date:20121119




On 19.11.2012 08:49, Einat wrote:

Hello,
How can I run a backward stepwise regression with part of the variables
fixed, while the others participate in the backward stepwise analysis?
Thank you, Einat


Read ?step and about its argument scope that can be a list with a 
lower component where you specify the minimal model.


Uwe Ligges





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Re: [R] loop to subtract arrays / error


Hello,

Try the following.

Xjj - matrix(nrow = 101, ncol = 1000)
for (i in 1:dim(Vsimr)[2]) {
Xjj[, i] - Vobsr - Vsimr[, i]
}

Hope this helps,

Rui Barradas
Em 19-11-2012 01:41, iembry escreveu:

Hi everyone, I am having trouble with creating a loop to subtract arrays.

In R, this is what I have done:

Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) # see data
below
Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,) # see
data below
Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus Vobsr
is 101x1 double matrix (column 1 is date information)
Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus Vsimr
is 101x1000 double matrix (column 1 is date information)
Vobsr - Vsimr

Error in Vobsr - Vsimr : non-conformable arrays

Thus I attempted to create the loop below to perform the subtraction
operation for each of the 1000 columns.


dim(Vsimr)[2]

[1] 1000


for (i in 1:dim(Vsimr)[2]) {

 Xjj - Vobsr - Vsimr[,i]
}
Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

How can I subtract each column of Vsimr from the single column of Vobsr over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.

How can I subtract each column of Vsimr from the single column of Vobsr over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.


I am including some of the data from the files that I am operating on below:
1 column of Observed_Flow.csv
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
78.184
87.179
97.862
110.23
124.29
140.08
157.57
176.76
197.63
220.18
244.4
270.31
297.88
327.14
358.09
390.71
425.03
461.03
498.72
538.09
579.16
621.91
666.35
712.48
760.29
809.8
860.99
913.87
968.44
1024.7
1082.6
1142.3
1203.6
1266.6
1331.3
1397.7
1465.7
1535.5
1606.9
1680.1
1754.9
1831.4
1907.1
1981.9
2055.9
2129
2201.3
2272.7
2343.3
2413.1
2482
2550.1
2617.3
2683.7
2749.2
2813.9
2877.8
2940.8
3003
3064.3
3124.8
3184.4
3243.2
3301.1
3358.2
3414.5
3469.9
3524.4
3578.2
3631
3683.1
3734.3
3784.6
3834.1
3882.8
3930.6
3977.6
4023.7
4069
4113.4
4157
4199.8
4241.7
4282.7
4323
4362.3
4400.9
4438.6
4475.4
4511.4
4546.6


2 columns of 1000 columns of 1000Samples_Vsim.csv
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
24.86 85.94
22.01 96.65
19.65 109.23
17.8 123.67
16.46 139.96
15.61 158.12
15.27 178.14
15.43 200.02
16.1 223.75
17.27 249.35
18.94 276.81
21.11 306.13
23.79 337.31
26.97 370.34
30.65 405.24
34.84 442
39.52 480.62
44.71 521.1
50.41 563.44
56.61 607.64
63.31 653.7
70.51 701.62
78.21 751.4
86.42 803.04
95.13 856.53
104.35 911.89
114.06 969.11
124.28 1028.2
135.01 1089.1
146.23 1151.9
157.96 1216.6
170.19 1283.1
182.93 1351.5
196.16 1421.7
209.9 1493.8
224.15 1567.8
238.89 1643.6
254.14 1721.3
269.89 1800.8
286.15 1882.2
302.91 1965.5
320.17 2050.6
337.18 2134.8
353.93 2218.1
370.44 2300.4
386.69 2381.8
402.7 2462.3
418.45 2541.8
433.95 2620.4
449.2 2698.1
464.2 2774.9
478.94 2850.7
493.44 2925.6
507.68 2999.5
521.67 3072.6
535.41 3144.7
548.9 3215.8
562.14 3286.1
575.12 3355.4
587.86 3423.8
600.34 3491.2
612.57 3557.7
624.55 3623.3
636.28 3688
647.76 3751.7
658.98 3814.5
669.96 3876.4
680.68 3937.3
691.15 3997.3
701.37 4056.4
711.34 4114.6
721.06 4171.8
730.52 4228.1
739.74 4283.4
748.7 4337.9
757.41 4391.4
765.87 4443.9
774.08 4495.6
782.04 4546.3
789.74 4596.1
797.2 4644.9
804.4 4692.8
811.35 4739.8
818.05 4785.9
824.5 4831
830.7 4875.2
836.64 4918.5
842.33 4960.8
847.78 5002.2
852.97 5042.7
857.91 5082.3



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Re: [R] read.table




On 19.11.2012 08:07, li1127217ye wrote:

read.table(data.txt,header=TRUE,

colClasses=c(character,character,numeric,character,
numeric,numeric))
 a   b   c d e f
1 SPX LSZ 100 C 0 34.40
2 SPX LSZ 100 P 0  1.30
3 SPX LSZ 105 C 0 30.30
4 SPX LSZ 105 P 0  1.85
5 SPX LSZ 110 C 0 26.40

It's right result! header=TRUE,not header=T. I don't know why not.



Because you have a T in your workspace that is not an alias for TRUE?

Uwe Ligges




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Re: [R] read.table

2012-11-19 Thread PIKAL Petr

Hi

 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of li1127217ye
 Sent: Monday, November 19, 2012 8:08 AM
 To: r-help@r-project.org
 Subject: Re: [R] read.table

  read.table(data.txt,header=TRUE,
 colClasses=c(character,character,numeric,character,
 numeric,numeric))
 a   b   c d e f
 1 SPX LSZ 100 C 0 34.40
 2 SPX LSZ 100 P 0  1.30
 3 SPX LSZ 105 C 0 30.30
 4 SPX LSZ 105 P 0  1.85
 5 SPX LSZ 110 C 0 26.40

 It's right result! header=TRUE,not header=T. I don't know why not.

You probably has some T variable defined in your workspace. What is result if 
you write

T

in your console.
Regards
Petr

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 https://stat.ethz.ch/mailman/listinfo/r-help
 PLEASE do read the posting guide http://www.R-project.org/posting-
 guide.html
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Re: [R] loop to subtract arrays / error


Hello,

Or simpler, since Vobsr only has one column:

Xjj - as.vector(Vobsr) - Vsimr

Hope this helps,

Rui Barradas
Em 19-11-2012 10:05, Rui Barradas escreveu:

Hello,

Try the following.

Xjj - matrix(nrow = 101, ncol = 1000)
for (i in 1:dim(Vsimr)[2]) {
Xjj[, i] - Vobsr - Vsimr[, i]
}

Hope this helps,

Rui Barradas
Em 19-11-2012 01:41, iembry escreveu:
Hi everyone, I am having trouble with creating a loop to subtract 
arrays.


In R, this is what I have done:
Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) # 
see data

below
Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,) # 
see

data below
Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus 
Vobsr

is 101x1 double matrix (column 1 is date information)
Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus 
Vsimr

is 101x1000 double matrix (column 1 is date information)
Vobsr - Vsimr

Error in Vobsr - Vsimr : non-conformable arrays

Thus I attempted to create the loop below to perform the subtraction
operation for each of the 1000 columns.


dim(Vsimr)[2]

[1] 1000


for (i in 1:dim(Vsimr)[2]) {

 Xjj - Vobsr - Vsimr[,i]
}
Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

How can I subtract each column of Vsimr from the single column of 
Vobsr over

the 1000 columns present?

I would like to thank each of you in advance for your assistance.

How can I subtract each column of Vsimr from the single column of 
Vobsr over

the 1000 columns present?

I would like to thank each of you in advance for your assistance.


I am including some of the data from the files that I am operating on 
below:

1 column of Observed_Flow.csv
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
78.184
87.179
97.862
110.23
124.29
140.08
157.57
176.76
197.63
220.18
244.4
270.31
297.88
327.14
358.09
390.71
425.03
461.03
498.72
538.09
579.16
621.91
666.35
712.48
760.29
809.8
860.99
913.87
968.44
1024.7
1082.6
1142.3
1203.6
1266.6
1331.3
1397.7
1465.7
1535.5
1606.9
1680.1
1754.9
1831.4
1907.1
1981.9
2055.9
2129
2201.3
2272.7
2343.3
2413.1
2482
2550.1
2617.3
2683.7
2749.2
2813.9
2877.8
2940.8
3003
3064.3
3124.8
3184.4
3243.2
3301.1
3358.2
3414.5
3469.9
3524.4
3578.2
3631
3683.1
3734.3
3784.6
3834.1
3882.8
3930.6
3977.6
4023.7
4069
4113.4
4157
4199.8
4241.7
4282.7
4323
4362.3
4400.9
4438.6
4475.4
4511.4
4546.6


2 columns of 1000 columns of 1000Samples_Vsim.csv
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
24.86 85.94
22.01 96.65
19.65 109.23
17.8 123.67
16.46 139.96
15.61 158.12
15.27 178.14
15.43 200.02
16.1 223.75
17.27 249.35
18.94 276.81
21.11 306.13
23.79 337.31
26.97 370.34
30.65 405.24
34.84 442
39.52 480.62
44.71 521.1
50.41 563.44
56.61 607.64
63.31 653.7
70.51 701.62
78.21 751.4
86.42 803.04
95.13 856.53
104.35 911.89
114.06 969.11
124.28 1028.2
135.01 1089.1
146.23 1151.9
157.96 1216.6
170.19 1283.1
182.93 1351.5
196.16 1421.7
209.9 1493.8
224.15 1567.8
238.89 1643.6
254.14 1721.3
269.89 1800.8
286.15 1882.2
302.91 1965.5
320.17 2050.6
337.18 2134.8
353.93 2218.1
370.44 2300.4
386.69 2381.8
402.7 2462.3
418.45 2541.8
433.95 2620.4
449.2 2698.1
464.2 2774.9
478.94 2850.7
493.44 2925.6
507.68 2999.5
521.67 3072.6
535.41 3144.7
548.9 3215.8
562.14 3286.1
575.12 3355.4
587.86 3423.8
600.34 3491.2
612.57 3557.7
624.55 3623.3
636.28 3688
647.76 3751.7
658.98 3814.5
669.96 3876.4
680.68 3937.3
691.15 3997.3
701.37 4056.4
711.34 4114.6
721.06 4171.8
730.52 4228.1
739.74 4283.4
748.7 4337.9
757.41 4391.4
765.87 4443.9
774.08 4495.6
782.04 4546.3
789.74 4596.1
797.2 4644.9
804.4 4692.8
811.35 4739.8
818.05 4785.9
824.5 4831
830.7 4875.2
836.64 4918.5
842.33 4960.8
847.78 5002.2
852.97 5042.7
857.91 5082.3



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Re: [R] function customization




On 16.11.2012 09:55, maxbre wrote:

Given my reproducible example:

new.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.007,
0.21, 0.077, 0.01, 0.033, 0.05, 0.014), LR = c(FALSE, FALSE,
TRUE, FALSE, TRUE, FALSE, FALSE, FALSE, TRUE, FALSE, TRUE, TRUE
), group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L,
2L, 2L), .Label = c(1, 2), class = factor)), .Names = c(TEC,
LR, group), row.names = c(NA, -12L), class = data.frame)

And this few lines of code:

library(NADA)

out-with(new.ex,
   cenfit(TEC, LR, group)
 )
out

giving the following output:

 n n.cen median   mean sd
group=1 6 2  0.033 0.05827778 0.08357853
group=2 6 3  0.033 0.0698 0.07925407



The output is produces from the survfit object by
 survival:::print.survfit
just adapt the code for your need and add the columns you like.

Uwe Ligges




I would like to add one more result for each group to the above output,
namely “sum”, computed as the product of “n” times “mean”

This is pretty much similar (a slight variation) to a question I posted
earlier in:
http://r.789695.n4.nabble.com/How-to-modify-a-S4-function-in-the-package-NADA-td4649586.html

But in this case I have some problems in modifying the cenfit() function
dealing with group as a factor

My objective is to modify the original function cenfit() so that to get also
the computation of “sum” as product of “n” times “mean”

For some reasons I can not properly understand, I’m not able to successfully
modify my earlier attempt (which was not accounting for groups)

mycenfit - function(x) {
  s = summary(x)
  c(n = nrow(s), n.cen = nrow(s) - sum(s$n.event), median = median(x),
  mean = mean(x)[[mean]], sd = sd(x), sum=mean(x)[[mean]]*length(x))
}

how to change it in order to properly deal with groups?

Thank you for any help

max



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Re: [R] Simulation with cpm package




On 13.11.2012 15:45, Christopher Desjardins wrote:

Hi,
I am running the following code based on the cpm vignette's code. I believe
the code is syntactically correct but it just seems to hang R. I can get
this to run if I set the sims to 100 but with 2000 it just hangs. Any ideas
why?


No: Works for me and completes within 90 minutes.

Uwe Ligges



Thanks,
Chris

library(cpm)
cpmTypes - c(Kolmogorov-Smirnov,Mann-Whitney,Cramer-von-Mises)
changeMagnitudes - c(1, 2, 4, 5)
changeLocations - c(50,100,300)
sims - 2000
ARL0 - 500
startup - 20
results - list()
for (cpmType in cpmTypes) {
   results[[cpmType]] - matrix(numeric(length(changeMagnitudes) *
  length(changeLocations)), nrow =
length(changeMagnitudes))
   for (cm in 1:length(changeMagnitudes)) {
 for (cl in 1:length(changeLocations)) {
   print(sprintf(cpm:%s magnitude::%s location:%s,
 cpmType, changeMagnitudes[cm], changeLocations[cl]))
   temp - numeric(sims)
   for (s in 1:sims) {
 x -c(rchisq(changeLocations[cl], df=3), rchisq(2000,

df=changeMagnitudes[cm]))
 temp[s] -detectChangePoint(x, cpmType,
 ARL0=ARL0,
startup=startup)$detectionTime
   }
   results[[cpmType]][cm,cl] - mean(temp[temp  changeLocations[cl]]) -
 changeLocations[cl]
 }
   } }

[[alternative HTML version deleted]]

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Re: [R] Interpretation of davies.test() in segmented package

2012-11-19 Thread Vito Muggeo


dear Greg,

 It does not, however give me
 Pr(|t|) for the break point coefficients.
Of course a test H_0: breakpoint=0 is meaningless..

 I need to answer the question
 H:0 Beta0=Beta with a certainty metric,
sorry, who is your Beta0?

davies.test() tests the hypothesis H0: leftSlope=rightSlope which 
implies diffSlope=0 and then the breakpoint does not exist. K in 
davies.test() means the number of evaluation points used to compute the 
approximate p-value..


Please contact me off list if you need more details (given detailed 
questions)


vito



Il 16/11/2012 20.57, Greg Cohn ha scritto:

My data:
I have raw data points that form a logit style curve as if they were a time
series. Which is to say they form 3 distinct lines with 3 distinct slopes
in backwards z pattern.  A certain class of my data looks essentially flat
to the eye with marginal oscillation. What is important to me is the x
value at which the state change is occurring, in other words, the break
point

Use of segmented():
Segmented does a very good job of capturing the breakpoints and fitting
three distinct slopes, i.e. linear models. It does not, however give me
Pr(|t|) for the break point coefficients. I need to answer the question
H:0 Beta0=Beta with a certainty metric, i.e. probability statistic. This is
especially important for my, flat looking data class.

davies.test() question:
davies.test() only excepts lm() or glm() objects as input. If I run
segmented to find 1 breakpoint instead of 2, I get a totally bogus answer.
Without knowing the breakpoints, how is this test able to assess the proper
breakpiont? It appears to only give 1 best breakpoint, which is not
consistent with the breakpoints found by segmented(). Also, is K the number
of breakpoints or the number of iterations that it evaluates the breakpoint?


Thanks in advance.



lmfit-glm(TotRad_KW~HRRPUA_kWm2,data=d1)
davies.test(lmfit,seg.Z=~HRRPUA_kWm2,k=1000,alternative=less,
beta0=0,dispersion=NULL)

Davies' test for a change in the slope

data:  Model =  gaussian , link = identity
formula = TotRad_KW ~ HRRPUA_kWm2
segmented variable = HRRPUA_kWm2
`Best' at = 561.205, n.points = 1000, p-value  2.2e-16
alternative hypothesis: less




segments - segmented(lmfit, seg.Z=~HRRPUA_kWm2,psi=c(475,550))
summary(segments)

***Regression Model with Segmented Relationship(s)***

Call:
segmented.glm(obj = lmfit, seg.Z = ~HRRPUA_kWm2, psi = c(475,
 550))

Estimated Break-Point(s):
 Est. St.Err
psi1.HRRP 430.2  4.087
psi2.HRRP 484.6  3.077

t value for the gap-variable(s) V:  0 0

Meaningful coefficients of the linear terms:
 Estimate Std. Error t value Pr(|t|)
(Intercept) -38.6993   274.7666  -0.141   0.8891
HRRPUA_kWm2   1.4297 0.7472   1.914   0.0668 .
U1.HRRP  42.2884 4.7696   8.866   NA
U2.HRRP -40.5897 4.7123  -8.614   NA
---
Signif. codes:  0 *** 0.001 ** 0.01 * 0.05 . 0.1   1
(Dispersion parameter for gaussian family taken to be 6934.706)

Null deviance: 70776718  on 31  degrees of freedom
Residual deviance:   180302  on 26  degrees of freedom
AIC: 377.19

Convergence attained in 2 iterations with relative change -1.662839e-14




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Università di Palermo
viale delle Scienze, edificio 13
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Re: [R] Download a file from url

2012-11-19 Thread Suzen, Mehmet

you can use RCurl for web connections from Duncan Lang, see its paper:
http://www.omegahat.org/RCurl/RCurlJSS.pdf

getURL there will solve your problem.

