Aug 30, 2013, at 10:26 AM, Gerard Smits g_sm...@verizon.net wrote: Hi All, This is a variant of a problem I posted yesterday (see below) where I found I had a large gap between my N= and he number I had evaluated using .(x). I seem to have trouble with newlines in a main title. I find now that all works as expected (no unsightly gap between my N= and the value, if all of the title is put on the same line. Whenever I try the newline, I run into problems. Below, I have one example that gives me no syntax errors, but simply does not print the information after the \n (the N= xxx) part. Any help appreciated. Thanks, Gerard PS using R 3.0.0 ss-n(m18_das28*b_dascore) par(oma=c(2,2,2,2)) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=as.expression(bquote(paste(Baseline xyz with Month 18 DAS28\n)), bquote(paste((N=,.(ss),, xlab=Baseline xyz, ylab=Month 18 DAS28, legend.plot=F) Prior, related post: On Aug 29, 2013, at 2:00 PM, Gerard Smits g_sm...@verizon.net wrote: Hi All, I'm using R 3.0.0. I'm trying to add the sample size of the paired data (calculated by a function n(), which returns a value of 70, correctly). My main title works fine except that the '70' appears far to the right on the line as in: at Month 18 (N= 70) Is there a way of left justifying the result of .(ss)? or some other way of removing with whitespace between n= and 70?. Thanks for any suggestions. Gerard library (car) data-read.csv(//users//smits//r_work//data.csv, header = TRUE) attach(data); ## ss-n(m18_das28*b_score) scatterplot(m18_das28~b_score, jitter=list(x=1, y=1), grid=F, smooth=F, las=1, pch=c(1), col='blue', main=bquote(paste(Hypothesis 9.4.1\nBaseline XYZ with Disease Activity (DAS28)\nat Month 18 (N=,.(ss),))), xlab=Baseline XYZ, ylab=Month 18 DAS28, legend.plot=F) You cannot use newlines in plotmath expressions. You will need to create each line of the plot title text separately using ?mtext instead of specifying 'main' in the plot call. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] calculate with different columns from different datasets
Thank you a lot A.K.! One more question: I'd like to compute the Spearman's rank correlation coefficients for V1 (from dat1) and V1 (from dat2) and so on... Do you know how to do that? I managed to write the Pearson's correlation product moment coefficient with your sapply-approach, but I have no idea how to rank it. Kind regards -- View this message in context: http://r.789695.n4.nabble.com/calculate-with-different-columns-from-different-datasets-tp4674918p4675034.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Memory usage bar plot
## Here is a plot. The input was parsed with Jim Holtman's code.
## The panel.dumbell is something I devised to show differences.
## Rich
input - readLines(textConnection(
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
-
453.9 MiB
=
Private + Shared = RAM used Program
96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
108.0 KiB + 12.5 KiB = 120.5 KiB klogd
124.0 KiB + 17.0 KiB = 141.0 KiB hidd
116.0 KiB + 30.0 KiB = 146.0 KiB acpid
124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
144.0 KiB + 15.0 KiB = 159.0 KiB gpm
136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
--
453.9 MiB
=))
# keep only the data
input - input[grepl('=', input)]
# separate into groups
grps - split(input, cumsum(grepl(= RAM, input)))
# parse the data (not sure if there is also 'MiB')
parsed - lapply(grps, function(.grp){
# parse ignoring first and last lines
.data - sub(.*= ([^ ]+) ([^ ]+)\\s+(.*), \\1 \\2 \\3
, .grp[2:(length(.grp) - 1L)]
)
# return matrix
do.call(rbind, strsplit(.data, ' '))
})
parsed
tmp1 - do.call(rbind, lapply(parsed, function(x) data.frame(x)))
names(tmp1) - c(RamUsed, units, Program)
tmp1$Time - factor(rep(1:2, each=7))
tmp1$RamUsed - as.numeric(tmp1$RamUsed)
library(lattice)
dotplot(Program ~ RamUsed, groups=Time, data=tmp1)
## this is silly. Let me construct a more interesting example with
different values at each time.
tmp1$RamUsed[8:14] - tmp1$RamUsed[1:7] + 10*(sample(1:7))
tmp1
dotplot(Program ~ RamUsed, groups=Time, data=tmp1,
auto.key=list(title=Time, border=TRUE, columns=2))
panel.dumbell - function(x, y, ..., lwd=1) {
n - length(x)/2
panel.segments(x[1:n], as.numeric(y)[n+(1:n)], x[n+(1:n)],
as.numeric(y)[n+(1:n)], lwd=lwd)
panel.dotplot(x, y, ...)
