Re: [R] reviewer comment

2013-03-15 Thread Mohamed Lajnef
Thanks John for your reply.

the  reviewer comment:

asymmetric distribution could affect Principal
Component Analysis results,  symmetry of distribution should be
tested. Authors should also indicate if outliers were observed and
consequently excluded because they could affect factors

My question: what does it mean asymmetry distribution could affect PCA  ? and 
also outliers could affect factors?

sorry for this not R-help question.

Best regards

M



Le 15/03/13 14:05, John Kane a écrit :
> No idea of what sentence.  R-help strips any html and only provides a text 
> message so all formatting has been lost.  I think the question is not really 
> an R-help question but if you resubmit the post you need to show the sentence 
> in question in another way.
>
> John Kane
> Kingston ON Canada
>
>
>> -Original Message-
>> From: mohamed.laj...@inserm.fr
>> Sent: Fri, 15 Mar 2013 11:26:45 +0100
>> To: r-help@r-project.org
>> Subject: [R] reviewer comment
>>
>> Could someone explain me this sentence reviewer below in blod underlined,
>>
>> Authors should try to be more detailed in the description of analyses:
>> some of the details reported in the "Principal components analysis"
>> paragraph (Results) should be moved here.
>> Because a highly_/*asymmetric distribution could affect Principal
>> Component Analysis results,  symmetry of distribution should be
>> tested. Authors should also indicate if outliers were observed and
>> consequently excluded because they could affect factors*/_
>>
>> Any help would be greatly appreciated!
>>
>> Regards
>> ML
>>
>> --
>> 
>> Mohamed Lajnef,IE INSERM U955 eq 15#
>> P?le de Psychiatrie #
>> H?pital CHENEVIER  #
>> 40, rue Mesly  #
>> 94010 CRETEIL Cedex FRANCE #
>> mohamed.laj...@inserm.fr   #
>> tel : 01 49 81 32 79#
>> Sec : 01 49 81 32 90   #
>> fax : 01 49 81 30 99   #
>> 
>>
>>
>>  [[alternative HTML version deleted]]
>>
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>> and provide commented, minimal, self-contained, reproducible code.
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[R] reviewer comment

2013-03-15 Thread Mohamed Lajnef
Could someone explain me this sentence reviewer below in blod underlined,

Authors should try to be more detailed in the description of analyses:
some of the details reported in the "Principal components analysis"
paragraph (Results) should be moved here.
Because a highly_/*asymmetric distribution could affect Principal
Component Analysis results,  symmetry of distribution should be
tested. Authors should also indicate if outliers were observed and
consequently excluded because they could affect factors*/_

Any help would be greatly appreciated!

Regards
ML

-- 
########
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P?le de Psychiatrie#
H?pital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
mohamed.laj...@inserm.fr   #
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Sec : 01 49 81 32 90   #
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Re: [R] p-values from lm()

2012-06-14 Thread Mohamed Lajnef
Hi David,


summary(res)$coefficients[,"Pr(>|z|)"] or  summary(res)$coefficients[,4]

M
Regrads

Le 14/06/12 12:44, David Studer a écrit :
> Hi everyone!
>
> Can anyone tell me, how to obtain p.values from a linear model?
>
> Example:
> mod1<-lm(dV~iV1+iV2)
>
> Now, I can get the coefficients with mod1$coef
> But how can I get p-values? ($p.values seems to work with cor.test() only)
>
> Thank you!
>
>   [[alternative HTML version deleted]]
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


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[R] Draw vertical line segments between pairs of points

2012-02-02 Thread Mohamed Lajnef
Dear all,

How to add /*vertical*/ lines above bar graph to display p-values  ( 
between pairs of points )?

Regards
ML

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[R] Multicollinearty in logistic regression models

2011-12-15 Thread Mohamed Lajnef
Dear All,

Is there a method to diagnostic multicollinearty in logistic regression 
models  like vif indicator in linear regression ( variance inflation 
Factor ...) ?

Thank you in advance
M

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Re: [R] R packages for pathway analysis?

2011-12-09 Thread Mohamed Lajnef
Hi Karena,

Ask your question in bionconductor mailing list  
(bioconduc...@r-project.org)

Regards
ML

Le 09/12/11 03:36, karena a écrit :
> What R packages do you guys use for 'pathway analysis'? By 'pathway
> analysis', I mean to check the enrichment of certain genes in KEGG | GO
> pathways.
>
> Thank you,
>
> Karena
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/R-packages-for-pathway-analysis-tp4175258p4175258.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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>


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Re: [R] Help to get fitted and residual value

2011-12-07 Thread Mohamed Lajnef
Hi arunkumar;

fit$residuals , fit$fitted  to extract respectively  residuals and 
fitted values