On Sun, Nov 18, 2012 at 2:07 AM, veepsirtt veepsi...@gmail.com wrote:
 Hi R,
 I installed wget and tried to download the file from this
 http://nseindia.com/content/equities/scripvol/datafiles/16-11-2012-TO-16-11-2012ACCEQN.csv
 but it fails.
 How to get it using wget?
 thanks
 veepsirtt


 #Define Working Directory, where files would be saved
 setwd('G:/NIFTY')

 #Define start and end dates, and convert them into date format
 startDate = as.Date(2011-01-05, order=ymd)
 endDate = as.Date(2011-02-01, order=ymd)

 f - tempfile()

 downloadfilename=paste(ACC, EQN, sep = )
 temp =

 #Generate URL
 http://nseindia.com/content/equities/scripvol/datafiles/16-11-2012-TO-16-11-2012ACCEQN.csv

 myURL = 
 paste(http://nseindia.com/content/equities/scripvol/datafiles/;,
 as.character(startDate, %d-%m-%Y), -TO-, as.character(endDate,
 %d-%m-%Y), downloadfilename, .csv, sep = )


   download.file(myURL, f, method='wget', extra=-U 'Mozilla/5.0 (X11; Linux)
 Gecko Firefox/5.0')

 temp - read.csv(f, sep = ,)
 head(temp)









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 http://r.789695.n4.nabble.com/Download-a-file-from-url-tp4642985p4649907.html
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[R] Introductory R course: 6-8.12.2012

2012-11-19 Thread Dr. Pablo E. Verde

Dear list members,

 

Apologies for cross-posting, there are some places available in an introductory 
R course.

Please, find below the information of this R training course.

 

If you have any question, don't hesitate to contact me.

 

Best regards,

 

Pablo

 

++
Course: Statistical Analysis with R
Where: Linux Hotel, Essen-Horst, Germany
When:  6-7-8 December 2012

 

Instructor: Dr. Pablo E. Verde
++
*Target audience*


This course is for data analysts who are familiar with classical statistics 
statistical 

software (e.g. SAS, SPSS) and they want to get a working knowledge in R. This is

a 3 days intensive training course with 8 hours per day including lecturing and 
exercises.
The course presentation is practical with many worked examples.
Lectures are given in English. Discussions can be in English, German or Spanish.
++

 

Day 1: Introduction to the R system


* Introduction to R 

* Data management with R 
* Graphical methods with R
* Classical statistical procedures with R

* Introduction to the R language

 

Day 2: Statistical modeling with R


* Linear regression and ANOVA

* Generalized linear models: logistic regression, loglinear model, etc.

* Issues in statistical modeling

* Computer simulation and model checking

 

 Day 3: Advance statistical modeling with R


* Modeling time to event data and survival analysis

* Introduction to mixed linear modeling

* Applications of generalized linear mixed effects models

* Own projects

 ++
Costs:
Public sector and commercial: 1088,85 Euro (three days course, incl full board
and VAT) additionally accommodation (on demand) 63,13 Euro shared double room
per night (incl. VAT) or 138,03  single room per night (inc. VAT).

 

Student: 
870 Euro (three days course - incl full board, shared double room 
[single room on demand] and VAT). Some of the courses are frequently fully 
booked. So please notice that you may have to try several times, until you
get a spare place.

 

For more information, please contact:  i...@linuxhotel.de
++

[[alternative HTML version deleted]]

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Re: [R] How to retrieve data from a matrix

2012-11-19 Thread ginger

Thank you very much, It works!



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Re: [R] loop to subtract arrays / error

2012-11-19 Thread Irucka Embry

Hi Rui Barradas, how are you?

Thank-you very much. That worked perfectly.

Irucka Embry


-Original Message- 
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/19/2012 4:05:11 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] loop to subtract arrays / error

Hello,

Try the following.

Xjj - matrix(nrow = 101, ncol = 1000)
for (i in 1:dim(Vsimr)[2]) {
 Xjj[, i] - Vobsr - Vsimr[, i]
}

Hope this helps,

Rui Barradas
Em 19-11-2012 01:41, iembry escreveu:
 Hi everyone, I am having trouble with creating a loop to subtract
arrays.

 In R, this is what I have done:
 Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) #
see data
 below
 Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,) #
see
 data below
 Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus
Vobsr
 is 101x1 double matrix (column 1 is date information)
 Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus
Vsimr
 is 101x1000 double matrix (column 1 is date information)
 Vobsr - Vsimr
 Error in Vobsr - Vsimr : non-conformable arrays

 Thus I attempted to create the loop below to perform the subtraction
 operation for each of the 1000 columns.

 dim(Vsimr)[2]
 [1] 1000

 for (i in 1:dim(Vsimr)[2]) {
 Xjj - Vobsr - Vsimr[,i]
 }
 Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

 How can I subtract each column of Vsimr from the single column of
Vobsr over
 the 1000 columns present?

 I would like to thank each of you in advance for your assistance.

 How can I subtract each column of Vsimr from the single column of
Vobsr over
 the 1000 columns present?

 I would like to thank each of you in advance for your assistance.


 I am including some of the data from the files that I am operating on
below:
 1 column of Observed_Flow.csv
 81.071
 73.187
 66.991
 62.482
 59.662
 58.529
 59.085
 61.328
 65.259
 70.878
 78.184
 87.179
 97.862
 110.23
 124.29
 140.08
 157.57
 176.76
 197.63
 220.18
 244.4
 270.31
 297.88
 327.14
 358.09
 390.71
 425.03
 461.03
 498.72
 538.09
 579.16
 621.91
 666.35
 712.48
 760.29
 809.8
 860.99
 913.87
 968.44
 1024.7
 1082.6
 1142.3
 1203.6
 1266.6
 1331.3
 1397.7
 1465.7
 1535.5
 1606.9
 1680.1
 1754.9
 1831.4
 1907.1
 1981.9
 2055.9
 2129
 2201.3
 2272.7
 2343.3
 2413.1
 2482
 2550.1
 2617.3
 2683.7
 2749.2
 2813.9
 2877.8
 2940.8
 3003
 3064.3
 3124.8
 3184.4
 3243.2
 3301.1
 3358.2
 3414.5
 3469.9
 3524.4
 3578.2
 3631
 3683.1
 3734.3
 3784.6
 3834.1
 3882.8
 3930.6
 3977.6
 4023.7
 4069
 4113.4
 4157
 4199.8
 4241.7
 4282.7
 4323
 4362.3
 4400.9
 4438.6
 4475.4
 4511.4
 4546.6


 2 columns of 1000 columns of 1000Samples_Vsim.csv
 81.07 81.07
 73.19 73.19
 65.81 67.16
 58.93 63
 52.55 60.7
 46.68 60.25
 41.31 61.67
 36.44 64.95
 32.08 70.08
 28.22 77.08
 24.86 85.94
 22.01 96.65
 19.65 109.23
 17.8 123.67
 16.46 139.96
 15.61 158.12
 15.27 178.14
 15.43 200.02
 16.1 223.75
 17.27 249.35
 18.94 276.81
 21.11 306.13
 23.79 337.31
 26.97 370.34
 30.65 405.24
 34.84 442
 39.52 480.62
 44.71 521.1
 50.41 563.44
 56.61 607.64
 63.31 653.7
 70.51 701.62
 78.21 751.4
 86.42 803.04
 95.13 856.53
 104.35 911.89
 114.06 969.11
 124.28 1028.2
 135.01 1089.1
 146.23 1151.9
 157.96 1216.6
 170.19 1283.1
 182.93 1351.5
 196.16 1421.7
 209.9 1493.8
 224.15 1567.8
 238.89 1643.6
 254.14 1721.3
 269.89 1800.8
 286.15 1882.2
 302.91 1965.5
 320.17 2050.6
 337.18 2134.8
 353.93 2218.1
 370.44 2300.4
 386.69 2381.8
 402.7 2462.3
 418.45 2541.8
 433.95 2620.4
 449.2 2698.1
 464.2 2774.9
 478.94 2850.7
 493.44 2925.6
 507.68 2999.5
 521.67 3072.6
 535.41 3144.7
 548.9 3215.8
 562.14 3286.1
 575.12 3355.4
 587.86 3423.8
 600.34 3491.2
 612.57 3557.7
 624.55 3623.3
 636.28 3688
 647.76 3751.7
 658.98 3814.5
 669.96 3876.4
 680.68 3937.3
 691.15 3997.3
 701.37 4056.4
 711.34 4114.6
 721.06 4171.8
 730.52 4228.1
 739.74 4283.4
 748.7 4337.9
 757.41 4391.4
 765.87 4443.9
 774.08 4495.6
 782.04 4546.3
 789.74 4596.1
 797.2 4644.9
 804.4 4692.8
 811.35 4739.8
 818.05 4785.9
 824.5 4831
 830.7 4875.2
 836.64 4918.5
 842.33 4960.8
 847.78 5002.2
 852.97 5042.7
 857.91 5082.3



 --
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tml
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Re: [R] loop to subtract arrays / error

2012-11-19 Thread Irucka Embry

Hi Rui, thank-you. 

That was simple and worked great.

Irucka 

-Original Message- 
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/19/2012 4:13:24 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] loop to subtract arrays / error

Hello,

Or simpler, since Vobsr only has one column:

Xjj - as.vector(Vobsr) - Vsimr

Hope this helps,

Rui Barradas
Em 19-11-2012 10:05, Rui Barradas escreveu:
 Hello,

 Try the following.

 Xjj - matrix(nrow = 101, ncol = 1000)
 for (i in 1:dim(Vsimr)[2]) {
 Xjj[, i] - Vobsr - Vsimr[, i]
 }

 Hope this helps,

 Rui Barradas
 Em 19-11-2012 01:41, iembry escreveu:
 Hi everyone, I am having trouble with creating a loop to subtract 
 arrays.

 In R, this is what I have done:
 Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) # 
 see data
 below
 Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep
=,) # 
 see
 data below
 Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus
 Vobsr
 is 101x1 double matrix (column 1 is date information)
 Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus
 Vsimr
 is 101x1000 double matrix (column 1 is date information)
 Vobsr - Vsimr
 Error in Vobsr - Vsimr : non-conformable arrays

 Thus I attempted to create the loop below to perform the subtraction
 operation for each of the 1000 columns.

 dim(Vsimr)[2]
 [1] 1000

 for (i in 1:dim(Vsimr)[2]) {
 Xjj - Vobsr - Vsimr[,i]
 }
 Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

 How can I subtract each column of Vsimr from the single column of 
 Vobsr over
 the 1000 columns present?

 I would like to thank each of you in advance for your assistance.

 How can I subtract each column of Vsimr from the single column of 
 Vobsr over
 the 1000 columns present?

 I would like to thank each of you in advance for your assistance.


 I am including some of the data from the files that I am operating
on 
 below:
 1 column of Observed_Flow.csv
 81.071
 73.187
 66.991
 62.482
 59.662
 58.529
 59.085
 61.328
 65.259
 70.878
 78.184
 87.179
 97.862
 110.23
 124.29
 140.08
 157.57
 176.76
 197.63
 220.18
 244.4
 270.31
 297.88
 327.14
 358.09
 390.71
 425.03
 461.03
 498.72
 538.09
 579.16
 621.91
 666.35
 712.48
 760.29
 809.8
 860.99
 913.87
 968.44
 1024.7
 1082.6
 1142.3
 1203.6
 1266.6
 1331.3
 1397.7
 1465.7
 1535.5
 1606.9
 1680.1
 1754.9
 1831.4
 1907.1
 1981.9
 2055.9
 2129
 2201.3
 2272.7
 2343.3
 2413.1
 2482
 2550.1
 2617.3
 2683.7
 2749.2
 2813.9
 2877.8
 2940.8
 3003
 3064.3
 3124.8
 3184.4
 3243.2
 3301.1
 3358.2
 3414.5
 3469.9
 3524.4
 3578.2
 3631
 3683.1
 3734.3
 3784.6
 3834.1
 3882.8
 3930.6
 3977.6
 4023.7
 4069
 4113.4
 4157
 4199.8
 4241.7
 4282.7
 4323
 4362.3
 4400.9
 4438.6
 4475.4
 4511.4
 4546.6


 2 columns of 1000 columns of 1000Samples_Vsim.csv
 81.07 81.07
 73.19 73.19
 65.81 67.16
 58.93 63
 52.55 60.7
 46.68 60.25
 41.31 61.67
 36.44 64.95
 32.08 70.08
 28.22 77.08
 24.86 85.94
 22.01 96.65
 19.65 109.23
 17.8 123.67
 16.46 139.96
 15.61 158.12
 15.27 178.14
 15.43 200.02
 16.1 223.75
 17.27 249.35
 18.94 276.81
 21.11 306.13
 23.79 337.31
 26.97 370.34
 30.65 405.24
 34.84 442
 39.52 480.62
 44.71 521.1
 50.41 563.44
 56.61 607.64
 63.31 653.7
 70.51 701.62
 78.21 751.4
 86.42 803.04
 95.13 856.53
 104.35 911.89
 114.06 969.11
 124.28 1028.2
 135.01 1089.1
 146.23 1151.9
 157.96 1216.6
 170.19 1283.1
 182.93 1351.5
 196.16 1421.7
 209.9 1493.8
 224.15 1567.8
 238.89 1643.6
 254.14 1721.3
 269.89 1800.8
 286.15 1882.2
 302.91 1965.5
 320.17 2050.6
 337.18 2134.8
 353.93 2218.1
 370.44 2300.4
 386.69 2381.8
 402.7 2462.3
 418.45 2541.8
 433.95 2620.4
 449.2 2698.1
 464.2 2774.9
 478.94 2850.7
 493.44 2925.6
 507.68 2999.5
 521.67 3072.6
 535.41 3144.7
 548.9 3215.8
 562.14 3286.1
 575.12 3355.4
 587.86 3423.8
 600.34 3491.2
 612.57 3557.7
 624.55 3623.3
 636.28 3688
 647.76 3751.7
 658.98 3814.5
 669.96 3876.4
 680.68 3937.3
 691.15 3997.3
 701.37 4056.4
 711.34 4114.6
 721.06 4171.8
 730.52 4228.1
 739.74 4283.4
 748.7 4337.9
 757.41 4391.4
 765.87 4443.9
 774.08 4495.6
 782.04 4546.3
 789.74 4596.1
 797.2 4644.9
 804.4 4692.8
 811.35 4739.8
 818.05 4785.9
 824.5 4831
 830.7 4875.2
 836.64 4918.5
 842.33 4960.8
 847.78 5002.2
 852.97 5042.7
 857.91 5082.3



 -- 
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http://r.789695.n4.nabble.com/loop-to-subtract-arrays-error-tp4650001.ht
ml
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 PLEASE do read the posting guide 
 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.

 __
 R-help@r-project.org mailing list
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 http://www.R-project.org/posting-guide.html
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[R] internal cluster quality indexes

2012-11-19 Thread eliza botto


Dear useRs,
 
I wanted to know something about the Index.G2 and Index.G3 which Calculate G2 
and G2 internal cluster quality indexes. 
i tried to find material from internet but it seems that the file have been 
removed. Is it good to have higher values of these indexes or lower?
 
thanks in advance
eliza 
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] list.files, recursively

2012-11-19 Thread Jannis

Thanks, Rui and Jim, for your replies. I tried to post this question to 
r.devel but its admin told me that the question rather belongs into 
r-help. Thanks, Jim, for your suggestion. I have already constructed 
something similar. I posted my question to suggest modifying the 
function so that not everybody has to program his own workaroung.


So, lets see whether this convinces the people reponsible for these base 
functions.



Cheers
Jannis

On 18.11.2012 19:02, Rui Barradas wrote:

Hello,

I believe that's a question for r-devel, but good point. It's
docummented that in non-recursive calls to list.files subdirectory names
are always included. (With a typo, There always are instead of They
always are.)

Rui Barradas
Em 18-11-2012 17:20, Jannis escreveu:

Dear R developers,


as far as I understand the manual of list.files(), there is only a way
to exclude directories from the returned vector if you use list.files
recursively. In non recursive mode, there seems to be no way of
excluding directories (the include dirs argument does not seem to have
any effect). Would it not be more intuitive and practical to allow the
switching off of directory names in both cases?


Thanks a lot
Jannis

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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Re: [R] function customization

2012-11-19 Thread maxbre

thank you all for the great  help, in particular to dennis murphy

in order to close the thread I'm posting here the final solution to my
question

new.ex-structure(list(TEC = c(0.21, 0.077, 0.06, 0.033, 0.014, 0.007, 
   0.21, 0.077, 0.01, 0.033, 0.05, 0.014), 
   LR = c(FALSE, FALSE,TRUE, FALSE, TRUE, FALSE, FALSE,
FALSE, TRUE, FALSE, TRUE, TRUE),
   group = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 2L,2L, 2L), 
 .Label = c(1, 2), class =
factor)), 
 .Names = c(TEC,LR, group),
row.names = c(NA, -12L), class = data.frame) 


library(NADA)


out-with(new.ex,
cenfit(TEC, LR, group)
)



#method 1
out - with(new.ex, cenfit(TEC, LR, group))
res - as.data.frame(do.call(rbind, mean(out) ))
res$n - out@survfit$n   # see str(out) to discover why
res$sum - with(res, n * mean)
res


#method 2
library(plyr)
res2 - ldply(mean(out), rbind)
names(res2)[-1] - names(mean(out)[[1]])
res2 - mutate(res2, n = out@survfit$n, sum = mean * n)
res2







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Re: [R] loop to subtract arrays / error

2012-11-19 Thread Irucka Embry

Hi Arun, thanks for your assistance. That worked as well.

Irucka

-Original Message- 
From: arun [smartpink...@yahoo.com]
Sent: 11/19/2012 7:22:09 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] loop to subtract arrays / error

HI,
May be this helps:

Vobsr-read.table(text=
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
,sep=,header=FALSE)
Vsimr=read.table(text=
81.07 81.07
73..19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
,sep=,header=FALSE)
 Vsimr1-as.matrix(Vsimr)
 sapply(split(Vsimr1,col(Vsimr1)),function(x) x-as.matrix(Vobsr))
#1  2
 #[1,]  -0.001 -0.001
 #[2,]   0.003  0.003
 #[3,]  -1.181  0.169
 #[4,]  -3.552  0.518
 #[5,]  -7.112  1.038
 #[6,] -11.849  1.721
 #[7,] -17.775  2.585
 #[8,] -24.888  3.622
 #[9,] -33.179  4.821
#[10,] -42.658  6.202

#if you are using data.frame()
res2-sapply(Vsimr,function(x) x-Vobsr[,1])
 head(res2,3)
# V1 V2
#[1,] -0..001 -0.001
#[2,]  0.003  0.003
#[3,] -1.181  0.169
#or just
res3-Vsimr-Vobsr[,1]
 head(res3,3)
#  V1 V2
#1 -0.001 -0.001
#2  0.003  0.003
#3 -1.181  0.169

A.K.