}
dotplot(Program ~ RamUsed, groups=Time, data=tmp1,
auto.key=list(title=Time, border=TRUE, columns=2),
panel=panel.dumbell,
par.settings=list(superpose.symbol=list(pch=19)),
)
On Fri, Aug 30, 2013 at 9:44 AM, jim holtman jholt...@gmail.com wrote:
Here is how to parse the data and put it into groups. Not sure what
the 'timing' of each group is since not time information was given.
Also not sure is there is an 'MiB' qualifier on the data, but you have
the matrix of data which is easy to do with as you want.
input - readLines(textConnection(
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ -
+ 453.9 MiB
+
+ =
+ Private + Shared = RAM used Program
+
+ 96.0 KiB + 11.5 KiB = 107.5 KiB uuidd
+ 108.0 KiB + 12.5 KiB = 120.5 KiB klogd
+ 124.0 KiB + 17.0 KiB = 141.0 KiB hidd
+ 116.0 KiB + 30.0 KiB = 146.0 KiB acpid
+ 124.0 KiB + 29.5 KiB = 153.5 KiB hald-addon-storage
+ 144.0 KiB + 15.0 KiB = 159.0 KiB gpm
+ 136.0 KiB + 26.5 KiB = 162.5 KiB pam_timestamp_check
+ --
+ 453.9 MiB
+ =))
# keep only the data
input - input[grepl('=', input)]
# separate into groups
grps - split(input, cumsum(grepl(= RAM, input)))
# parse the data (not sure if there is also 'MiB')
parsed - lapply(grps, function(.grp){
+ # parse ignoring first and last lines
+ .data - sub(.*= ([^ ]+) ([^ ]+)\\s+(.*), \\1 \\2 \\3
+ , .grp[2:(length(.grp) - 1L)]
+ )
+ # return matrix
+ do.call(rbind, strsplit(.data, ' '))
+ })
parsed
$`1`
[,1][,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB hald-addon-storage
[6,] 159.0 KiB gpm
[7,] 162.5 KiB pam_timestamp_check
$`2`
[,1][,2] [,3]
[1,] 107.5 KiB uuidd
[2,] 120.5 KiB klogd
[3,] 141.0 KiB hidd
[4,] 146.0 KiB acpid
[5,] 153.5 KiB
Re: [R] problem in while loop?
Thanks Berend.
Yes that is right. I should get 5 values(results) of x_e because I have
five values of X. I wonder how can I fix it?
On 30/08/2013 13:13, Berend Hasselman wrote:
On 30-08-2013, at 09:44, Jonssonamen.alya...@bordeaux.inra.fr wrote:
I have three datasets that I want to compute the errors between them using
linear regression.for this, I want to iterate to reach certain criteria for
the calibration. if changes become smaller than eps the iteration is
successful, hence stop and write parameters into cal:eps=0.1 if number
of iterations is itermax the iteration failed, hence stop and fill cal
with missing value itermax=400
So I tried this code:
x= c(5,2,4,2,1)
y= c(5,3,4,6,9)
z= c(5,8,4,7,3)
itermax=400
get initial calibration parameters, here we assume that:x is the reference
dataset offset x_a=0, slope x_b=1 the other two datasets y, z are
calibrated to x using a simple linear regression
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))#calculate error of x
iter- 0
while(((x_e-x) 0.1) (iter itermax)) {
iter- 0 ##start iteration
x = x_e
res=lm(x~y)
y_a=coef(res)[1] ; y_b=coef(res)[2]
res1=lm(x~z)
z_a=coef(res1)[1] ; z_b=coef(res1)[2]
y_t = y/y_b - y_a/y_b # calibrate y
z_t = z/z_b - z_a/z_b #calibrate z
x_e = sqrt(mean((x-y_t)*(x-z_t)))
iter- iter + 1 # increase iteration counter
}
But I got the same result for X_e before and after the loop:
x_e
[1] 6.454089
I tried your code and got this error message:
Error in model.frame.default(formula = x ~ y, drop.unused.levels = TRUE) :
variable lengths differ (found for 'y')
Calls: lm - eval - eval - Anonymous - model.frame.default
And looking at your code: x is a vector, x_e is a scalar and in the while loop
you are assigning x_e to x so x is then a scalar.