Regrads
Mohamed
Le 07/12/11 07:20, arunkumar a écrit :
> Data<- read.csv("C:/FE and RE.csv")
> Formula="Y~X2+X3+X4 + X5+X6
> fit=lm(formula=Formula,data=Data)
>
> My sample Data
>
> State Year Y X2 X3 X4 X5 X6
> S2 1960 27.8 397.5 42.2 50.7 78.3 65.8
> S1 1960 29.9 413.3 38.1 52 79.2 66.9
> S2 1961 29.8 439.2 40.3 54 79.2 67.8
> S1 1961 30.8 459.7 39.5 55.3 79.2 69.6
>
> Is there any way to fitted and residual for the new set of data. or should i
> create formula manually and pass the values individually
>
>
> S2 1962 31.2 492.9 37.3 54.7 77.4 68.7
> S1 1962 33.3 528.6 38.1 63.7 80.2 73.6
> S2 1963 35.6 560.3 39.3 69.8 80.4 76.3
>
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Help-to-get-fitted-and-residual-value-tp4167808p4167808.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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>


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H?pital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
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Re: [R] quick ANOVA question

2011-11-29 Thread Mohamed Lajnef
Hi,

Anova compare the means of more than two groups, in your case anova test 
H0 hypothisis

H0:  means(control)=mean(tread1)=mean(tread2)=mean(tread3), if there is 
at least one différence,
you can identify the mean difference between groups by using post hoc 
test to testing your B suggestion.

Regrads
ML

Le 29/11/11 11:49, syrvn a écrit :
> Hello,
>
>
> imagine the following experimental design:
>
> Group - Value
>
> control - 5
> control - 6
> control - 5
>
> treated1 - 8
> treated1 - 9
> treated1 - 9
>
> treated2 - 15
> treated2 - 16
> treated2 - 15
>
> treated3 - 25
> treated3 - 30
> treated3 - 28
>
>
> What I like to know is if I apply an ANOVA to this data and choose the
> control group as the reference group (using the relevel function)
> what groups exactly are compared? Are only all treated groups 1, 2, 3 tested
> against the control group or are all possible combinations tested?
>
> What is compared?
>
> A) Testing: control vs treated1; control vs treated2; control vs treated3;
>
> or
>
> B) Testing: control vs treated1; control vs treated2; control vs treated3;
>
>  treated1 vs treated2; treated1 vs treated3; treated2 vs
> treated3;
>
>
> Cheers
>
>
>
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/quick-ANOVA-question-tp4118724p4118724.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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>


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Re: [R] Combining histograms and distribution curve

2011-11-29 Thread Mohamed Lajnef
Hi,
Try lines function to get a distribution cruve:

X<-c(8, 9, 7, 6, 7, 8, 5, 4, 16, 12, 11, 10, 8, 7, 9 )
 > hist(X, prob=TRUE)
 > lines(density(X))

Regrads
ML

Le 29/11/11 11:04, gwanme...@aol.com a écrit :
> Dear Sir,
>
> I am reviewing a group of patients who had a surgical procedure. The
> histogram of the age frequency has a bimodal distribution. I am trying to
> emphasise this by adding a distribution curve on the histogram.
>
> Let's say the ages are 8, 9, 7, 6, 7, 8, 5, 4, 16, 12, 11, 10, 8, 7, 9  for
> example. How can I plot a histogram with a distribution curve on  it?
>
> Your help is highly appreciated.
>
> Ivo
>   [[alternative HTML version deleted]]
>
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>


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H?pital CHENEVIER  #
40, rue Mesly  #
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Re: [R] how to get AIC BIC information using funtion lm

2011-11-28 Thread Mohamed Lajnef
Hi,

Try the AIC and BIC function.  ?AIC for help

Regrads
ML

Le 28/11/11 08:28, arunkumar a écrit :
> Hi
> how to get  AIC BIC information using funtion lm ?
>
> Please help me
>
> Data<- read.csv("C:/FE and RE.csv")
> Formula="Y~X2+X3+X4
> lm(formula=Formula,data=Data)
>
> My sample Data
>
> State Year Y X2 X3 X4 X5 X6
> S2 1960 27.8 397.5 42.2 50.7 78.3 65.8
> S1 1960 29.9 413.3 38.1 52 79.2 66.9
> S2 1961 29.8 439.2 40.3 54 79.2 67.8
> S1 1961 30.8 459.7 39.5 55.3 79.2 69.6
> S2 1962 31.2 492.9 37.3 54.7 77.4 68.7
> S1 1962 33.3 528.6 38.1 63.7 80.2 73.6
> S2 1963 35.6 560.3 39.3 69.8 80.4 76.3
> S1 1963 36.4 624.6 37.8 65.9 83.9 77.2
> S2 1964 36.7 666.4 38.4 64.5 85.5 78.1
>
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/how-to-get-AIC-BIC-information-using-funtion-lm-tp4114401p4114401.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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>