- Original Message -
From: iembry iruc...@mail2world.com
To: r-help@r-project.org
Cc: 
Sent: Sunday, November 18, 2012 8:41 PM
Subject: [R] loop to subtract arrays / error

Hi everyone, I am having trouble with creating a loop to subtract
arrays.

In R, this is what I have done:
 Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) # see
data
 below
 Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,) #
see
 data below

 Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus
Vobsr
 is 101x1 double matrix (column 1 is date information)
 Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus
Vsimr
 is 101x1000 double matrix (column 1 is date information)

 Vobsr - Vsimr
Error in Vobsr - Vsimr : non-conformable arrays

Thus I attempted to create the loop below to perform the subtraction
operation for each of the 1000 columns.

 dim(Vsimr)[2]
[1] 1000

 for (i in 1:dim(Vsimr)[2]) {
Xjj - Vobsr - Vsimr[,i]
}
Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

How can I subtract each column of Vsimr from the single column of Vobsr
over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.

How can I subtract each column of Vsimr from the single column of Vobsr
over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.


I am including some of the data from the files that I am operating on
below:
1 column of Observed_Flow.csv
81..071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
78.184
87.179
97.862
110.23
124.29
140.08
157.57
176.76
197.63
220.18
244.4
270.31
297.88
327.14
358.09
390.71
425.03
461.03
498.72
538.09
579.16
621.91
666.35
712.48
760.29
809.8
860.99
913.87
968.44
1024.7
1082.6
1142.3
1203.6
1266.6
1331.3
1397.7
1465.7
1535.5
1606.9
1680.1
1754.9
1831.4
1907.1
1981.9
2055.9
2129
2201.3
2272.7
2343.3
2413.1
2482
2550.1
2617.3
2683.7
2749.2
2813.9
2877.8
2940.8
3003
3064.3
3124.8
3184.4
3243.2
3301.1
3358.2
3414.5
3469.9
3524.4
3578.2
3631
3683.1
3734.3
3784.6
3834.1
3882.8
3930.6
3977.6
4023.7
4069
4113.4
4157
4199.8
4241.7
4282.7
4323
4362.3
4400.9
4438.6
4475.4
4511.4
4546.6


2 columns of 1000 columns of 1000Samples_Vsim.csv
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
24.86 85.94
22.01 96.65
19.65 109.23
17.8 123.67
16.46 139.96
15.61 158.12
15.27 178.14
15.43 200.02
16.1 223.75
17.27 249.35
18.94 276.81
21.11 306.13
23.79 337.31
26.97 370.34
30.65 405.24
34.84 442
39.52 480.62
44.71 521.1
50.41 563.44
56.61 607.64
63.31 653.7
70.51 701.62
78.21 751.4
86.42 803.04
95.13 856.53
104.35 911.89
114.06 969.11
124.28 1028.2
135.01 1089.1
146.23 1151.9
157.96 1216.6
170.19 1283.1
182.93 1351.5
196.16 1421.7
209.9 1493.8
224.15 1567.8
238.89 1643.6
254.14 1721.3
269.89 1800.8
286.15 1882.2
302.91 1965.5
320.17 2050.6
337.18 2134.8
353.93 2218.1
370.44 2300.4
386.69 2381.8
402.7 2462.3
418.45 2541.8
433.95 2620.4
449.2 2698.1
464.2 2774.9
478.94 2850.7
493.44 2925.6
507.68 2999.5
521.67 3072.6
535.41 3144.7
548.9 3215.8
562.14 3286.1
575.12 3355.4
587.86 3423.8
600.34 3491.2
612.57 3557.7
624.55 3623.3
636.28 3688
647.76 3751.7
658.98 3814.5
669.96 3876.4
680.68 3937.3
691.15 3997.3
701.37 4056.4
711.34 4114.6
721.06 4171.8
730.52 4228.1
739.74 4283.4
748.7 4337.9
757.41 4391.4
765.87 4443.9
774.08 4495.6
782.04 4546.3
789.74 4596..1
797.2 4644.9
804.4 4692.8
811.35 4739.8
818.05 4785.9
824.5 4831
830.7 4875.2
836.64 4918.5
842.33 4960.8
847.78 5002.2
852.97 5042.7
857.91 5082.3



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Re: [R] Calculateing means

2012-11-19 Thread S Ellison

 
 I have a data matrix with 570 columns containing 95 (samples) 
 with 6 replicates each.
 How can I calculate the mean of the replicates for 95 samples?

Write a function that calculates the sample means for a vector of 95 
observations and then use apply() to apply that function to the whole matrix.

S

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Re: [R] simple linear regression with proportion data

2012-11-19 Thread S Ellison

 

 -Original Message-
  Can I use simple linear regression when I have proportion data for 
  both dependent and independent variables?  Or, should I use beta 
  regression analysis?  Or any suggestion?
  
 
   The distribution of the independent variable is irrelevant 
 (in some circumstances it matters whether it is measured 
 without error or not).

Agreed that if you just want a line that goes somewhere near the data you can 
do pretty much anything.
But don't those circumstances you referred to include 'any time you want an 
unbiased estimate of the slope or a reliable standard error on coefficients'?

S

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[R] Classification methods - which one?

2012-11-19 Thread Peter Kupfer

Dear all,
i searched for some classification methods and I have no glue if i took the 
right once.
My problem: I have a matrix with 17000 rows and 33 colums (genes and patients). 
The patients are grouped into 3 diseases.
No I want to classify the patients and for sure i want to know which rows are 
more helpful for the classification than others. 

I tried SVM and random forest. Do you think this are the right classification 
methods? Maybe there are some hints you can give me. I am more familiar with 
the Bioconductor packages. Furthermore: This is/was not my field of study in 
the past but I want to understand it and I am willing to deal with this field.
Would be amazing if one of the (more) mathematical people can give me a hint. 
Thanks and all the best

Peter


PS: I can upload my underlying data if somebody is interested

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Re: [R] side by side boxplots

2012-11-19 Thread lbp

I want to plot the results of multiple paired comparison in a side-by-side, in 
opposite direction with the standard error, graph. Would you please give me 
some help of howto. Thanks. Charlie.   




lbp
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Re: [R] additive interaction for a dichotomous dependent variable (i.e. risk difference)

2012-11-19 Thread wouterjohannes

Hi,

No, I never got any response. I used an article by Knol to solve the issue. Ik
hope this is useful for you too.

Best regards,

Wouter

Knol, M.J., van, d.T., Grobbee, D.E., Numans, M.E., Geerlings, M.I., 2007.
Estimating interaction on an additive scale between continuous determinants in
a logistic regression model. International Journal of Epidemiology 36,
-1118.

Wouter Peyrot
Arts in opleiding tot psychiater

GGZ inGeest, dienst onderzoek, onderzoek
Locatie Valeriuskliniek, Valeriusplein 9, Postbus 74077, 1070 BB Amsterdam
T (020) 788 5425
www.ggzingeest.nlhttp://www.ggzingeest.nl

GGZ inGeest partner van VUmc

Van: sarahw [via R] [mailto:ml-node+s789695n464968...@n4.nabble.com]
Verzonden: vrijdag 16 november 2012 2:39
Aan: Peyrot, Wouter
Onderwerp: Re: additive interaction for a dichotomous dependent variable (i.e.
risk difference)

Did you ever get a response to this or resolve this yourself?

Many thanks!

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Re: [R] Stepwise analysis with fixed variables

2012-11-19 Thread Einat

Thank you for the quick reply.

Two more questions:
1. For example, if this is my code:

RegModel =
lm(glucose~sex+BMI+height+weight+education+ses,weight=w_without_non_response)
summary(RegModel)
step(RegModel, direction =backward,scope=list(lower=?,upper=?))

and I want the sex and height variables to be fixed, but the rest of the
variables to go into the backward analysis, how should I write the scope
function?

2.How can I add an alpha level to the step function as a criterion for the
backward regression analysis?

Thank you very much,

Einat



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[R] RQuantlib - Convertible Bond Pricing

2012-11-19 Thread colaiutachambers

Hi everyone,
I’m working on my Master’s Degree thesis about the pricing of C.B. trying to
do that with “R”.
I read the paper “RQuantLib: Interfacing QuantLib from R” and now I’m
matching several market price (taken from Bloomberg or Deutsche Bank
database) with “R” output.
Could you help me to understand the parameters within these functions?
Let me show you one of the problem that I met in the attached files.
As you could see in the image, just shifting the Conv.Ratio from 1 to 10 all
the curves in the plot get flattened.

First http://imageshack.us/photo/my-images/19/cr10plot.png/  
Second http://imageshack.us/photo/my-images/641/cr1plot.png/  
Script http://imageshack.us/photo/my-images/801/cr1m.png/  


Best regards.

Gabriele Carrarini
BTG Pactual  
London Berkeley Square House





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Re: [R] R-Square in WLS

2012-11-19 Thread frespider

Hi Peter,

why you are involving -1 with this concept?  Can you explain more please

Cheers
Date: Sun, 18 Nov 2012 23:28:26 -0800
From: ml-node+s789695n4650012...@n4.nabble.com
To: frespi...@hotmail.com
Subject: Re: R-Square in WLS

On Nov 18, 2012, at 21:32 , Thomas Lumley wrote:

 On Fri, Nov 16, 2012 at 4:48 PM, frespider [hidden email] wrote:

 Hi,

 I am fitting a weighted least square regression and trying to compute

 SSE,SST and SSReg but I am not getting SST = SSReg + SSE and I dont know

 what I am coding wrong. Can you help please?

 For a start, you need to replace your mu and muZ by weighted means.

The -1 in the model formulas also suggests that there will be problems even in 
the non-weighted case. The addition formula for SSDs works for successive model 
reductions, so it is required that the span of the design matrix X contains the 
vector of all ones.

-thomas

 [snip]

 ## Y = Log(Z) Scale

 Yhat - X%*%bhat # predicted values

 mu - mean(Y)

 To - Y - mu

 Er - Y - Yhat

 Re - Yhat - mu

 lgSST - sum(Weights*(To)^2)# log SST

 lgSSE - sum(Weights*(Er)^2)# log SSE

 lgSSR - sum(Weights*(Re)^2)# log SSR

 lgR-sq - lgSSR/lgSST

 ###  Z Scale

 ##

 Z - exp(Y)

 muZ - mean(Z)

 Zhat - exp(Yhat+0.5*Sigma2)

 ToZ - Z-muZ

 ErZ - Z - Zhat

 ReZ - Zhat - muZ

 SST - sum(Weights*(ToZ)^2)  # SST

 SSE - sum(Weights*(ErZ)^2)  # SSE

 SSR - sum(Weights*(ReZ)^2)  # SSR

 Rsq - SSR/SST

 I don't understand what is wrong with the code.  The sum square regression

 plus the sum square error do not add up to the sum square total in both the

 Y scale and Z scale.  Y is a normal distribution and Z is log normally

 distributed.  Where is the error?

 Also, is there a way to calculate the weighted sum square?

   -thomas

 -- 

 Thomas Lumley

 Professor of Biostatistics

 University of Auckland

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Solbjerg Plads 3, 2000 Frederiksberg, Denmark

Phone: (+45)38153501

Email: [hidden email]  Priv: [hidden email]

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[R] lattice: defining grouping variable only for the upper/lower panel with splom

2012-11-19 Thread AnjaM

Using the mtcars dataset, how to define the grouping variable to be valid
only for the upper or lower panel?

The following doesn't work:

# Code start

splom(~data.frame(mpg, disp, hp, drat, wt, qsec),
  data=mtcars, pscales=0,
  auto.key=list(columns=3),
  upper.panel = function(...){
panel.grid(...)
panel.xyplot(groups=cyl,...)
  }
)

# Code end



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[R] Biologist R learner

2012-11-19 Thread andrew

I am a Biologist and a beginner, please help me to solve this: please
anyone..its my homework and I dont have a clue abt R yet..
   Write a function which does the following tasks:

(a) Calculates minimum and maximum value of a given argument x.
(b) If x is positive, some new vector gets the value of TRUE, and FALSE
otherwise.
(c) Creates a vector where the i:th and (i-1):th values of x are always
summed. First value of
the new vector has the same value as the first component of x. Use the
created function to
some vector x to show that the function works.




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Re: [R] loop to subtract arrays / error

HI,
May be this helps:

Vobsr-read.table(text=
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
,sep=,header=FALSE)
Vsimr=read.table(text=
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
,sep=,header=FALSE)
 Vsimr1-as.matrix(Vsimr)
 sapply(split(Vsimr1,col(Vsimr1)),function(x) x-as.matrix(Vobsr))
#    1  2
 #[1,]  -0.001 -0.001
 #[2,]   0.003  0.003
 #[3,]  -1.181  0.169
 #[4,]  -3.552  0.518
 #[5,]  -7.112  1.038
 #[6,] -11.849  1.721
 #[7,] -17.775  2.585
 #[8,] -24.888  3.622
 #[9,] -33.179  4.821
#[10,] -42.658  6.202

#if you are using data.frame()
res2-sapply(Vsimr,function(x) x-Vobsr[,1])
 head(res2,3)
# V1 V2
#[1,] -0.001 -0.001
#[2,]  0.003  0.003
#[3,] -1.181  0.169
#or just
res3-Vsimr-Vobsr[,1]
 head(res3,3)
#  V1 V2
#1 -0.001 -0.001
#2  0.003  0.003
#3 -1.181  0.169

A.K.








- Original Message -
From: iembry iruc...@mail2world.com
To: r-help@r-project.org
Cc: 
Sent: Sunday, November 18, 2012 8:41 PM
Subject: [R] loop to subtract arrays / error

Hi everyone, I am having trouble with creating a loop to subtract arrays.

In R, this is what I have done:
 Vobsr - read.csv(Observed_Flow.csv, header = TRUE, sep =,) # see data
 below
 Vsimr - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,) # see
 data below

 Vobsr - as.matrix(Vobsr[,-1]) # remove column 1 from analysis thus Vobsr
 is 101x1 double matrix (column 1 is date information)
 Vsimr - as.matrix(Vsimr[,-1]) # remove column 1 from analysis thus Vsimr
 is 101x1000 double matrix (column 1 is date information)

 Vobsr - Vsimr
Error in Vobsr - Vsimr : non-conformable arrays

Thus I attempted to create the loop below to perform the subtraction
operation for each of the 1000 columns.

 dim(Vsimr)[2]
[1] 1000

 for (i in 1:dim(Vsimr)[2]) {
    Xjj - Vobsr - Vsimr[,i]
}
Xjj is a 101x1 double matrix rather than a 101X1000 double matrix

How can I subtract each column of Vsimr from the single column of Vobsr over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.

How can I subtract each column of Vsimr from the single column of Vobsr over
the 1000 columns present?

I would like to thank each of you in advance for your assistance.


I am including some of the data from the files that I am operating on below:
1 column of Observed_Flow.csv
81.071
73.187
66.991
62.482
59.662
58.529
59.085
61.328
65.259
70.878
78.184
87.179
97.862
110.23
124.29
140.08
157.57
176.76
197.63
220.18
244.4
270.31
297.88
327.14
358.09
390.71
425.03
461.03
498.72
538.09
579.16
621.91
666.35
712.48
760.29
809.8
860.99
913.87
968.44
1024.7
1082.6
1142.3
1203.6
1266.6
1331.3
1397.7
1465.7
1535.5
1606.9
1680.1
1754.9
1831.4
1907.1
1981.9
2055.9
2129
2201.3
2272.7
2343.3
2413.1
2482
2550.1
2617.3
2683.7
2749.2
2813.9
2877.8
2940.8
3003
3064.3
3124.8
3184.4
3243.2
3301.1
3358.2
3414.5
3469.9
3524.4
3578.2
3631
3683.1
3734.3
3784.6
3834.1
3882.8
3930.6
3977.6
4023.7
4069
4113.4
4157
4199.8
4241.7
4282.7
4323
4362.3
4400.9
4438.6
4475.4
4511.4
4546.6


2 columns of 1000 columns of 1000Samples_Vsim.csv
81.07 81.07
73.19 73.19
65.81 67.16
58.93 63
52.55 60.7
46.68 60.25
41.31 61.67
36.44 64.95
32.08 70.08
28.22 77.08
24.86 85.94
22.01 96.65
19.65 109.23
17.8 123.67
16.46 139.96
15.61 158.12
15.27 178.14
15.43 200.02
16.1 223.75
17.27 249.35
18.94 276.81
21.11 306.13
23.79 337.31
26.97 370.34
30.65 405.24
34.84 442
39.52 480.62
44.71 521.1
50.41 563.44
56.61 607.64
63.31 653.7
70.51 701.62
78.21 751.4
86.42 803.04
95.13 856.53
104.35 911.89
114.06 969.11
124.28 1028.2
135.01 1089.1
146.23 1151.9
157.96 1216.6
170.19 1283.1
182.93 1351.5
196.16 1421.7
209.9 1493.8
224.15 1567.8
238.89 1643.6
254.14 1721.3
269.89 1800.8
286.15 1882.2
302.91 1965.5
320.17 2050.6
337.18 2134.8
353.93 2218.1
370.44 2300.4
386.69 2381.8
402.7 2462.3
418.45 2541.8
433.95 2620.4
449.2 2698.1
464.2 2774.9
478.94 2850.7
493.44 2925.6
507.68 2999.5
521.67 3072.6
535.41 3144.7
548.9 3215.8
562.14 3286.1
575.12 3355.4
587.86 3423.8
600.34 3491.2
612.57 3557.7
624.55 3623.3
636.28 3688
647.76 3751.7
658.98 3814.5
669.96 3876.4
680.68 3937.3
691.15 3997.3
701.37 4056.4
711.34 4114.6
721.06 4171.8
730.52 4228.1
739.74 4283.4
748.7 4337.9
757.41 4391.4
765.87 4443.9
774.08 4495.6
782.04 4546.3
789.74 4596.1
797.2 4644.9
804.4 4692.8
811.35 4739.8
818.05 4785.9
824.5 4831
830.7 4875.2
836.64 4918.5
842.33 4960.8
847.78 5002.2
852.97 5042.7
857.91 5082.3



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[R] Biologist R learner problem!!!help pls

2012-11-19 Thread Anna23

. QUse the built in dataset called iris in this task.
(a) Calculate the result of following formula separately in every species
for all of the
numerical variables: log(x)/x.
(b) Calculate trimmed mean for each of the numerical variables using
apply–function. Choose
your own trimming percentage




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[R] help on matrix column removal based on another matrix results

2012-11-19 Thread iembry

Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)

This is what I what I am running in R:
 source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}

 Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
 Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
 Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
 Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis

 NSEr - NSEr(Vobs,Vsim);
 write.table(NSEr, NSEr.csv, sep =,)

 NSErr - t(matrix(NSEr))

 ## select the behavioural simulations and discard the rest
 Vsim - Vsim[NSErr  0.6]
 write.table(Vsim, Vsim.csv, sep =,)

**Vsim becomes numeric[42016] rather than a double matrix of 101x416. 

What is the proper way to remove the columns in Vsim where the NSEr for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416. 