Berend
--
Amen Alyaari, UPMC
PhD student
Unit of Functional Ecology Environmental Physics [EPHYSE]
National Institute of Agricultural Research [INRA].
71, Avenue Edouard Bourlaux
33140 Villenave d'Ornon
Téléphone : +33(0) 5 57 12 24 27
Fax : +33 (0)5 57 12 24 20
FRANCE
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] mean
Hi, In case, some of the list elements are vectors, this procedure would not work. a1- list(c(3,5),4,5:6) as.numeric(a1) #Error: (list) object cannot be coerced to type 'double' The OP didn't provide any info as to how the data looks like. So, these are just assumptions. mean(unlist(a1)) #if the overall mean is needed. #[1] 4.6 #or mean of each list element sapply(a1,mean) #[1] 4.0 4.0 5.5 #or if it includes data.frame or matrix. a2- list(4, as.data.frame(matrix(1:10,5,2)),4:6) a3- list(4, matrix(1:10,5,2),4:6) mean(unlist(a2)) #[1] 5.285714 mean(unlist(a3)) #[1] 5.285714 A.K. - Original Message - From: Albyn Jones jo...@reed.edu To: agnes69 fes...@gredeg.cnrs.fr Cc: r-help@r-project.org Sent: Friday, August 30, 2013 11:59 AM Subject: Re: [R] mean It would be easier to diagnose the problem if you included an example illustrating exactly what you did. I'll guess: a - list(3,4,5) mean(a) [1] NA Warning message: In mean.default(a) : argument is not numeric or logical: returning NA mean(as.numeric(a)) [1] 4 But that's just a guess, as I don't know the actual contents of your list! albyn On Fri, Aug 30, 2013 at 5:18 AM, agnes69 fes...@gredeg.cnrs.fr wrote: When I try to apply mean to a list, I get the answer : argument is not numeric or logical: returning NA Could you help me? (I am a very beginner) -- View this message in context: http://r.789695.n4.nabble.com/mean-tp4674999.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] writing spdiags function for R
I am posting here the brilliant solutions, gently provided by Prof JC Nash
nashjc at uottawa.ca, to me; so that people struggling in the future with
the same issue can find a way through.
FYI, compared to the original Matlab implementation: 1) it does not handle
the case with more than one input, and
2) (m n) matrices give the B matrix columns in a different order, but the
d vector of indices will also be changed
accordingly, so the set of columns is OK, just ordered differently cif. JN.
Copied below the .R version of the spdiags code, a Fortran implementation of
it, and an R wrapper to run the .f
thanks again JN, your help was really invaluable :)
#
R version
#
spdiagsj - function(A) { # A is a matrix
m - dim(A)[[1]]
n - dim(A)[[2]]
k - min(m, n) # length of diagonals
Bdata-NULL # start with nothing in B matrix (as vector)
jb-0 # column index of last column saved for B
d-NULL # index vector of diagonals from A
# d contains 0 for the principal diagonal, -i for i'th lower
# diagonal (prefaced with zeros), +j for j'th upper diagonal
# (suffixed by zeros)
q-(m-1)+n # There are m-1 subdiagonals and n-1 superdiagonals + main
diagonal
if (m n) { # tall matrix
Adata - as.vector(t(A)) # convert to vector BY ROWS
} else { # fat or square matrix (m = n)
Adata - as.vector(A) # convert to vector BY COLUMNS
}
# Augment the data with columns of zeros fore and aft
Adata-c(rep(0,(k-1)*k), Adata, rep(0,(k-1)*k))
cat(Augmented Adata with ,length(Adata), elements:\n)
print(Adata)
for (i in 1:q) {
tv - c(Adata[[k*(i-1)+1]], rep(0,k-1)) # top element of augmented
column
# plus enough zeros to pad it out (some zeros may be replaced
below)