Thank-you again.



Below is the rest of the code in R:
 ## normalise Qsim and compute the quantiles
 NSEr - NSEr[NSEr  0.6]
 write.table(NSEr, NSEr_great_0.6.csv, sep =,)
 NSEr - NSEr - 0.6
 write.table(NSEr, NSEr_minus0.6.csv, sep =,)
 NSEr - NSEr/sum(NSEr)
 write.table(NSEr, NSEr_normalized.csv, sep =,)
 #NSEr = sum(NSEr)

 limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
 c(0.05,0.95), normwt=F)



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[R] R SNA: Creating a adjacency matrix containing all actors but only values of a subset

2012-11-19 Thread Daniel Hain

1 down vote favorite

my problem is the following:

I am using the R SNA package for social network analysis. Lets say, my
starting point is an edgelist with the following characteristics. Every row
contains a firm name, the ID of a project they are involved and further
characteristics, let's say the projects year. Firms can be in several
projects, and one project can consist of a cooperation of more than one
firm. Example:

Name Project Year
AA 1 2003
AB 1 2003
AB 2 2003
AB 3 2004
AC 2 2003
AC 4 2005

For the network analysis i need a adjacency matrix with all firms a srow and
column header, which i construct as follows:

grants.edgelist - read.csv(00-composed.csv, header = TRUE, sep = ;,
quote=\, dec=,, fill = TRUE, comment.char=)
grants.2mode - table(grants.edgelist) # cross tabulate - 2-mode
sociomatrix
grants.adj - grants.2mode%*%t(grants.2mode) # Adjacency matrix as
product of the 2-mode sociomatrix`
---
Now my problem: I want to run a netlm Regression on the adjacency matrix,
where i test how the network in one given year explains the network in the
next year. However, therefore i wanted to subset the grants.edgelist in a
set for (lets say) 2003 and 2005 only. However, i figured out that not all
firms are in projects every year, and therefore the corresponding adjacency
matrix has different rows and columns.

Now my question: How could i obtain a adjacency matrix containing all firms
in row and column header, but their intersection set on zero expect of the
year i want to observe. I hope it is clear what i mean.

Thank you very much in advance. This problem is driving m crazy today!

Best wishes

Daniel

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Re: [R] Stepwise regression scope: all interacting terms (.^2)

2012-11-19 Thread Mark Ebbert

David, thanks for the feedback!

Steve, thanks for the direction! I have heard and read some about Dr. Harrell's
work but somehow had missed the term penalized logistic regression. That was
helpful for finding more specific sources to follow Dr. Harrell's (and other's)
suggestions. I may have more questions in the near future.

On Nov 16, 2012, at 3:32 PM, Steve Lianoglou wrote:

Hi Mark,

To put some context to David's response below, you can search the list
archives for times when people ask about stepwise regression. You can
get started here:

http://search.gmane.org/search.php?group=gmane.comp.lang.r.generalquery=stepwise+penalized

The long and short of it is that you are almost always encouraged to
use some regularization/penalized model instead of this stepwise
approach. Frank Harrell, in particular, is generally quite vocal
against stepwise regression -- I'm actually surprised he hasn't chimed
in by now, but maybe he's getting a bit tired of fighting the good
fight -- or, it's close to the holiday and he's taking a break ;-)

Anyway ... HTH,

-steve

On Fri, Nov 16, 2012 at 4:13 PM, David Winsemius dwinsem...@comcast.net
wrote:

On Nov 16, 2012, at 12:16 PM, Mark Ebbert wrote:

I haven't heard anything on this question. Is there something fundamentally
wrong with my question? Any feedback is appreciated.

Perhaps failure to read this sig at the bottom of every posted message to
rhelp?

PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

Mark
On Nov 15, 2012, at 8:13 AM, Mark T. W. Ebbert wrote:

Dear Gurus,

Thank you in advance for your assistance. I'm trying to understand scope
better when performing stepwise regression using step.

From the help page of step:
If scope is a single formula, it specifies the upper component, and the
lower model is empty.

I have a model with a binary response variable and 10 predictor variables.
When I perform stepwise regression I define scope=.^2 to allow
interactions between all terms.

I generally avoid answering questions about stepwise regression, because
most of them do not include sufficient background material to justify that
strategy. Yours certainly did not.

But I am missing something. When I perform stepwise regression (both
directions) on the main model (y~x1+x2+…+x10) the method returns quickly
with an answer; however, when I define all interactions in the main model
(y~x1+x2+…+x10+x1:x2+x1:x3+…) and then perform stepwise regression
(backward only) it runs so long I have to kill it.

So here's my question: what is the difference between scope=.^2 on the
additive (proper term?) model and defining all interactions and doing
backward regression? My understanding is that .^2 is supposed to allow all
interactions!

Well, I would have guessed all two-way interactions (all 45 of them in your
case) would be included and then successively reduce until you got to your
specified (arbitrary and most likely incorrectly set) endpoint.) I think the
help page Details section is unclear on this point. I do not think that the
120 potential three-way interactions are part of the scope in that instance,
but it should be easy enough for you to test that possibility.

--
David Winsemius, MD
Alameda, CA, USA

--
Steve Lianoglou
Graduate Student: Computational Systems Biology
| Memorial Sloan-Kettering Cancer Center
| Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

[R] Question about Package 'sampleSection' for IRT model

2012-11-19 Thread Kamontip Srihaset



Dear All,

I am Ph.D student at Chulalongkorn University in Thailand, I want to use 
Package 'sampleSection' to estimate missing data which generate under IRT 
model(3-PL);

n-500 ## number of examinee
I-20 ## number of items
num.imp-5 ##number of imputations
p.missing-c(0.09, 0.01) #prob of missing
theta-sort(rnorm(n,0,1)) #ability
a-rnorm(I,0.5,0.1) #discrimination
b-rnorm(I,0,1) #difficulty
c-runif(I,0,0.25) #guess
Only item 1 have missing data. If the response to items 1 was a 1 (correct), 
the probability of missingwas 1%. If the
response was a 0 (incorrect), the probability of missing was 9%. Thus, the 
probability of missing was linked to the response of items itself (an unknown 
characteristic in real missing data situations).

I don't know how to apply function 'Heckman-style selection models' for this 
case, becase all my variables are unobserved.
Could you please tell me how to estimate data under my situation.
I'am looking forward your advice.

Sincerely yours,
Kamontip Srihaset



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[R] Ben Bolker's '‘emdbook’ Package , rbetabinom

2012-11-19 Thread arun4

Hello, 
I am using rbetabinom ( to generate beta binomial random variables) function
available in the emdbookpackage written by Professor. Ben Bolker for my
research study.
I have no questions with this function. However, I am looking for the
theoretical method/algorithm  of the function rbetabinom  . 
Morris (1997), American Naturalist 150:299-327 is given as the reference in
the package, But I couldn't fund any theoretical methods to generate beta
binomial random variables in this article.

I would like to kind request you to suggest me any journal paper or document
to study the methods of generating  beta binomial random variables used to
write the function rbetabinom.

Thank you very much. 

Sincerely,
Arun.





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Re: [R] RMySQL install on windows

2012-11-19 Thread Jeffrey Horner

Sorry for not taking care of this... If anyone would like to take over
maintainership of RMySQL I'm sure the R community would greatly appreciate
it. I just don't have the time these days.

Jeff

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Re: [R] help on matrix column removal based on another matrix results


Hello,

Try

Vsim[] - Vsim[NSErr  0.6, ]


Hope this helps,

Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:

Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)

This is what I what I am running in R:

source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R)

NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}


Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr  0.6]
write.table(Vsim, Vsim.csv, sep =,)

**Vsim becomes numeric[42016] rather than a double matrix of 101x416.

What is the proper way to remove the columns in Vsim where the NSEr for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.

Thank-you again.



Below is the rest of the code in R:

## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr  0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)



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Re: [R] Biologist R learner

2012-11-19 Thread Berend Hasselman


On 19-11-2012, at 15:32, andrew wrote:

 I am a Biologist and a beginner, please help me to solve this: please
 anyone..its my homework and I dont have a clue abt R yet..
   Write a function which does the following tasks:
 
 (a) Calculates minimum and maximum value of a given argument x.
 (b) If x is positive, some new vector gets the value of TRUE, and FALSE
 otherwise.
 (c) Creates a vector where the i:th and (i-1):th values of x are always
 summed. First value of
 the new vector has the same value as the first component of x. Use the
 created function to
 some vector x to show that the function works.
 

Please read the posting guide:  http://www.r-project.org/posting-guide.html
R-help is not for homework.
Read the introductory manual An Introduction to R (links are on the page for 
the posting guide.)

Berend

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Re: [R] Biologist R learner problem!!!help pls

2012-11-19 Thread Berend Hasselman


On 19-11-2012, at 15:48, Anna23 wrote:

 . QUse the built in dataset called iris in this task.
 (a) Calculate the result of following formula separately in every species
 for all of the
 numerical variables: log(x)/x.
 (b) Calculate trimmed mean for each of the numerical variables using
 apply–function. Choose
 your own trimming percentage
 

Please read the posting guide:  http://www.r-project.org/posting-guide.html
R-help is not for homework.
Read the introductory manual An Introduction to R (links are on the page for 
the posting guide.)

Berend

__
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Re: [R] help on matrix column removal based on another matrix results


Sorry, the comma is in the wrong place, it should be

Vsim[] - Vsim[ , NSErr  0.6]

Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:

Hello,

Try

Vsim[] - Vsim[NSErr  0.6, ]


Hope this helps,

Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:

Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)

This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R) 


NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}


Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr  0.6]
write.table(Vsim, Vsim.csv, sep =,)

**Vsim becomes numeric[42016] rather than a double matrix of 101x416.

What is the proper way to remove the columns in Vsim where the NSEr 
for that

column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.

Thank-you again.



Below is the rest of the code in R:

## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr  0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)



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http://www.R-project.org/posting-guide.html

and provide commented, minimal, self-contained, reproducible code.


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http://www.R-project.org/posting-guide.html

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Re: [R] manipulating longitudinal data in r

2012-11-19 Thread Michael Dewey


At 17:13 18/11/2012, Jeff Newmiller wrote:
Michael, this comment doesn't seem appropriate to the question, 
since the sample data is a ragged array that requires the addition 
of NAs to fit into a wide format.


It does now but we do not know whether that was how it started life. 
My remarks were intended as a general comment, not necessarily a 
solution to the OP's specific problem.



---
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---
Sent from my phone. Please excuse my brevity.

Michael Dewey i...@aghmed.fsnet.co.uk wrote:

At 08:56 17/11/2012, Kemi Racheal wrote:
Dear list member,

I have the following data example
ke - data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1,1,2))

I want to add another variable b, such that the max of 'a' by id is
returned
i.e data ke becomes
ke -
data.frame(patid=c(1,1,1,2,3,3),a=c(1,2,2,1,1,2),b=c(2,2,2,1,2,2))

Any help will be appreciated.

Dear Kemi
It is often easier to do some sorts of manipulations on the wide
format of the data. I appreciate that you can always do it both ways.


Oluwakemi


 [[alternative HTML version deleted]]

Michael Dewey
i...@aghmed.fsnet.co.uk
http://www.aghmed.fsnet.co.uk/home.html

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Re: [R] survfit number of variables != number of variable names

2012-11-19 Thread Terry Therneau


I can't reproduce the problem.

Tell us what version of R and what version of the survival package.
Create a reproducable example.  I don't know if some variables are numeric and some are 
factors, how/where the surv object was defined, etc.


Terry Therneau



On 11/17/2012 05:00 AM, r-help-requ...@r-project.org wrote:

This works ok:


  cox = coxph(surv ~ bucket*(today + accor + both) + activity, data = data)
  fit = survfit(cox, newdata=data[1:100,])

but using strata leads to problems:


  cox.s = coxph(surv ~  bucket*(today + accor + both) + strata(activity),
  data = data)
  fit.s = survfit(cox.s, newdata=data[1:100,])

Error in model.frame.default(data = data[1:100, ], formula = ~bucket +  :
   number of variables != number of variable names

Note that the following give rise to the same error:


  fit.s = survfit(cox.s, newdata=data)

Error in model.frame.default(data = data, formula = ~bucket + today +  :
   number of variables != number of variable names

but if I use data implicitly, all is working fine:

  fit.s = survfit(cox.s)

Any idea on how I could solve this?

Best, and thank you,

ge


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[R] expand time period

2012-11-19 Thread dms

I'd like to expand the following data to perform a daily time series.
It should cover from '2012-07-01' to '2012-10-06' with the values I
actually have being the mean from one point measurement to another. Does
anyone has a clue to perform this task.


structure(list(Date.beg = structure(c(15635, 15617, 15615, 15610,
15609, 15605, 15604, 15601, 15593, 15593, 15586, 15581, 15580,
15577, 15572, 15565, 15552, 15540, 15530, 15516), class = Date),
Date.end = structure(c(15619, 15619, 15616, 15615, 15610,
15607, 15604, 15602, 15595, 15594, 15587, 15582, 15581, 15579,
15572, 15567, 15554, 15541, 15533, 15517), class = Date),
Pollster = structure(c(1L, 2L, 1L, 2L, 1L, 2L, 4L, 1L, 2L,
1L, 1L, 2L, 1L, 4L, 1L, 2L, 2L, 1L, 3L, 1L), .Label = c(Datafolha,
Ibope, Veritá, Vox Populi), class = factor), Serra.PSDB = c(24,
22, 23, 19, 22, 17, 17, 21, 19, 20, 21, 20, 22, 22, 27, 26,
26, 30, 31.4, 31), Russomanno.PRB = c(23, 22, 25, 27, 30,
34, 34, 35, 35, 32, 35, 31, 31, 31, 31, 26, 25, 26, 17.7,
24), Haddad.PT = c(20, 22, 19, 18, 18, 18, 17, 15, 15, 17,
16, 16, 14, 14, 8, 9, 6, 7, 9.5, 6), Chalita.PMDB = c(11,
11, 11, 10, 9, 7, 5, 8, 6, 8, 7, 5, 7, 5, 6, 5, 5, 6, 4.3,
6), Others = c(8, 7, 7, 4, 7, 6, 4, 5, 6, 7, 9, 5, 8, 6,
10, 12, 15, 15, 8.6, 16), Null = c(8, 8, 8, NA, 8, 10, 10,
NA, 13, 9, 8, 12, 10, NA, 10, 12, 14, 11, 13.1, 11), Undecided = c(6,
8, 6, NA, 6, 8, 13, NA, 6, 7, 4, 9, 7, 13, 6, 10, 9, 6, 13.3,
5), Round = structure(c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 2L), .Label = c(year before,
off campaign, first, second), class = factor), Stage =
structure(c(2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c(0, 1, 2), class = factor),
Serra.Haddad = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(,
1, 0), class = factor), N = c(3959, 1204, 2099, 1204,
1799, 1204, 2000, 1802, 1001, 1221, 1078, 1001, 1069, 1200,
1077, 805, 805, 1075, 1331, 1081), Err = c(2, 3, 2, 3, 2,
3, 2.2, 2, 3, 3, 3, 3, 3, 2.8, 3, 3, 3, 3, 2.7, 3)), .Names =
c(Date.beg,
Date.end, Pollster, Serra.PSDB, Russomanno.PRB, Haddad.PT,
Chalita.PMDB, Others, Null, Undecided, Round, Stage,
Serra.Haddad, N, Err), row.names = c(6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25), class = data.frame)

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[R] How to subset my data and at the same time keep the balance?

2012-11-19 Thread Eddie Smith

Hi guys,

I have 1000 rows of a dataset. In my analysis, I need 70% of the data,
run my analysis and then use the remaining 30% to test my model.

Could anybody kindly help me on this?

Cheers

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Re: [R] How to subset my data and at the same time keep the balance?


Hello,

See the following example.

x - matrix(rnorm(2000), ncol = 2)

idx - sample(nrow(x), 0.7*nrow(x))
x2 - x[idx, ]
nrow(x2)  # 700

x3 - x[-idx, ]
nrow(x3)  # 300

Hope this helps,

Rui Barradas
Em 19-11-2012 17:16, Eddie Smith escreveu:

Hi guys,

I have 1000 rows of a dataset. In my analysis, I need 70% of the data,
run my analysis and then use the remaining 30% to test my model.

Could anybody kindly help me on this?

Cheers

__
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and provide commented, minimal, self-contained, reproducible code.


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Re: [R] How to subset my data and at the same time keep the balance?

2012-11-19 Thread Sarah Goslee

I'm not sure what you mean by balance, but you can use sample() to
randomly order the values 1:1000, then use the first 700 as row
indices for the first set, and the last 300 as the test set.

Sarah

On Mon, Nov 19, 2012 at 12:16 PM, Eddie Smith eddie...@gmail.com wrote:
 Hi guys,

 I have 1000 rows of a dataset. In my analysis, I need 70% of the data,
 run my analysis and then use the remaining 30% to test my model.

 Could anybody kindly help me on this?

 Cheers
--
Sarah Goslee
http://www.functionaldiversity.org

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[R] how to get bootstrap estimates

2012-11-19 Thread Tania Patiño

Hello all, could you explain me, how to get bootstrap estimates if i have
the next data:

scor
   mec vec alg ana sta
1   77  82  67  67  81
2   63  78  80  70  81
3   75  73  71  66  81
4   55  72  63  70  68
5   63  63  65  70  63
6   53  61  72  64  73
7   51  67  65  65  68
8   59  70  68  62  56
9   62  60  58  62  70
10  64  72  60  62  45
11  52  64  60  63  54
12  55  67  59  62  44
13  50  50  64  55  63
14  65  63  58  56  37
15  31  55  60  57  73
16  60  64  56  54  40
17  44  69  53  53  53
18  42  69  61  55  45
19  62  46  61  57  45
20  31  49  62  63  62
21  44  61  52  62  46
22  49  41  61  49  64
23  12  58  61  63  67
24  49  53  49  62  47
25  54  49  56  47  53
26  54  53  46  59  44
27  44  56  55  61  36
28  18  44  50  57  81
29  46  52  65  50  35
30  32  45  49  57  64
31  30  69  50  52  45
32  46  49  53  59  37
33  40  27  54  61  61
34  31  42  48  54  68
35  36  59  51  45  51
36  56  40  56  54  35
37  46  56  57  49  32
38  45  42  55  56  40
39  42  60  54  49  33
40  40  63  53  54  25
41  23  55  59  53  44
42  48  48  49  51  37
43  41  63  49  46  34
44  46  52  53  41  40
45  46  61  46  38  41
46  40  57  51  52  31
47  49  49  45  48  39
48  22  58  53  56  41
49  35  60  47  54  33
50  48  56  49  42  32
51  31  57  50  54  34
52  17  53  57  43  51
53  49  57  47  39  26
54  59  50  47  15  46
55  37  56  49  28  45
56  40  43  48  21  61
57  35  35  41  51  50
58  38  44  54  47  24
59  43  43  38  34  49
60  39  46  46  32  43
61  62  44  36  22  42
62  48  38  41  44  33
63  34  42  50  47  29
64  18  51  40  56  30
65  35  36  46  48  29
66  59  53  37  22  19
67  41  41  43  30  33
68  31  52  37  27  40
69  17  51  52  35  31
70  34  30  50  47  36
71  46  40  47  29  17
72  10  46  36  47  39
73  46  37  45  15  30
74  30  34  43  46  18
75  13  51  50  25  31
76  49  50  38  23   9
77  18  32  31  45  40
78   8  42  48  26  40
79  23  38  36  48  15
80  30  24  43  33  25
81   3   9  51  47  40
82   7  51  43  17  22
83  15  40  43  23  18
84  15  38  39  28  17
85   5  30  44  36  18
86  12  30  32  35  21
87   5  26  15  20  20
88   0  40  21   9  14
and I have the next errors:

ro 12 = ro (mec,vec)
ro 34 = ro (alg,ana)
ro 35 = ro (alg,sta)
ro 45 = ro (ana,sta)
ro 14 = ro (mec,ana)

Thank you,

Tania

[[alternative HTML version deleted]]

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Re: [R] How to subset my data and at the same time keep the balance?

HI,
May be this helps:
dat1-read.table(text=
  V1 V2
1 5 10
2 6  3
3 8  4
4 9 20
5 15 30
6 25 40
7 2  4
8 3  1
9 1  5
10 8 10
,header=TRUE)
dat2-dat1[sample(NROW(dat1),NROW(dat1)*(1-0.3)),] #70% of data
dat2$newcol-TRUE
 dat1$newcol1-TRUE
 dat4-merge(dat1,dat2,by=c(V1,V2),all=TRUE)
 dat5-dat4[is.na(dat4$newcol),][,1:2]  #remaining 30%
 dat5
#  V1 V2
#2  2  4
#4  5 10
#8  9 20
A.K.



- Original Message -
From: Eddie Smith eddie...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Monday, November 19, 2012 12:16 PM
Subject: [R] How to subset my data and at the same time keep the balance?

Hi guys,

I have 1000 rows of a dataset. In my analysis, I need 70% of the data,
run my analysis and then use the remaining 30% to test my model.

Could anybody kindly help me on this?

Cheers

__
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.


__
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[R] generated list element names

2012-11-19 Thread Sam Steingold

How can I create lists with element names created on the fly?

--8---cut here---start-8---
 list (foo = 10)
$foo
[1] 10

 list (foo = 10)
$foo
[1] 10

 list (paste(f,oo,sep=) = 10)
Error: unexpected '=' in list (paste(f,oo,sep=) =
--8---cut here---end---8---

I understand that tags in list() are not evaluated, but is there a more
elegant way than

--8---cut here---start-8---
 z - list(10)
 names(z) - paste(f,oo,sep=)
 z
$foo
[1] 10
--8---cut here---end---8---

thanks!

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://www.memritv.org
http://thereligionofpeace.com http://truepeace.org
Unix roulette: `dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM`

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Re: [R] generated list element names

2012-11-19 Thread jim holtman

How about this (if you don't like writing two lines, encapsulate it in
a function):

 x - list(10)
 names(x) - paste('f', 'oo', sep = '')
 str(x)
List of 1
 $ foo: num 10



On Mon, Nov 19, 2012 at 1:07 PM, Sam Steingold s...@gnu.org wrote:
 How can I create lists with element names created on the fly?

 --8---cut here---start-8---
 list (foo = 10)
 $foo
 [1] 10

 list (foo = 10)
 $foo
 [1] 10

 list (paste(f,oo,sep=) = 10)
 Error: unexpected '=' in list (paste(f,oo,sep=) =
 --8---cut here---end---8---

 I understand that tags in list() are not evaluated, but is there a more
 elegant way than

 --8---cut here---start-8---
 z - list(10)
 names(z) - paste(f,oo,sep=)
 z
 $foo
 [1] 10
 --8---cut here---end---8---

 thanks!

 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 
 11.0.11103000
 http://www.childpsy.net/ http://www.memritv.org
 http://thereligionofpeace.com http://truepeace.org
 Unix roulette: `dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM`

 __
 R-help@r-project.org mailing list
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 PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.



-- 
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.

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Re: [R] additive interaction for a dichotomous dependent variable (i.e. risk difference)

On Nov 19, 2012, at 1:57 AM, wouterjohannes wrote:

No, I never got any response. I used an article by Knol to solve the
issue. Ik hope this is useful for you too.

Best regards,

Wouter

Knol, M.J., van, d.T., Grobbee, D.E., Numans, M.E., Geerlings, M.I.,
2007. Estimating interaction on an additive scale between continuous
determinants in a logistic regression model. International Journal
of Epidemiology 36, -1118.

I'm a bit surprised that you did not use the binomial family with a
probit link. The help page for ?family mentions the probit link as
appropriate for normal CDF. It would seem possible to construct
competing models both with binomial errors and compare the deviance.

If you wanted to do this via embedding each model in a higher order
model where the degree of multipicativity was represented by a
parameter to be estimated, I remember that Breslow offered GLIM code
to perform that test in one of his two volumes. My copies are
temporarily in storage or I would have given a more specific citation.

N.E. Breslow N. E. Day (1980) Statistical Methods in Cancer
Research. Volume I - The Analysis of Case-Control Studies
N.E. Breslow N. E. Day (1987) Statistical Methods in Cancer
Research. Volume II - The Analysis of Cohort Studies

Wait ... I can do better. The entire chapter that I was remembering
appears to now be accessible online from the publisher's website:

http://www.iarc.fr/en/publications/pdfs-online/stat/sp82/SP82_vol2-4.pdf

Did you ever get a response to this or resolve this yourself?

Many thanks!

If you reply to this email, your message will be added to the
discussion below:

http://r.789695.n4.nabble.com/additive-interaction-for-a-dichotomous-dependent-variable-i-e-risk-difference-tp4635842p4649687.html

David Winsemius, MD
Alameda, CA, USA

Re: [R] generated list element names

2012-11-19 Thread Sam Steingold

 * jim holtman wubyg...@tznvy.pbz [2012-11-19 13:14:05 -0500]:

 How about this (if you don't like writing two lines, encapsulate it in
 a function):

 x - list(10)
 names(x) - paste('f', 'oo', sep = '')
 str(x)
 List of 1
  $ foo: num 10


I am sorry, how is this different from my second snippet (except that
you use x and I use z and you use single quotes in paste and I use
double quotes)?


 On Mon, Nov 19, 2012 at 1:07 PM, Sam Steingold s...@gnu.org wrote:
 How can I create lists with element names created on the fly?

 --8---cut here---start-8---
 list (foo = 10)
 $foo
 [1] 10

 list (foo = 10)
 $foo
 [1] 10

 list (paste(f,oo,sep=) = 10)
 Error: unexpected '=' in list (paste(f,oo,sep=) =
 --8---cut here---end---8---

 I understand that tags in list() are not evaluated, but is there a more
 elegant way than

 --8---cut here---start-8---
 z - list(10)
 names(z) - paste(f,oo,sep=)
 z
 $foo
 [1] 10
 --8---cut here---end---8---

 thanks!

 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 
 11.0.11103000
 http://www.childpsy.net/ http://www.memritv.org
 http://thereligionofpeace.com http://truepeace.org
 Unix roulette: `dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM`

 __
 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.

-- 
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000
http://www.childpsy.net/ http://memri.org http://truepeace.org http://ffii.org
http://think-israel.org http://jihadwatch.org http://palestinefacts.org
The only time you have too much fuel is when you're on fire.

__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] generated list element names

2012-11-19 Thread William Dunlap

If you have a list and want to add a new (or replace a)
named component use
   myList[[compName]] - compValue
as in
   myList - list()
   compName - Incr
   compValue - function(x) x + 1
   myList[[compName]] - compValue

If you want to make a new list-with-names from scratch try
   structure(list(1, cat, function(x)x+1), names=c(One,Pet,Increment))
(structure() is a general way to make an object and add attributes to
it in one statement.)

Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com


 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
 Behalf
 Of Sam Steingold
 Sent: Monday, November 19, 2012 10:07 AM
 To: r-help@r-project.org
 Subject: [R] generated list element names
 
 How can I create lists with element names created on the fly?
 
 --8---cut here---start-8---
  list (foo = 10)
 $foo
 [1] 10
 
  list (foo = 10)
 $foo
 [1] 10
 
  list (paste(f,oo,sep=) = 10)
 Error: unexpected '=' in list (paste(f,oo,sep=) =
 --8---cut here---end---8---
 
 I understand that tags in list() are not evaluated, but is there a more
 elegant way than
 
 --8---cut here---start-8---
  z - list(10)
  names(z) - paste(f,oo,sep=)
  z
 $foo
 [1] 10
 --8---cut here---end---8---
 
 thanks!
 
 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 
 11.0.11103000
 http://www.childpsy.net/ http://www.memritv.org
 http://thereligionofpeace.com http://truepeace.org
 Unix roulette: `dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM`
 
 __
 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.

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Re: [R] survfit number of variables != number of variable names

2012-11-19 Thread Georges Dupret

-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Hi Terry,

I attached a small data set to this email. This is what I get (I
restricted the formula to avoid NA's):

 surv = with(small, Surv(time=absence, event=(censored==FALSE))) 
 (cox.s = coxph(surv ~  bucket*(today) + strata(activity), data =
 small))
Call:
coxph(formula = surv ~ bucket * (today) + strata(activity), data = small)


   coef exp(coef) se(coef)  zp
bucket5750.4526 1.5720.740  0.612 0.54
todayTRUE   -0.0886 0.9150.676 -0.131 0.90
bucket575:todayTRUE -0.1670 0.8460.794 -0.210 0.83

Likelihood ratio test=2.32  on 3 df, p=0.509  n= 100, number of
events= 100
 fit = survfit(cox.s, newdata=small[1:50,])
Error in model.frame.default(data = small[1:50, ], formula = ~bucket +  :
  number of variables != number of variable names

 

also:

 R.version
   _
platform   x86_64-redhat-linux-gnu
arch   x86_64
os linux-gnu
system x86_64, linux-gnu
status
major  2
minor  15.1
year   2012
month  06
day22
svn rev59600
language   R
version.string R version 2.15.1 (2012-06-22)
nickname   Roasted Marshmallows

package ‘survival’ is version 2.36-14

and finally, variable absence is numeric, bucket  activity are
factors and all other variables are logical.

I tested the same formula without 'strata' and I had no problem.

Best and thank you,

ge

On 11/19/2012 09:01 AM, Terry Therneau wrote:
 I can't reproduce the problem.
 
 Tell us what version of R and what version of the survival
 package. Create a reproducable example.  I don't know if some
 variables are numeric and some are factors, how/where the surv
 object was defined, etc.
 
 Terry Therneau
 
 
 
 On 11/17/2012 05:00 AM, r-help-requ...@r-project.org wrote:
 This works ok:
 
 cox = coxph(surv ~ bucket*(today + accor + both) + activity,
 data
 = data)
 fit = survfit(cox, newdata=data[1:100,])
 but using strata leads to problems:
 
 cox.s = coxph(surv ~  bucket*(today + accor + both) +
 strata(activity),
 data = data) fit.s = survfit(cox.s, newdata=data[1:100,])
 Error in model.frame.default(data = data[1:100, ], formula =
 ~bucket +  : number of variables != number of variable names
 
 Note that the following give rise to the same error:
 
 fit.s = survfit(cox.s, newdata=data)
 Error in model.frame.default(data = data, formula = ~bucket +
 today +  : number of variables != number of variable names
 
 but if I use data implicitly, all is working fine:
 fit.s = survfit(cox.s)
 Any idea on how I could solve this?
 
 Best, and thank you,
 
 ge
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Re: [R] How to subset my data and at the same time keep the balance?

2012-11-19 Thread Eddie Smith

Thanks a lot! I got some ideas from all the replies and here is the final one.

newdata

select - sample(nrow(newdata), nrow(newdata) * .7)
data70 - newdata[select,]  # select
write.csv(data70, data70.csv, row.names=FALSE)

data30 - newdata[-select,]  # testing
write.csv(data30, data30.csv, row.names=FALSE)

Cheers

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Re: [R] lattice: defining grouping variable only for the upper/lower panel with splom

2012-11-19 Thread ilai

On Mon, Nov 19, 2012 at 5:42 AM, AnjaM a.miren...@gmail.com wrote:

 Using the mtcars dataset, how to define the grouping variable to be valid
 only for the upper or lower panel?

 The following doesn't work:

 # Code start


Almost :

splom(~data.frame(mpg, disp, hp, drat, wt, qsec),
   data=mtcars, pscales=0,
   auto.key=list(columns=3),
   upper.panel = function(...,groups){
 panel.grid(...)
 panel.xyplot(groups=mtcars$cyl,...)
   }
 )




 # Code end



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[R] Help vectorizing data generation for IRT graded response model

2012-11-19 Thread drj571

Hello, 

I have code that will generate data for a 5 category IRT graded response
model.  However, the code could be improved through vectorizing.  Here is
the code below:

###Inputs
N - 100 #Number of people taking test
n - 10 #Number of items
nCat - 5  #number of categories
###Generate Item parameters for 10 items
a.1 - rlnorm(n, .25, .5)
b.1 - matrix(0, n, (nCat - 1))
theta - rnorm(N, mean = 0, sd =1)
###Generate threhold parameters for b.1 GRM with 10 items there are 4
thresholds for 5 categories
###Using this method is for b dist as N(0,1) since b.1[,1] mean is -.6
###and b.1[,2] to b.1[,4] are all .2
b.1[, 1] - rnorm(n, -.6, 1)
for(j in 1:n) {
b.1[j, 2] - b.1[j,1] + runif(1, .5, .9)
b.1[j, 3] - b.1[j,2] + runif(1, .5, .9)
b.1[j, 4] - b.1[j,3] + runif(1, .5, .9)
}

###This code simulates participants taking a test and generates the 5
category item responses
p - array(0,c(N,n,nCat))
pstar - array(1,c(N,n,nCat))
u - matrix(0,N,n)
for (i in 1:N) {
for (j in 1:n) {
#Draw a random number to determine categories
r - runif(1, 0, 1)
for (k in 2:nCat) {
pstar[i, j, k] - 1 / (1 + exp(-a.1[j] * (theta[i] - b.1[j,
(k-1)])))
p[i,j,(k-1)] - pstar[i, j, (k-1)] - pstar[i, j, k]
}
p[i, j, nCat] - pstar[i, j, 5] #probability of last category or
higher is that category

if (r = p[i, j, 1]) {
u[i,j] - 1
} else
if (r = p[i,j,1] + p[i,j,2]) {
u[i,j] - 2   
} else
if (r = p[i,j,1] + p[i,j,2] + p[i,j,3]) {
u[i,j] - 3
} else
if (r = p[i,j,1] + p[i,j,2] + p[i,j,3] + p[i,j,4])
{
u[i,j] -4
} else
if (r = 1) {
u[i,j] -5
}
}
}


Obviously, that is really long and hairy.  I am wondering what would be the
best way to write this more compactly.  In conjunction with a colleague, I
have a solution for a binary IRT model with response categories 0,1.

Here is that code:
N - 100
n - 10
a.1 - rlnorm(n, .25, .5)
b.1 - rnorm(n, 0, 1)
theta - rnorm(N, 0, 1)

###Function to generate 2PL data
dichEq - function(a.1, b.1, theta) {
1/(1 + exp(-a.1 * (theta -b.1)))
}
probVal - mapply(FUN = function(x,y) dichEq(x,y,theta), a.1, b.1)
u - apply(probVal, 2, function(x) rbinom(length(x), 1, x))

Can someone provide some guidance on how to generalize this to the ordered
category case?