# i.e., start at 1, then m+1, 2*m+1 etc.
qx-min((q-i), (k-1)) #
cat( qx=,qx,\n)
if (qx 0) { # qx will be 0 when we are at last
superdiagonal,i.e., i == q
for (j in 1:qx) { # get the rest of the diagonal elements
tv[[j+1]] - Adata[[k*(i+j-1)+j+1]]
}
}
if (any(tv != 0)){ # check for non-zeros, if there are, then save
jb-jb+1 # next column of B
d-c(d,(i-k)) # record the index
Bdata-c(Bdata, tv) # save the diagonal as column of B in
vector form
}
}
if (m n) d - -d # reset index
cat(Bdata:); print(Bdata)
B - matrix(Bdata, nrow=k, byrow=FALSE) # convert to matrix form
result-list(B=B, d=d)
}
cat(Matlab example 1\n)
dta - c(0, 5, 0, 10, 0, 0, 0, 0, 6, 0, 11, 0, 3, 0, 0, 7, 0, 12, 1, 4, 0,
0, 8, 0, 0, 2, 5, 0, 0, 9)
A1 - matrix(dta, nrow=5, ncol=6, byrow=TRUE)
print(A1)
res1-spdiagsj(A1)
print(res1)
tmp-readline(Next)
cat(Matlab example 2\n)
n-10 # choose 10 for an example
A2-matrix(rep(0, n*n), nrow=n, ncol=n)
for (i in 1:n) {
for (j in 1:n) {
if (i == j) A2[i, j] - -2
if ( (i == (j-1)) || (i == (j+1))) A2[i,j] - 1
}
}
print(A2)
res2-spdiagsj(A2)
print(res2)
tmp-readline(Next)
cat(Matlab example 3\n)
dta3 - c(11, 0, 13, 0, 0, 22, 0, 24, 0, 0, 33, 0, 41, 0, 0, 44,
0, 52, 0, 0, 0, 0, 63, 0, 0, 0, 0, 74)
A3 - matrix(dta3, nrow=7, ncol=4, byrow=TRUE)
print(A3)
res3-spdiagsj(A3)
print(res3)
tmp-readline(Next)
cat(try transpose\n)
A3T-t(A3)
print(A3T)
res3T-spdiagsj(A3T)
print(res3T)
tmp-readline(Next)
cat(Example 5B \n)
dta5b1-c(6, 0, 13, 0, 0, 0, 7, 0, 14, 0, 1, 0, 8, 0, 15, 0, 2, 0, 9, 0, 0,
0, 3, 0, 10)
A5b1-matrix(dta5b1, nrow=5, ncol=5, byrow=TRUE)
print(A5b1)
res5b1-spdiagsj(A5b1)
print(res5b1)
#
Fortran version
#
subroutine jspd(m, n, k, Adata, jb, Bdata, d, tv, na, nb, nd)
C Central part of spdiags for R
C m and n are row and column sizes of A (underlying matrix)
C jb will be number of returned diagonals
C returns jb, Bdata, d
integer m, n, na, nb, nd, jb, d(nd)
integer i, j, k, kend, q, mn, kk1, js, je, qx
double precision Adata(na), Bdata(nb), tv(k)
LOGICAL not0
C k = min(m, n)
C ?? check if k=1
C Bdata-NULL # start with nothing in B matrix (as vector)
jb = 0
kend=0
C column index of last column saved for B
C d = NULL # index vector of diagonals from A
C # d contains 0 for the principal diagonal, -i for i'th lower
C # diagonal (prefaced with zeros), +j for j'th upper diagonal
C # (suffixed by zeros)
q = (m-1)+n
C There are m-1 subdiagonals and n-1 superdiagonals + main diagonal
C assume we have already built Adata for tall or fat matrix
C # Augment the data with columns of zeros fore and aft
mn = m*n
C print *,Original Adata
C print
Re: [R] Validating data type
It sounds like that column of data is not of type date at all. You cannot have one element of a column different from the rest of the column. In a data.frame you can have different types of data in different columns but not in the same column. Where mydata is your data.frame do : str(mydata) This will give you a listing of the type of data in each column in your data.frame. My guess would be that R has read in that column as character or factor. Just because it looks like a date on the screen does not mean it is one. You probably will have to convert it to a date. See ?as.Date for one way to do this. You might also want to have a look at the lubridate package. For further reference https://github.com/hadley/devtools/wiki/Reproducibility http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example In particular for a question like yours, supplying some sample data using dput() would have really helped. If you are still having a problem do : dput(mydata) and paste the output into the email. The reader can then paste the data into their version of R and see exactly what you are working with. For large datasets usually a sample amoutnt will do , so dput(head(mydata, 100) for example will supply 100 rows of data. Below is a example of a data.frame in dput format. Just copy and paste it into R and you will have a new date.frame John Kane Kingston ON Canada ##dput file=== dat1 - structure(list(xx = structure(c(5L, 6L, 10L, 9L, 17L, 10L, 15L, 16L, 5L, 14L, 5L, 7L, 17L, 6L, 11L, 8L, 5L, 3L, 1L, 17L, 7L, 10L, 5L, 