Thanks, 

Jared



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Re: [R] generated list element names

2012-11-19 Thread jim holtman

I missed the last snipet; just saw the first.  So you have your
solution.  If you want a function, try:

 f.newList - function(x,name){.x - list(x);names(.x) - name;.x}
 f.newList(10, paste('f', 'oo', sep = ''))
$foo
[1] 10


On Mon, Nov 19, 2012 at 1:32 PM, Sam Steingold s...@gnu.org wrote:
 * jim holtman wubyg...@tznvy.pbz [2012-11-19 13:14:05 -0500]:

 How about this (if you don't like writing two lines, encapsulate it in
 a function):

 x - list(10)
 names(x) - paste('f', 'oo', sep = '')
 str(x)
 List of 1
  $ foo: num 10


 I am sorry, how is this different from my second snippet (except that
 you use x and I use z and you use single quotes in paste and I use
 double quotes)?


 On Mon, Nov 19, 2012 at 1:07 PM, Sam Steingold s...@gnu.org wrote:
 How can I create lists with element names created on the fly?

 --8---cut here---start-8---
 list (foo = 10)
 $foo
 [1] 10

 list (foo = 10)
 $foo
 [1] 10

 list (paste(f,oo,sep=) = 10)
 Error: unexpected '=' in list (paste(f,oo,sep=) =
 --8---cut here---end---8---

 I understand that tags in list() are not evaluated, but is there a more
 elegant way than

 --8---cut here---start-8---
 z - list(10)
 names(z) - paste(f,oo,sep=)
 z
 $foo
 [1] 10
 --8---cut here---end---8---

 thanks!

 --
 Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 
 11.0.11103000
 http://www.childpsy.net/ http://www.memritv.org
 http://thereligionofpeace.com http://truepeace.org
 Unix roulette: `dd if=/dev/urandom of=/dev/kmem bs=1 count=1 seek=$RANDOM`

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 11.0.11103000
 http://www.childpsy.net/ http://memri.org http://truepeace.org http://ffii.org
 http://think-israel.org http://jihadwatch.org http://palestinefacts.org
 The only time you have too much fuel is when you're on fire.



-- 
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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.

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[R] Plot Area Dimensions

2012-11-19 Thread Richard James

Dear colleagues,

I wish to create a figure with 6 plots arranged vertically with no spacing
between them as they all have a common x-axis. 

However, using the code below I'm unable to get the plot area the same size
for each plot. 

The bottom plot with the x-axis label is smaller than the others, as is the
top plot which has larger margins. 

How can I get the plot region the same size for all 6 plots, whislt still
having a large enough margin for the x-axis label on the bottom plot?

y-rnorm(1:100)
x-rnorm(1:100)
par(mfrow=c(6,1))
par(mar=c(0,5,2,5))
plot(y~x, xlab=, xaxt=n, ylab=y)
par(mar=c(0,5,0,5))
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
par(mar=c(4,5,0,5))
plot(y~x, xlab=x, ylab=y)

Regards

Richard



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Re: [R] survfit number of variables != number of variable names

2012-11-19 Thread Carina Salt

Hi - I've seen a similar issue going on with survfit when using strata in
the model, although I get a different error message from ge.  If it helps
to track down the problem (rather than confusing things further) here is
some code that should reproduce the issue I've seen.  I'm running R 2.15.2,
with survival version 2.36-14 .

Cheers,
Carina

##

library(survival)

# use aml dataset from survival  create 3 imaginary possible stratas -
strat3 has 3 levels, strat4 has 4 levels, strat5 has 5 levels
aml$strat3-as.factor(rep(c(1:3),length=nrow(aml)))
aml$strat4-as.factor(rep(c(1:4),length=nrow(aml)))
aml$strat5-as.factor(rep(c(1:5),length=nrow(aml)))

# create a counting process format dataset from aml - call this aml2
aml2-survSplit(aml,cut=c(5,10,50),end=time,start=start,event=status,episode=i,id=indiv)

# create a dataset of 4 'new' observations - call this aml.new
aml.new-aml[1:4,]
aml.new$time-c(30,50,70,100)
aml.new$status-1
aml.new$x[1:4]-c(rep(Maintained,2),rep(Nonmaintained,2))
aml.new$strat3[1:4]-1
aml.new$strat4[1:4]-1
aml.new$strat5[1:4]-1

# create a counting process format dataset from aml.new - call this aml2.new
aml2.new-survSplit(aml.new,cut=c(5,10,50),end=time,start=start,event=status,episode=i,id=indiv)

# First a model using no strata - survfit works fine on new dataset
myModel-coxph(Surv(start, time, status) ~ x,data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# Now a model using strata = strat4 (which has 4 levels) - survfit again
works fine on new dataset (which has 4 new individuals)
myModel-coxph(Surv(start, time, status) ~ x+strata(strat4),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# Now a model using strata = strat3 (which has 3 levels) - survfit works
here too
myModel-coxph(Surv(start, time, status) ~ x+strata(strat3),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# Now a model using strata = strat5 (which has 5 levels) - survfit now does
not work, with error saying
# Error in survfitcoxph.fit(y, x, wt, x2, risk, newrisk, strata, se.fit,  :
#  'names' attribute [5] must be the same length as the vector [4]
myModel-coxph(Surv(start, time, status) ~ x+strata(strat5),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# Now recreate aml.new but with 3 rather than 4 'new' observations
rm(aml.new)
aml.new-aml[1:3,]
aml.new$time-c(30,50,70)
aml.new$status-1
aml.new$x[1:3]-c(rep(Maintained,2),rep(Nonmaintained,1))
aml.new$strat3[1:3]-1
aml.new$strat4[1:3]-1
aml.new$strat5[1:3]-1

# create a counting process format dataset from aml.new
aml2.new-survSplit(aml.new,cut=c(5,10,50),end=time,start=start,event=status,episode=i,id=indiv)


# Survfit on model using strat3 still works
myModel-coxph(Surv(start, time, status) ~ x+strata(strat3),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# But Survfit on model using strat4 doesn't work now
myModel-coxph(Surv(start, time, status) ~ x+strata(strat4),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))

# Survfit on strat5 model doesn't work either
myModel-coxph(Surv(start, time, status) ~ x+strata(strat5),data = aml2)
plot(survfit(myModel,aml2.new,id=indiv))







On 19 November 2012 17:01, Terry Therneau thern...@mayo.edu wrote:

 I can't reproduce the problem.

 Tell us what version of R and what version of the survival package.
 Create a reproducable example.  I don't know if some variables are numeric
 and some are factors, how/where the surv object was defined, etc.

 Terry Therneau




 On 11/17/2012 05:00 AM, r-help-requ...@r-project.org wrote:

 This works ok:

cox = coxph(surv ~ bucket*(today + accor + both) + activity, data =
 data)
   fit = survfit(cox, newdata=data[1:100,])

 but using strata leads to problems:

cox.s = coxph(surv ~  bucket*(today + accor + both) +
 strata(activity),
   data = data)
   fit.s = survfit(cox.s, newdata=data[1:100,])

 Error in model.frame.default(data = data[1:100, ], formula = ~bucket +  :
number of variables != number of variable names

 Note that the following give rise to the same error:

fit.s = survfit(cox.s, newdata=data)

 Error in model.frame.default(data = data, formula = ~bucket + today +  :
number of variables != number of variable names

 but if I use data implicitly, all is working fine:

   fit.s = survfit(cox.s)

 Any idea on how I could solve this?

 Best, and thank you,

 ge


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and

Re: [R] Biologist R learner

2012-11-19 Thread chuck.01

Please take the advice of Berend if you ever want to get help here.  Also,
you will need to do some basic and initial work yourself; read what he
suggested, use Google to search keywords, and see great help like this:

http://cran.r-project.org/doc/contrib/Short-refcard.pdf

also, use the built in help pages by using a question mark like this:

?min
?max
?ifelse
?cumsum

Good luck; don't let a sharp learning curve stop you from learning R.



Anna23 wrote
 I am a Biologist and a beginner, please help me to solve this: please
 anyone..its my homework and I dont have a clue abt R yet..
Write a function which does the following tasks:
 
 (a) Calculates minimum and maximum value of a given argument x.
 (b) If x is positive, some new vector gets the value of TRUE, and FALSE
 otherwise.
 (c) Creates a vector where the i:th and (i-1):th values of x are always
 summed. First value of
 the new vector has the same value as the first component of x. Use the
 created function to
 some vector x to show that the function works.





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Re: [R] survfit number of variables != number of variable names

2012-11-19 Thread Georges Dupret

Hi!

In answer to:


I noticed that you were using what might be called an externally  
created Surv object. I have a memory that Terry Therneau has  
criticized that practice. I cannot remember if it was in exactly this  
situation but I might ask if setting up the model as:

cox = coxph(Surv(stime, event) ~ bucket*(today + accor + both) +  
activity, data = data)

... might give the survival machinery a better handle on where  
everything might be found. 


I tried to create the Surv object internally but I face the same issue:

 (cox.s = coxph(Surv(time=absence, event=(censored==FALSE)) ~ 
 bucket*(today) + strata(activity), data = small))
Call:
coxph(formula = Surv(time = absence, event = (censored == FALSE)) ~ 
bucket * (today) + strata(activity), data = small)

   coef exp(coef) se(coef)  zp
bucket5750.4526 1.5720.740  0.612 0.54
todayTRUE   -0.0886 0.9150.676 -0.131 0.90
bucket575:todayTRUE -0.1670 0.8460.794 -0.210 0.83

Likelihood ratio test=2.32  on 3 df, p=0.509  n= 100, number of events= 100 
 fit = survfit(cox.s, newdata=small[1:50,])
Error in model.frame.default(data = small[1:50, ], formula = ~bucket +  : 
  number of variables != number of variable names

Best, and thank you for the suggestion.

ge



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[R] xts plot behavior

2012-11-19 Thread swiss_guy

*Hi

I have problem with plot.xts .
I try to subset some data in a xts time series.*



*subseting works fore more that one event*

*But I receive nothing, If I try to get one event *



I'm happy for every hint!
Thanks!





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Re: [R] generated list element names



On Nov 19, 2012, at 10:46 AM, William Dunlap wrote:


If you have a list and want to add a new (or replace a)
named component use
  myList[[compName]] - compValue
as in
  myList - list()
  compName - Incr
  compValue - function(x) x + 1
  myList[[compName]] - compValue

If you want to make a new list-with-names from scratch try
  structure(list(1, cat, function(x)x+1),  
names=c(One,Pet,Increment))

(structure() is a general way to make an object and add attributes to
it in one statement.)


I'm guessing that Sam wanted to see:

myList - list()
myList[[ paste0(fo, o) ]] - 10

 myList
$foo
[1] 10

Or:

structure(list(10), names=paste0(fo, o) )
$foo
[1] 10

(At least that's my guess from the context of the question.)




Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com



-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org 
] On Behalf

Of Sam Steingold
Sent: Monday, November 19, 2012 10:07 AM
To: r-help@r-project.org
Subject: [R] generated list element names

How can I create lists with element names created on the fly?

--8---cut here---start-8---

list (foo = 10)

$foo
[1] 10


list (foo = 10)

$foo
[1] 10


list (paste(f,oo,sep=) = 10)

Error: unexpected '=' in list (paste(f,oo,sep=) =
--8---cut here---end---8---

I understand that tags in list() are not evaluated, but is there a  
more

elegant way than

--8---cut here---start-8---

z - list(10)
names(z) - paste(f,oo,sep=)
z

$foo
[1] 10
--8---cut here---end---8---

thanks!

--
Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X  
11.0.11103000

http://www.childpsy.net/ http://www.memritv.org
http://thereligionofpeace.com http://truepeace.org
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David Winsemius, MD
Alameda, CA, USA

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Re: [R] Classification methods - which one?

2012-11-19 Thread Peter Kupfer

Dear Max, 
first: Thanks a lot for your suggestion and the open words about methods in 
real life. I guess: Thats my problem.
Regarding my analysis: Yes, thats the problem and I have to coerce to do this 
analysis regarding lack of time to start something/other methods. 
So you suggest Linear Discriminant Analysis. Is there a special packages you 
recommend? Nearest Shrunken Centroids i checked with the package PAMR 
(http://www-stat.stanford.edu/~tibs/PAM/Rdist/doc/readme.html)
The example works fine but I guess i have to many rows (or in this case genes) 
for the analysis. My main problem is that i cannot reduce the amount of the 
genes because some of the bosses want to compare the output of classification 
methods with a ruled-based algorithm which works with all genes (after P/A 
calls and an alternative CDF) on the array. So an reduction of the 17 000 genes 
is only possible in a limited way (around 7000 genes after some pre-processing 
steps).
For all tips and suggestions I am more than happy.
Best
Peter



Am 19.11.2012 um 16:36 schrieb Max Kuhn mxk...@gmail.com:

 My suggestion is not to do any predictive modeling. Basically, the
 data doesn't support a sensible and reproducible model. Yes, the
 literature is saturated with this type of analysis but almost none of
 the examples have any utility in real life.
 
 Stick to differential expression analysis, investigate the results
 statistically and biologically then design a prospective experiment
 with a specific set of genes and a more refined measurement system.
 
 If you are doing this analysis to learn something from the data (as
 opposed to generating accurate predictions), a predictive model is one
 of the worst ways of going about it.
 
 If you are coerced to do this analysis, stick to linear methods
 (regularized LDA, nearest shrunken centroids, etc) that are less
 likely to over-fit and bias yourself towards those that have embedded
 feature selection.
 
 Max
 
 
 On Mon, Nov 19, 2012 at 10:16 AM, Peter Kupfer peter.kup...@me.com wrote:
 Dear all,
 i searched for some classification methods and I have no glue if i took the 
 right once.
 My problem: I have a matrix with 17000 rows and 33 colums (genes and 
 patients). The patients are grouped into 3 diseases.
 No I want to classify the patients and for sure i want to know which rows 
 are more helpful for the classification than others.
 
 I tried SVM and random forest. Do you think this are the right 
 classification methods? Maybe there are some hints you can give me. I am 
 more familiar with the Bioconductor packages. Furthermore: This is/was not 
 my field of study in the past but I want to understand it and I am willing 
 to deal with this field.
 Would be amazing if one of the (more) mathematical people can give me a hint.
 Thanks and all the best
 
 Peter
 
 
 PS: I can upload my underlying data if somebody is interested
 
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 -- 
 
 Max

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Re: [R] simple linear regression with proportion data

On 12-11-19 10:18 AM, S Ellison wrote:
 
 -Original Message-
 Can I use simple linear regression when I have proportion data
 for both dependent and independent variables?  Or, should I use
 beta regression analysis?  Or any suggestion?
 
 
 The distribution of the independent variable is irrelevant (in some
 circumstances it matters whether it is measured without error or
 not).

 Agreed that if you just want a line that goes somewhere near the data
 you can do pretty much anything. But don't those circumstances you
 referred to include 'any time you want an unbiased estimate of the
 slope or a reliable standard error on coefficients'?

  Did you see that I wrote distribution of the **independent**
variable above?

  Ben

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Re: [R] Plot Area Dimensions

2012-11-19 Thread Richard M. Heiberger

I think this task would be easier in lattice

library(lattice)
xyplot(y + y + y + y + y + y ~ x, outer=TRUE, layout=c(1,6),
  strip=FALSE, strip.left=TRUE,
  ylab=6 copies of the Y variable,
  main=put an interesting title here)

Six different y variables instead of six copies of the same would give a
more
interesting plot:

tmp - data.frame(matrix(rnorm(700), 100, 7,
  dimnames=list(1:100,
c(x,y1,y2,y3,y4,y5,y6
xyplot(y1 + y2 + y3 + y4 + y5 + y6 ~ x, data=tmp,
  outer=TRUE, layout=c(1,6),
  strip=FALSE, strip.left=TRUE,
  ylab=6 levels of the y response,
  main=put an interesting title here)


Rich

On Mon, Nov 19, 2012 at 10:51 AM, Richard James
richard.j.coo...@uea.ac.ukwrote:

 Dear colleagues,

 I wish to create a figure with 6 plots arranged vertically with no spacing
 between them as they all have a common x-axis.

 However, using the code below I'm unable to get the plot area the same size
 for each plot.

 The bottom plot with the x-axis label is smaller than the others, as is the
 top plot which has larger margins.

 How can I get the plot region the same size for all 6 plots, whislt still
 having a large enough margin for the x-axis label on the bottom plot?

 y-rnorm(1:100)
 x-rnorm(1:100)
 par(mfrow=c(6,1))
 par(mar=c(0,5,2,5))
 plot(y~x, xlab=, xaxt=n, ylab=y)
 par(mar=c(0,5,0,5))
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 par(mar=c(4,5,0,5))
 plot(y~x, xlab=x, ylab=y)

 Regards

 Richard



 --
 View this message in context:
 http://r.789695.n4.nabble.com/Plot-Area-Dimensions-tp4650051.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Error in Sweave but not underlying script

2012-11-19 Thread Bush, Daniel P. DPI

No, that did not resolve the issue, but thanks for the suggestion.

Daniel Bush | School Finance Consultant 
School Financial Services | Wis. Dept. of Public Instruction 
daniel.bush -at- dpi.wi.gov | 608-267-9212


-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com] 
Sent: Friday, November 16, 2012 1:51 PM
To: Bush, Daniel P. DPI
Cc: 'r-help@r-project.org'
Subject: Re: [R] Error in Sweave but not underlying script

On 16/11/2012 2:26 PM, Bush, Daniel P. DPI wrote:
 I'm trying to use Sweave to create a dynamic report of a variety of financial 
 data checks. I have an .R code file to pull the data from a database, 
 manipulate and filter it, and create individual data frames for each test. My 
 Sweave .RNW document then calls that file with source() to generate the data 
 for the report. The .R file works fine on its own, but when I run it from 
 within the Sweave document I get the following error message:

 Error in .subset(x, j) : only 0's may be mixed with negative 
 subscripts

 Again, the .R code works perfectly well on its own--I only get the error when 
 calling it through Sweave. Is there some quirk to Sweave that certain 
 functions don't work properly?