15L, 15L, 16L, 17L, 14L, 8L, 13L, 12L, 13L, 18L, 9L, 5L, 2L, 1L, 16L, 1L, 1L, 1L, 16L, 4L, 10L, 1L, 18L, 18L, 14L, 13L, 4L), .Label = c(a, b, d, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t), class = factor), yy = c(0.332304663767243, -1.77867401940838, 0.828612337938625, 0.481702424196278, 0.0825987297345907, -1.40224568135063, -0.24334456876, 0.0865304079310024, -0.124012796374592, -0.0107544463484595, -0.542307211820575, 0.0129727866797914, -0.478553152291621, -1.63895681984396, 0.0911014618211326, -0.890215628553797, -1.42140590396317, 0.202337039384179, 1.30089052407852, 0.07517013402338, -0.807355878474237, 1.12978841894929, 0.154740986108198, 0.21209595540936, 0.65345449749952, 0.533479658343466, 0.665882552612018, -0.60572360781, -0.0971202279326936, -0.862179166296771, -0.977706435316816, 0.559634439503645, 0.0320050874597674, -1.65502174652502, 0.853046541850183, -0.801904205812903, -0.820335448022446, -0.912451936657161, 0.222469916395761, 0.0168002536713376, -0.218537143966283, 1.00191128410043, -0.430912734152427, -1.1327880971227, -0.664284053548425, 1.3082467197158, 1.46148850229679, -1.11954785811615, -1.61706514557631, 0.604530320200236)), .Names = c(xx, yy), row.names = c(NA, -50L), class = data.frame) ##===end dput file -Original Message- From: jeffj...@worldvision.org Sent: Thu, 29 Aug 2013 20:29:54 -0700 To: r-help@r-project.org Subject: [R] Validating data type I'm very new to R. I have a data file that I have read in via read.csv. I expect one of the columns to be of type date for example. However at least one value in that column is not of date type. I know this because another program I am trying to process the file with is erroring, yet it doesn't tell me what row/value is erroring. Does R have a way to: treat column x as date type, and print out all values/row numbers do not conform to that type for that specified column? Many thanks! Jeff [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. TRY FREE IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if5 Capture screenshots, upload images, edit and send them to your friends through IMs, post on Twitter®, Facebook®, MySpace™, LinkedIn® – FAST! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] delete repeated values in array
Hi, May be this helps: rle(try)$values #[1] 1 2 3 1 2 4 #or aggregate(try,list(cumsum(c(1,abs(diff(try),unique)[,2] #[1] 1 2 3 1 2 4 #or res-tapply(try,cumsum(c(1,abs(diff(try,head,1) attr(res,dimnames)-NULL res #[1] 1 2 3 1 2 4 A.K. I am trying to delete repeated values in an array. try - c(1,1,1,1,1,2,2,2,3,3,3,1,1,1,2,2,4,4,4) what I want back is: 1 2 3 1 2 4 so unique() doesn't work for my purposes as it would give me: 1 2 3 4 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error: XML content does not seem to be XML: ''
Hi there, I am attempting to access an Allegro Graph site, using SPARQL function in RStudio and am getting this error message Error: XML content does not seem to be XML: ''. However, the same query that I am using within the SPARQL function can get back the results when directly from the browser. Please advise. Cheers, Anil [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Basic Error I'm sure
Hi, I'm just starting to learn how to use R and I'm trying to create a histogram with 7 breaks. This is my code so far: dat=read.table(titanic.csv,header=TRUE,sep=,,na.string=.) age=dat$Age breaks=seq(min(age),max(age),length=7) hist(age,breaks,freq=FALSE) I don't know why, but on the fourth line it always gives this error: Error in if (from == to) rep.int(from, length.out) else as.vector(c(from, : missing value where TRUE/FALSE needed What am I doing wrong? I know the data itself does have a lot of NA values for age, would this be the problem? Thanks, CZ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] does rstudio-server support named configuration
I appreciate your time, thank you in advance :) I'd like to start 2 rstudio-server in one Linux machine, one daemon listen port 8787 running R-2.15, on daemon listen port 8788 running R-3.01. Can I have named rstudio-server configurations so I can start rstudio-server like this : :rstudio-server --config-file = R-2.15.conf start, it runs R-2.15 listening port 8787 :rstudio-server --config-file = R-3.01.conf start, it runs R-3.01 listening port 8788 Ipython-notebook support named configuration, does rstudio-server have this feature ? Cheers, Choulin. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] for loop of a geometric sequence
so I have to create a for loop of the geometric sequence h(x,n)=1+x+x^2+x^3^4...x^n. I know that it would be easier to simply vectorize the sequence to x^(0:n), but I am required to make the loop, and I can't wrap my brain around how to loop it because the equation is so simple. -- View this message in context: http://r.789695.n4.nabble.com/for-loop-of-a-geometric-sequence-tp4675035.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Basic Error I'm sure
Hi Casey, Yes, if there are missing values than you don't strictly speaking know what the minimum value is. You need to tell min() and max() to exclude missing, i.e., breaks=seq(min(age, na.rm=TRUE),max(age, na.rm=TRUE),length=7) Best, Ista On Fri, Aug 30, 2013 at 9:41 PM, Casey Zhang casey1...@gmail.com wrote: Hi, I'm just starting to learn how to use R and I'm trying to create a histogram with 7 breaks. This is my code so far: dat=read.table(titanic.csv,header=TRUE,sep=,,na.string=.) age=dat$Age breaks=seq(min(age),max(age),length=7) hist(age,breaks,freq=FALSE) I don't know why, but on the fourth line it always gives this error: Error in if (from == to) rep.int(from, length.out) else as.vector(c(from, : missing value where TRUE/FALSE needed What am I doing wrong? I know the data itself does have a lot of NA values for age, would this be the problem? Thanks, CZ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] does rstudio-server support named configuration
The Rstudio support forum is at http://support.rstudio.com Best, Ista On Fri, Aug 30, 2013 at 12:41 PM, qiulin qiulin.w...@gmail.com wrote: I appreciate your time, thank you in advance :) I'd like to start 2 rstudio-server in one Linux machine, one daemon listen port 8787 running R-2.15, on daemon listen port 8788 running R-3.01. Can I have named rstudio-server configurations so I can start rstudio-server like this : :rstudio-server --config-file = R-2.15.conf start, it runs R-2.15 listening port 8787 :rstudio-server --config-file = R-3.01.conf start, it runs R-3.01 listening port 8788 Ipython-notebook support named configuration, does rstudio-server have this feature ? Cheers, Choulin. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Basic Error I'm sure
Hello, Check the result of min(c(1,2,3,4,NA,6)) And read ?min Hope this helps, Pascal 2013/8/31 Casey Zhang casey1...@gmail.com Hi, I'm just starting to learn how to use R and I'm trying to create a histogram with 7 breaks. This is my code so far: dat=read.table(titanic.csv,header=TRUE,sep=,,na.string=.) age=dat$Age breaks=seq(min(age),max(age),length=7) hist(age,breaks,freq=FALSE) I don't know why, but on the fourth line it always gives this error: Error in if (from == to) rep.int(from, length.out) else as.vector(c(from, : missing value where TRUE/FALSE needed What am I doing wrong? I know the data itself does have a lot of NA values for age, would this be the problem? Thanks, CZ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] does rstudio-server support named configuration
Hello, Here is the R-help mailing list. For Rstudio-sever support, please see http://support.rstudio.org/help/discussions Regards, Pascal 2013/8/31 qiulin qiulin.w...@gmail.com I appreciate your time, thank you in advance :) I'd like to start 2 rstudio-server in one Linux machine, one daemon listen port 8787 running R-2.15, on daemon listen port 8788 running R-3.01. Can I have named rstudio-server configurations so I can start rstudio-server like this : :rstudio-server --config-file = R-2.15.conf start, it runs R-2.15 listening port 8787 :rstudio-server --config-file = R-3.01.conf start, it runs R-3.01 listening port 8788 Ipython-notebook support named configuration, does rstudio-server have this feature ? Cheers, Choulin. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