No, it's a pretty standard evaluation environment.  However,  it may be running 
R without some functions that exist in your workspace when you source the 
script within an R session.  It's also possible (but doesn't seem likely) that 
RStudio is causing some problems; you could try running R CMD Sweave 
yourdoc.Rnw from the command line, outside of Rstudio, to see if that makes a 
difference.

Duncan Murdoch

 I am using the built-in Sweave function within RStudio 0.97.168.

 DB

 Daniel Bush
 School Finance Consultant
 School Financial Services
 Wisconsin Department of Public Instruction PO Box 7841 | Madison, WI 
 53707-7841 daniel.bush -at- dpi.wi.gov | sfs.dpi.wi.gov
 Ph: 608-267-9212 | Fax: 608-266-2840

 __
 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
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Re: [R] Error in Sweave but not underlying script

2012-11-19 Thread Duncan Murdoch


On 19/11/2012 3:27 PM, Bush, Daniel P. DPI wrote:

No, that did not resolve the issue, but thanks for the suggestion.


Here's another possibility:  your Sweave session may not be setting the 
same option defaults in startup code as your regular session. This one 
just bit me:  I normally work with options(stringsAsFactors = FALSE), 
because I don't like to have strings automatically converted to 
factors.  However, this option wasn't set when running Sweave externally.


Duncan Murdoch



Daniel Bush | School Finance Consultant
School Financial Services | Wis. Dept. of Public Instruction
daniel.bush -at- dpi.wi.gov | 608-267-9212


-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Friday, November 16, 2012 1:51 PM
To: Bush, Daniel P. DPI
Cc: 'r-help@r-project.org'
Subject: Re: [R] Error in Sweave but not underlying script

On 16/11/2012 2:26 PM, Bush, Daniel P. DPI wrote:
 I'm trying to use Sweave to create a dynamic report of a variety of financial 
data checks. I have an .R code file to pull the data from a database, manipulate 
and filter it, and create individual data frames for each test. My Sweave .RNW 
document then calls that file with source() to generate the data for the report. 
The .R file works fine on its own, but when I run it from within the Sweave 
document I get the following error message:

 Error in .subset(x, j) : only 0's may be mixed with negative
 subscripts

 Again, the .R code works perfectly well on its own--I only get the error when 
calling it through Sweave. Is there some quirk to Sweave that certain functions 
don't work properly?

No, it's a pretty standard evaluation environment.  However,  it may be running R without 
some functions that exist in your workspace when you source the script within an R 
session.  It's also possible (but doesn't seem likely) that RStudio is causing some 
problems; you could try running R CMD Sweave yourdoc.Rnw from the command 
line, outside of Rstudio, to see if that makes a difference.

Duncan Murdoch

 I am using the built-in Sweave function within RStudio 0.97.168.

 DB

 Daniel Bush
 School Finance Consultant
 School Financial Services
 Wisconsin Department of Public Instruction PO Box 7841 | Madison, WI
 53707-7841 daniel.bush -at- dpi.wi.gov | sfs.dpi.wi.gov
 Ph: 608-267-9212 | Fax: 608-266-2840

 __
 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.



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Re: [R] aggregate() runs out of memory

2012-11-19 Thread Sam Steingold

Thanks Steve,
what is the analogue of .N for min and max?
i.e., what is the data.table's version of
aggregate(infl$delay,by=list(infl$share.id),FUN=min)
aggregate(infl$delay,by=list(infl$share.id),FUN=max)
thanks!
Sam.

On Fri, Sep 14, 2012 at 3:40 PM, Steve Lianoglou
mailinglist.honey...@gmail.com wrote:
 Hi,

 On Fri, Sep 14, 2012 at 3:26 PM, Sam Steingold s...@gnu.org wrote:
 I have a large data.frame Z (2,424,185,944 bytes, 10,256,441 rows, 17 
 columns).
 I want to get the result of
 table(aggregate(Z$V1, FUN = length, by = list(id=Z$V2))$x)
 alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is
 24.3G, and no end in sight.
 both V1 and V2 are characters (not factors).
 Is there anything I could do to speed this up?
 Thanks.

 You might find you'll get a lot of mileage out of data.table when
 working with such large data.frames ...

 To get something close to what you're after, you can try:

 R library(data.table)
 R Z - as.data.table(Z)
 R setkeyv(Z, 'V2')
 R agg - Z[, list(count=.N), by='V2']

 From here you might

 R tab1 - table(agg$count)

 I think that'll get you where you want to be ... I'm ashamed to say
 that I haven't really done much w/ aggregate since I mostly have used
 plyr and data.table like stuff, so I might be missing your end goal --
 providing a reproducible example with a small data.frame from you can
 help here (for me at least).

 HTH,
 -steve

 --
 Steve Lianoglou
 Graduate Student: Computational Systems Biology
  | Memorial Sloan-Kettering Cancer Center
  | Weill Medical College of Cornell University
 Contact Info: http://cbio.mskcc.org/~lianos/contact



--
Sam Steingold http://sds.podval.org http://www.childpsy.net/

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Re: [R] aggregate() runs out of memory

2012-11-19 Thread Steve Lianoglou

Hi,

On Mon, Nov 19, 2012 at 1:25 PM, Sam Steingold s...@gnu.org wrote:
 Thanks Steve,
 what is the analogue of .N for min and max?
 i.e., what is the data.table's version of
 aggregate(infl$delay,by=list(infl$share.id),FUN=min)
 aggregate(infl$delay,by=list(infl$share.id),FUN=max)
 thanks!

It would be helpful if I could see a bit of your table (like
`head(infl)`, if it's not too big), but anyway: there is no real
analogue of min/max -- you just use them

For instance, if you want the min and max of `delay` within each group
defined by `share.id`, and let's assume `infl` is a data.frame, you
can do something like so:

R as.data.table(infl)
R setkey(infl, share.id)
R result - infl[, list(min=min(delay), max=max(delay)), by=share.id]

HTH,

-steve


-- 
Steve Lianoglou
Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
Contact Info: http://cbio.mskcc.org/~lianos/contact

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[R] Find the number of intersections according factor levels

2012-11-19 Thread li li

Dear all,
  I have two vectors x and y, both of which are of length 1000.
I created a factor as.factor(rep(1:100,each=10)).
  Now I want  to count, for each factor level, the number of intersections
between x and y. In other words, I would like to count the number of
intersections between x[1:10] and y[1:10], the number of
intersections between x[11:20] and y[11:20], the number of
intersections between x[21:30] and y[21:30] and so on.
I know I could use the match function or the %in%  function in a loop,
but would rather use some simpler way to do this, for example using tapply.
  Can anyone give a hint? Thanks a lot.
Hanna

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[R] Is it possible to be sponsored by R?

2012-11-19 Thread Christophe Genolini

Hi the list,

I am a member of the organizing comity of the French Statistics Association 
(SFdS)'s conference. We 
are looking for sponsors. Some software (SAS, RITME, ...) are represented. Do 
you know if there is 
any possibility to be sponsored by R (or by an association close to R)? Do you 
think I can ask to 
the R fondation?

Sincerely
Christophe



-- 
Christophe Genolini
Maître de conférences en bio-statistique
Vice président Communication interne et animation du campus
Université Paris Ouest Nanterre La Défense


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Re: [R] Plot Area Dimensions

2012-11-19 Thread David L Carlson

You can also use layout() with base graphics. This
example sets up a column of 14 strips and allocates
3 strips to the top and bottom graphs and 2 strips
To the four middle graphs. Using Richard's tmp dataframe:

layout(matrix(c(1, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5,
   6, 6, 6), 14, 1))
layout.show(6)
par(mar=c(0,5,2,5))
plot(y1~x, tmp, xlab=, xaxt=n, ylab=y)
par(mar=c(0,5,0,5))
plot(y2~x, tmp, xlab=, xaxt=n, ylab=y)
plot(y3~x, tmp, xlab=, xaxt=n, ylab=y)
plot(y4~x, tmp, xlab=, xaxt=n, ylab=y)
plot(y5~x, tmp, xlab=, xaxt=n, ylab=y)
par(mar=c(4,5,0,5))
plot(y6~x, tmp, xlab=x, ylab=y)

-
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352



 -Original Message-
 From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
 project.org] On Behalf Of Richard M. Heiberger
 Sent: Monday, November 19, 2012 2:26 PM
 To: Richard James
 Cc: r-help@r-project.org
 Subject: Re: [R] Plot Area Dimensions
 
 I think this task would be easier in lattice
 
 library(lattice)
 xyplot(y + y + y + y + y + y ~ x, outer=TRUE, layout=c(1,6),
   strip=FALSE, strip.left=TRUE,
   ylab=6 copies of the Y variable,
   main=put an interesting title here)
 
 Six different y variables instead of six copies of the same would give
 a
 more
 interesting plot:
 
 tmp - data.frame(matrix(rnorm(700), 100, 7,
   dimnames=list(1:100,
 c(x,y1,y2,y3,y4,y5,y6
 xyplot(y1 + y2 + y3 + y4 + y5 + y6 ~ x, data=tmp,
   outer=TRUE, layout=c(1,6),
   strip=FALSE, strip.left=TRUE,
   ylab=6 levels of the y response,
   main=put an interesting title here)
 
 
 Rich
 
 On Mon, Nov 19, 2012 at 10:51 AM, Richard James
 richard.j.coo...@uea.ac.ukwrote:
 
  Dear colleagues,
 
  I wish to create a figure with 6 plots arranged vertically with no
 spacing
  between them as they all have a common x-axis.
 
  However, using the code below I'm unable to get the plot area the
 same size
  for each plot.
 
  The bottom plot with the x-axis label is smaller than the others, as
 is the
  top plot which has larger margins.
 
  How can I get the plot region the same size for all 6 plots, whislt
 still
  having a large enough margin for the x-axis label on the bottom plot?
 
  y-rnorm(1:100)
  x-rnorm(1:100)
  par(mfrow=c(6,1))
  par(mar=c(0,5,2,5))
  plot(y~x, xlab=, xaxt=n, ylab=y)
  par(mar=c(0,5,0,5))
  plot(y~x, xlab=, xaxt=n, ylab=y)
  plot(y~x, xlab=, xaxt=n, ylab=y)
  plot(y~x, xlab=, xaxt=n, ylab=y)
  plot(y~x, xlab=, xaxt=n, ylab=y)
  par(mar=c(4,5,0,5))
  plot(y~x, xlab=x, ylab=y)
 
  Regards
 
  Richard
 
 
 
  --
  View this message in context:
  http://r.789695.n4.nabble.com/Plot-Area-Dimensions-tp4650051.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]]
 
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 guide.html
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[R] scatterplot3d and box3d

2012-11-19 Thread xgao

I created a 3d scatter plot and am trying to change the color of outer box
lines with box3d.

Anybody can help me to figure out how to do this?

My example is:

library(scatterplot3d)
x=seq(1:6)
y=seq(7:12)
z=x*2
scatterplot3d(x, y,z)

Thanks.

Xin

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[R] Coefficient of Variation, NA, Aggregate

2012-11-19 Thread Amanda Jones

Hello helpers,

I have a two part issue. FIRSTLY, I am attempting to write a function
for coefficient of variation, using

 co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where rowleyi 
 is my data set, which has multiple columns and rows of data.

This is not working because some of my columns have NAs. When I try to use

 co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it gives 
 me an error message:

Error in co.var(rowleyi$TL, na.rm = TRUE) :
  unused argument(s) (na.rm = TRUE)

I do not know what this means. How can I get this function to work?
SECONDLY, how can I then get that function to work within an
aggragate? Do I still use

aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies is 
the header for rows? This has worked for mean, std.error, sd, etc.

Thank you!
Amanda Jones

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[R] Help: Meta-analysis with metacor

2012-11-19 Thread Catherine Proulx


Trying to do a meta-analysis of correlations in R using the meta package; have 
tried several things and keep getting a similar error. Can anyone help explain 
the error?
 cor-c(-0.3018, 0.667, -3.8002, -0.607, -0.4885, -3.8002, -0.0701, 0.1348, 
 -0.9505, -0.5709, -0.6127, -1.2419, -0.1511, -0.1054) 
 n-c(3,4,3,3,3,3,16,36,30,9,3,3,30,4) library(meta) metacor(cor, n, 
 data=NULL, subset=NULL, sm=ZCOR, level=0.95, level.comb=level, 
 comb.fixed=TRUE, comb.random=TRUE, hakn=NULL, method.tau=DL, 
 tau.preset=NULL, TE.tau=NULL, method.bias=linreg, title=title, 
 complab=comparison, outclab=outcome)
Error in data.frame(subset = NULL, comb.fixed = TRUE, comb.random = TRUE,  :   
arguments imply differing number of rows: 0, 1
Thanks,
Catherine 
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[R] Performing gage RR study in R w/more than 2 factors

2012-11-19 Thread Matt Jacob

Hi everyone,

I'm fairly new to R, and I don't have a background in statistics, so
please bear with me. ;-)

I'm dealing with 2^k factorial designs, and I was just wondering if
there's any way to analyze more than two factors of a gage RR study in
R. For example, Minitab has an expanded gage RR function that lets
you include up to eight additional factors besides the usual two that
are present in gage studies (parts and operators). If I wanted to
include n additional random factors, is there a package or built-in
functionality that will allow me to do that?

I've been experimenting with the SixSigma package, and that has a ss.rr
method which works great---as long as your experiment only contains two
factors. I've also been using lmer from lme4 to fit a linear model of my
experiment, but the standard deviations generated by lmer don't match
what I'm seeing in Minitab. Since all my factors are random, the formula
I'm using looks like this:

vals ~ 1 + (1|f1) + (1|f2) + (1|f3) + (1|f1:f2) + (1|f1:f3) + (1|f2:f3)

What am I doing wrong, and how can I fix it?

Thanks,

Matt

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Re: [R] Import excel file

2012-11-19 Thread Martin Studer

Hi Cyril,

please let me know the following details:
- sessionInfo() output from R
- Version of XLConnect you are using
- Version of rJava you are using
- Version of Java you are using (complete output of java -version on the
command line)
- Value of the JAVA_HOME environment variable if it is set (e.g. output of
Sys.getenv(JAVA_HOME) in R)

Best regards,
Martin



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Re: [R] Is it possible to be sponsored by R?

Christophe Genolini cgenolin at u-paris10.fr writes:

 
 Hi the list,
 
 I am a member of the organizing comity of the French Statistics
 Association (SFdS)'s conference. We 
 are looking for sponsors. Some software (SAS, RITME, ...) 
 are represented. Do you know if there is 
 any possibility to be sponsored by R (or by an association close to R)?
 Do you think I can ask to 
 the R fondation?
 
 Sincerely
 Christophe

  I rather doubt it, but it depends what you're looking for. If you
want financial support, I suspect the answer is almost definitely not;
if you simply want some kind of symbolic support, then it's very
probably not.  The board of the R foundation is listed at
http://www.r-project.org/foundation/board.html (with only a postal
contact address ...)  You might post to r-de...@r-project.org, which
is read by some R-Core members (which overlaps with the membership of
the R Foundation).

  These are only my guesses, certainly not definitive.

  Ben Bolker

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Re: [R] Ben Bolker's '‘emdbook’ Package , rbetabinom

arun4 arun.ganesh2012 at gmail.com writes:

 I am using rbetabinom ( to generate beta binomial random variables) function
 available in the emdbookpackage written by Professor. Ben Bolker for my
 research study.
 I have no questions with this function. However, I am looking for the
 theoretical method/algorithm  of the function rbetabinom  . 
 Morris (1997), American Naturalist 150:299-327 is given as the reference in
 the package, But I couldn't fund any theoretical methods to generate beta
 binomial random variables in this article.
 
 I would like to kind request you to suggest me any journal paper or document
 to study the methods of generating  beta binomial random variables used to
 write the function rbetabinom.

  I answered this off-list (suggesting that the OP look at the
source code of the function, which is very simple).

  The book gives other references to the beta-binomial, including
Crowder (1978), Reeve and Murdoch (1985), and Hatfield (1996), although
the latter two are ecological rather than statistical examples
and I doubt any of them discusses random deviate generation.

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Re: [R] Coefficient of Variation, NA, Aggregate

HI,
For the first part, may be this helps:
set.seed(5)
mat1-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
rowleyi-data.frame(mat1)
 co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
 apply(rowleyi,2,function(x) co.var(x))
#  X1   X2   X3   X4   X5 
#53.29387 49.53113 45.82576 35.35534 34.99271 
#or
sapply(rowleyi,function(x) co.var(x))
#  X1   X2   X3   X4   X5 
#53.29387 49.53113 45.82576 35.35534 34.99271 
A.K.





- Original Message -
From: Amanda Jones akjone...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Monday, November 19, 2012 4:01 PM
Subject: [R] Coefficient of Variation, NA, Aggregate

Hello helpers,

I have a two part issue. FIRSTLY, I am attempting to write a function
for coefficient of variation, using

 co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where rowleyi 
 is my data set, which has multiple columns and rows of data.

This is not working because some of my columns have NAs. When I try to use

 co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it gives 
 me an error message:

Error in co.var(rowleyi$TL, na.rm = TRUE) :
  unused argument(s) (na.rm = TRUE)

I do not know what this means. How can I get this function to work?
SECONDLY, how can I then get that function to work within an
aggragate? Do I still use

aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies is 
the header for rows? This has worked for mean, std.error, sd, etc.

Thank you!
Amanda Jones

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Re: [R] aggregate() runs out of memory


On Nov 19, 2012, at 1:25 PM, Sam Steingold wrote:

 Thanks Steve,
 what is the analogue of .N for min and max?

?seq

 i.e., what is the data.table's version of
 aggregate(infl$delay,by=list(infl$share.id),FUN=min)

 aggregate(infl$delay,by=list(infl$share.id),FUN=max)

 DT[, list( max(v)), by=x]
   x V1
1: a  3
2: b  6
3: c  9


 thanks!
 Sam.
 
 On Fri, Sep 14, 2012 at 3:40 PM, Steve Lianoglou
 mailinglist.honey...@gmail.com wrote:
 Hi,
 
 On Fri, Sep 14, 2012 at 3:26 PM, Sam Steingold s...@gnu.org wrote:
 I have a large data.frame Z (2,424,185,944 bytes, 10,256,441 rows, 17 
 columns).
 I want to get the result of
 table(aggregate(Z$V1, FUN = length, by = list(id=Z$V2))$x)
 alas, aggregate has been running for ~30 minute, RSS is 14G, VIRT is
 24.3G, and no end in sight.
 both V1 and V2 are characters (not factors).
 Is there anything I could do to speed this up?
 Thanks.
 
 You might find you'll get a lot of mileage out of data.table when
 working with such large data.frames ...
 
 To get something close to what you're after, you can try:
 
 R library(data.table)
 R Z - as.data.table(Z)
 R setkeyv(Z, 'V2')
 R agg - Z[, list(count=.N), by='V2']
 
 From here you might
 
 R tab1 - table(agg$count)
 
 I think that'll get you where you want to be ... I'm ashamed to say
 that I haven't really done much w/ aggregate since I mostly have used
 plyr and data.table like stuff, so I might be missing your end goal --
 providing a reproducible example with a small data.frame from you can
 help here (for me at least).
 
 HTH,
 -steve
 
 --
 Steve Lianoglou
 Graduate Student: Computational Systems Biology
 | Memorial Sloan-Kettering Cancer Center
 | Weill Medical College of Cornell University
 Contact Info: http://cbio.mskcc.org/~lianos/contact
 
 
 
 --
 Sam Steingold http://sds.podval.org http://www.childpsy.net/
 
 __
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David Winsemius, MD
Alameda, CA, USA

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Re: [R] scatterplot3d and box3d

2012-11-19 Thread Duncan Murdoch


On 12-11-19 12:20 PM, x...@genome.wustl.edu wrote:

I created a 3d scatter plot and am trying to change the color of outer box
lines with box3d.

Anybody can help me to figure out how to do this?

My example is:

library(scatterplot3d)
x=seq(1:6)
y=seq(7:12)
z=x*2
scatterplot3d(x, y,z)


See ?scatterplot3d.

Duncan Murdoch

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Re: [R] scatterplot3d and box3d

 xgao at genome.wustl.edu writes:

 
 I created a 3d scatter plot and am trying to change the color of outer box
 lines with box3d.
 
 Anybody can help me to figure out how to do this?
 
 My example is:
 
 library(scatterplot3d)
 x=seq(1:6)
 y=seq(7:12)
 z=x*2
 scatterplot3d(x, y,z)


 This is not going to be possible in the narrow sense: box3d is
from the rgl package, which uses a completely different graphics
device/protocol.  You can combine plot3d with box3d (or rgl.bbox)

By the way, seq(7:12) probably doesn't do what you think it does!

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Re: [R] Plot Area Dimensions

2012-11-19 Thread Greg Snow

The key is to not change the margins, set them once and stick with those
margins.  The next question then becomes how do I leave area at the
top/bottom for the title and common axis? to which the answer is Set
outer margins at the beginning.

Modifying your code:

y-rnorm(1:100)
x-rnorm(1:100)
par(mfrow=c(6,1), mar=c(0,5,0,5), oma=c(4,0,2,0))
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=, xaxt=n, ylab=y)
plot(y~x, xlab=x, ylab=y)


On Mon, Nov 19, 2012 at 8:51 AM, Richard James
richard.j.coo...@uea.ac.ukwrote:

 Dear colleagues,

 I wish to create a figure with 6 plots arranged vertically with no spacing
 between them as they all have a common x-axis.

 However, using the code below I'm unable to get the plot area the same size
 for each plot.

 The bottom plot with the x-axis label is smaller than the others, as is the
 top plot which has larger margins.

 How can I get the plot region the same size for all 6 plots, whislt still
 having a large enough margin for the x-axis label on the bottom plot?

 y-rnorm(1:100)
 x-rnorm(1:100)
 par(mfrow=c(6,1))
 par(mar=c(0,5,2,5))
 plot(y~x, xlab=, xaxt=n, ylab=y)
 par(mar=c(0,5,0,5))
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 plot(y~x, xlab=, xaxt=n, ylab=y)
 par(mar=c(4,5,0,5))
 plot(y~x, xlab=x, ylab=y)

 Regards

 Richard



 --
 View this message in context:
 http://r.789695.n4.nabble.com/Plot-Area-Dimensions-tp4650051.html
 Sent from the R help mailing list archive at Nabble.com.

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-- 
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com

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Re: [R] Coefficient of Variation, NA, Aggregate

HI,

Your example dataset is in unreadable format.  You could use dput().
set.seed(5)
 mat1-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
 rowleyi-data.frame(mat1)
  co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
 rowleyi-data.frame(subspecies=rep(LETTERS[1:2],2),rowleyi)


with(rowleyi,aggregate(cbind(X1,X2,X3,X4,X5),by=list(subspecies),function(x) 
co.var(x)))
  Group.1   X1    X2   X3   X4   X5
1   A   NA 70.710678   NA 20.20305 28.28427
2   B 56.56854  8.318903 60.60915 47.14045  0.0


With your aggregate()
aggregate(.~subspecies,data=rowleyi,co.var)
#  subspecies   X1   X2   X3   X4 X5
#1  B 56.56854 8.318903 60.60915 47.14045  0

A.K.




- Original Message -
From: Amanda Jones akjone...@gmail.com
To: r-help@r-project.org
Cc: 
Sent: Monday, November 19, 2012 4:01 PM
Subject: [R] Coefficient of Variation, NA, Aggregate

Hello helpers,

I have a two part issue. FIRSTLY, I am attempting to write a function
for coefficient of variation, using

 co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where rowleyi 
 is my data set, which has multiple columns and rows of data.

This is not working because some of my columns have NAs. When I try to use

 co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it gives 
 me an error message:

Error in co.var(rowleyi$TL, na.rm = TRUE) :
  unused argument(s) (na.rm = TRUE)

I do not know what this means. How can I get this function to work?
SECONDLY, how can I then get that function to work within an
aggragate? Do I still use

aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies is 
the header for rows? This has worked for mean, std.error, sd, etc.

Thank you!
Amanda Jones

__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.


__
R-help@r-project.org mailing list
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Re: [R] Coefficient of Variation, NA, Aggregate

2012-11-19 Thread Amanda Jones

Fantastic, thank you!

On Mon, Nov 19, 2012 at 3:44 PM, arun smartpink...@yahoo.com wrote:
 HI,

 Your example dataset is in unreadable format.  You could use dput().
 set.seed(5)
  mat1-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
  rowleyi-data.frame(mat1)
   co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
  rowleyi-data.frame(subspecies=rep(LETTERS[1:2],2),rowleyi)


 with(rowleyi,aggregate(cbind(X1,X2,X3,X4,X5),by=list(subspecies),function(x) 
 co.var(x)))
   Group.1   X1X2   X3   X4   X5
 1   A   NA 70.710678   NA 20.20305 28.28427
 2   B 56.56854  8.318903 60.60915 47.14045  0.0


 With your aggregate()
 aggregate(.~subspecies,data=rowleyi,co.var)
 #  subspecies   X1   X2   X3   X4 X5
 #1  B 56.56854 8.318903 60.60915 47.14045  0

 A.K.




 - Original Message -
 From: Amanda Jones akjone...@gmail.com
 To: r-help@r-project.org
 Cc:
 Sent: Monday, November 19, 2012 4:01 PM
 Subject: [R] Coefficient of Variation, NA, Aggregate

 Hello helpers,

 I have a two part issue. FIRSTLY, I am attempting to write a function
 for coefficient of variation, using

 co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where 
 rowleyi is my data set, which has multiple columns and rows of data.

 This is not working because some of my columns have NAs. When I try to use

 co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it 
 gives me an error message:

 Error in co.var(rowleyi$TL, na.rm = TRUE) :
   unused argument(s) (na.rm = TRUE)

 I do not know what this means. How can I get this function to work?
 SECONDLY, how can I then get that function to work within an
 aggragate? Do I still use

aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies 
is the header for rows? This has worked for mean, std.error, sd, etc.

 Thank you!
 Amanda Jones

 __
 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
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Re: [R] Coefficient of Variation, NA, Aggregate

HI,

No problem.

You got two NA in the previous example.
According to the coefficient of variaion documentation in R 
(http://hosho.ees.hokudai.ac.jp/~kubo/Rdoc/library/raster/html/cv.html)
Compute the coefficient of variation (expressed as a percentage). If there is 
only a single value, sd is NA and cv returns NA if aszero=FALSE (the default). 
However, if (aszero=TRUE), cv returns 0.
 
If I use another example:
set.seed(5)
 mat1-matrix(sample(c(1:10,NA),30,replace=TRUE),ncol=5)
rowleyi-data.frame(mat1)
co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
 rowleyi-data.frame(subspecies=rep(LETTERS[1:2],3),rowleyi)

with(rowleyi,aggregate(cbind(X1,X2,X3,X4,X5),by=list(subspecies),function(x) 
co.var(x)))
#  Group.1   X1   X2   X3   X4   X5
#1   A 28.28427 28.28427 25.0 52.67827 57.73503
#2   B 34.64102 61.97443 52.67827 51.50788   NA


#Other way: 
 do.call(cbind,lapply(lapply(lapply(rowleyi[,-1],function(x) 
data.frame(subspecies=rowleyi[,1],x)),function(x) 
x[complete.cases(x),]),function(x) aggregate(.~subspecies,data=x,co.var)))
#  X1.subspecies X1.x X2.subspecies X2.x X3.subspecies X3.x
#1 A 28.28427 A 28.28427 A 25.0
#2 B 34.64102 B 61.97443 B 52.67827
  X4.subspecies X4.x X5.subspecies X5.x
#1 A 52.67827 A 57.73503
#2 B 51.50788 B   NA
A.K.




- Original Message -
From: Amanda Jones akjone...@gmail.com
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org
Sent: Monday, November 19, 2012 5:50 PM
Subject: Re: [R] Coefficient of Variation, NA, Aggregate

Fantastic, thank you!

On Mon, Nov 19, 2012 at 3:44 PM, arun smartpink...@yahoo.com wrote:
 HI,

 Your example dataset is in unreadable format.  You could use dput().
 set.seed(5)
  mat1-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
  rowleyi-data.frame(mat1)
   co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
  rowleyi-data.frame(subspecies=rep(LETTERS[1:2],2),rowleyi)


 with(rowleyi,aggregate(cbind(X1,X2,X3,X4,X5),by=list(subspecies),function(x) 
 co.var(x)))
   Group.1       X1        X2       X3       X4       X5
 1       A       NA 70.710678       NA 20.20305 28.28427
 2       B 56.56854  8.318903 60.60915 47.14045  0.0


 With your aggregate()
 aggregate(.~subspecies,data=rowleyi,co.var)
 #  subspecies       X1       X2       X3       X4 X5
 #1          B 56.56854 8.318903 60.60915 47.14045  0

 A.K.




 - Original Message -
 From: Amanda Jones akjone...@gmail.com
 To: r-help@r-project.org
 Cc:
 Sent: Monday, November 19, 2012 4:01 PM
 Subject: [R] Coefficient of Variation, NA, Aggregate

 Hello helpers,

 I have a two part issue. FIRSTLY, I am attempting to write a function
 for coefficient of variation, using

 co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where 
 rowleyi is my data set, which has multiple columns and rows of data.

 This is not working because some of my columns have NAs. When I try to use

 co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it 
 gives me an error message:

 Error in co.var(rowleyi$TL, na.rm = TRUE) :
   unused argument(s) (na.rm = TRUE)

 I do not know what this means. How can I get this function to work?
 SECONDLY, how can I then get that function to work within an
 aggragate? Do I still use

aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies 
is the header for rows? This has worked for mean, std.error, sd, etc.

 Thank you!
 Amanda Jones

 __
 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.



__
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and provide commented, minimal, self-contained, reproducible code.

Re: [R] Performing gage RR study in R w/more than 2 factors

2012-11-19 Thread Bert Gunter

I believe that you need to consult a local statistician, as there are
likely way too many statistical issues here that you do not fully
understand. Alternatively, try posting to a statistical list like
stats.stackexchange.com, as I think most of your issues are primarily
statistical, not R related.

Cheers,
Bert

On Mon, Nov 19, 2012 at 11:12 AM, Matt Jacob m...@jacobmail.org wrote:
 Hi everyone,

 I'm fairly new to R, and I don't have a background in statistics, so
 please bear with me. ;-)

 I'm dealing with 2^k factorial designs, and I was just wondering if
 there's any way to analyze more than two factors of a gage RR study in
 R. For example, Minitab has an expanded gage RR function that lets
 you include up to eight additional factors besides the usual two that
 are present in gage studies (parts and operators). If I wanted to
 include n additional random factors, is there a package or built-in
 functionality that will allow me to do that?

 I've been experimenting with the SixSigma package, and that has a ss.rr
 method which works great---as long as your experiment only contains two
 factors. I've also been using lmer from lme4 to fit a linear model of my
 experiment, but the standard deviations generated by lmer don't match
 what I'm seeing in Minitab. Since all my factors are random, the formula
 I'm using looks like this:

 vals ~ 1 + (1|f1) + (1|f2) + (1|f3) + (1|f1:f2) + (1|f1:f3) + (1|f2:f3)

 What am I doing wrong, and how can I fix it?

 Thanks,

 Matt

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-- 

Bert Gunter
Genentech Nonclinical Biostatistics

Internal Contact Info:
Phone: 467-7374
Website:
http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm

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Re: [R] xts plot behavior

2012-11-19 Thread R. Michael Weylandt

http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example

RMW

On Mon, Nov 19, 2012 at 5:27 PM, swiss_guy
steven.stut...@swissglobal-am.com wrote:
 *Hi

 I have problem with plot.xts .
 I try to subset some data in a xts time series.*



 *subseting works fore more that one event*

 *But I receive nothing, If I try to get one event *



 I'm happy for every hint!
 Thanks!





 --
 View this message in context: 
 http://r.789695.n4.nabble.com/xts-plot-behavior-tp4650069.html
 Sent from the R help mailing list archive at Nabble.com.

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Re: [R] Coefficient of Variation, NA, Aggregate


Hello,

Just a note, you can (should?) have an argument na.rm in your function 
definition with a small modification, like this:


co.var - function(x,na.rm=TRUE) 100*(sd(x,na.rm=na.rm)/mean(x,na.rm=na.rm))

Then you can choose to use the default TRUE or not.

Hope this helps,

Rui Barradas
Em 19-11-2012 22:50, Amanda Jones escreveu:

Fantastic, thank you!

On Mon, Nov 19, 2012 at 3:44 PM, arun smartpink...@yahoo.com wrote:

HI,

Your example dataset is in unreadable format.  You could use dput().
set.seed(5)
  mat1-matrix(sample(c(1:9,NA),20,replace=TRUE),ncol=5)
  rowleyi-data.frame(mat1)
   co.var-function(x) 100*(sd(x,na.rm=TRUE)/mean(x,na.rm=TRUE))
  rowleyi-data.frame(subspecies=rep(LETTERS[1:2],2),rowleyi)


with(rowleyi,aggregate(cbind(X1,X2,X3,X4,X5),by=list(subspecies),function(x) 
co.var(x)))
   Group.1   X1X2   X3   X4   X5
1   A   NA 70.710678   NA 20.20305 28.28427
2   B 56.56854  8.318903 60.60915 47.14045  0.0


With your aggregate()
aggregate(.~subspecies,data=rowleyi,co.var)
#  subspecies   X1   X2   X3   X4 X5
#1  B 56.56854 8.318903 60.60915 47.14045  0

A.K.




- Original Message -
From: Amanda Jones akjone...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Monday, November 19, 2012 4:01 PM
Subject: [R] Coefficient of Variation, NA, Aggregate

Hello helpers,

I have a two part issue. FIRSTLY, I am attempting to write a function
for coefficient of variation, using


co.var - function(rowleyi) ( 100*sd(rowleyi)/mean(rowleyi) )  #where rowleyi 
is my data set, which has multiple columns and rows of data.

This is not working because some of my columns have NAs. When I try to use


co.var(rowleyi$TL, na.rm=TRUE)  #where TL is one of my column names, it gives 
me an error message:

Error in co.var(rowleyi$TL, na.rm = TRUE) :
   unused argument(s) (na.rm = TRUE)

I do not know what this means. How can I get this function to work?
SECONDLY, how can I then get that function to work within an
aggragate? Do I still use


aggregate(. ~ subspecies, data = rowleyi, CV, na.rm=TRUE) #where subspecies is 
the header for rows? This has worked for mean, std.error, sd, etc.

Thank you!
Amanda Jones

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[R] kinitr

2012-11-19 Thread Muhuri, Pradip (SAMHSA/CBHSQ)

Hello,

I am an Intro-level R and ggplot2 user and looking for resources to self teach 
dynamic report generation in R using knitr. Any advice would be highly 
appreciated.

Thanks,

Pradip

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[R] github

2012-11-19 Thread Muhuri, Pradip (SAMHSA/CBHSQ)


Hello,

I would like to learn how to set up Github/repository and upload/update files 
and am looking for Github for Dummies.  Any help will be appreciated.

Thanks,

Pradip

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Re: [R] kinitr

2012-11-19 Thread Pascal Oettli


Hello,

Why don't you search on Internet?

Regards,
Pascal


Le 20/11/2012 10:57, Muhuri, Pradip (SAMHSA/CBHSQ) a écrit :

Hello,

I am an Intro-level R and ggplot2 user and looking for resources to self teach 
dynamic report generation in R using knitr. Any advice would be highly 
appreciated.

Thanks,

Pradip

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Re: [R] github


On Nov 19, 2012, at 6:07 PM, Muhuri, Pradip (SAMHSA/CBHSQ) wrote:

 
 Hello,
 
 I would like to learn how to set up Github/repository and upload/update files 
 and am looking for Github for Dummies.  Any help will be appreciated.

Wrong list for this question.

-- 

David Winsemius, MD
Alameda, CA, USA

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Re: [R] survfit number of variables != number of variable names