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H?pital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
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Re: [R] changing date format in a dataframe

2011-11-16 Thread Mohamed Lajnef
Hi,

If the dates are the same format and the same length you can try this :

dates<-c("1/1/2009","2/1/2009","3/1/2009")

dates_new<-as.Date(paste(substr(dates,1,3),substr(dates,7,8),sep="/"),format="%d/%m/%y"),
 
you can change the format to "%m/%d/%y"

Regards
M


Le 16/11/11 15:12, arunkumar a écrit :
> Hi
>
> I have a data frame and i need to change the date format in it.
> my dataframe
>
> X Date
> 1 1/1/2009
> 2 2/1/2009
> 3 3/1/2009
>
> I need to change it to 2009-01-01
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/changing-date-format-in-a-dataframe-tp4076411p4076411.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>


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H?pital CHENEVIER  #
40, rue Mesly  #
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Re: [R] determine frequencies in a matrix by row

2011-06-08 Thread Mohamed Lajnef
Hi,

Try apply function:
apply(matrix,1,table)

Regards
M



Le 08/06/11 08:23, the_big_kowalski a écrit :
> Hi,
>
> I have a matrix
>   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> [1,]131131123 2
> [2,]131211122 1
> [3,]122321112 1
> [4,]321113121 2
> [5,]112211311 2
> and want to determine how many times 1s, 2s, and 3s occur per each row.
> I tried using 'table', but end up with frequencies for the whole matrix.
> Thanks in advance for your help
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/determine-frequencies-in-a-matrix-by-row-tp3581733p3581733.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
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>


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Re: [R] write gene_id in a bed file

2011-06-07 Thread Mohamed Lajnef
Hi Nanami,

you do not use the same file in export, href or new_CTTS ?

Regards
M



Le 07/06/11 14:42, ads pit a écrit :
> Hi all,
> I have build the following data frame
>   head(href)
> chr tx_start   tx_end g_id strand cds_start  cds_end exon_count
> 1 chr1  8384389  8404227 NM_001080397  +   8384389  8404073  8
> 2 chr1 16767166 16786584 NM_001145277  +  16767256 16785491  7
> 3 chr1 16767166 16786584 NM_001145278  +  16767256 16785385  8
> 4 chr1 16767166 16786584NM_018090  +  16767256 16785385  8
> 5 chr1 48998526 50489626NM_032785  -  48999844 50489468 14
> 6 chr1 33546713 33585995NM_052998  +  33547850 33585783 12
>
> Now when I'm trying to export it into a bed file I did:
> output_href<- write.table(new_CTTS, file="new_href.bed", quote=FALSE, sep =
> "\t", na="NA", row.names=FALSE, col.names=TRUE)
>
> The thing is when I look at the file I only have:
>   head(test)
>chr1 X564620 X564649 X564644 X565645  X94 X. X10
> 1 chr1  565369  565404  565371  566372  217  +   8
> 2 chr1  565463  565541  565480  566481 1214  +  15
> 3 chr1  565653  565697  565662  53 1031  +  28
> 4 chr1  565861  565922  565883  566884  316  +  12
> 5 chr1  566537  566573  566564  567565  119  +  11
> 6 chr1  567535  567579  567562  568563 2085  +  39
>
> but what I want is to include the corresponding gene ID as well in a last
> column. How can I do that?
>
> Best,
> Nanami
>
>   [[alternative HTML version deleted]]
>
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40, rue Mesly  #
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[R] Several Regression by combinations variables

2011-05-30 Thread Mohamed Lajnef
Dear All,
  I have 11 variables, and i would like generate  combinations of those 
variables by fours variables i,e 330 possibility (choose(11,4)).?
After that, make a regression analysis with this 330 possibility ?

is there a program ( or package) to do that ?

Any help would be appreciated

Regards
M

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[R] Fwd: transpose ?

2011-05-25 Thread Mohamed Lajnef
Dear All,
Sorry for the previous mail,suppose this data.frame
D


V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22
C  C  C   C   T   T   G   G   A   A   C   C   G   G   C   C
G  G  T   T   A   A   A   A   T   A   T   T   C   C   G   G
C  C  C   C   T   T   G   G   A   A   C   C   G   G   C   C





I would translate D as follow ( just for the first line)

C C T G A C G C
C C T G A C G C
(V8 under V7) (V9 under V10) ...

Any help would be appreciated

Regards
M

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[R] transpose ?

2011-05-25 Thread Mohamed Lajnef
Dear All,
Suppose this data.frame
D

V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22
C  C  C   C   T   T   G   G   A   A   C   C   G   G   C   C
G  G  T   T   A   A   A   A   T   A   T   T   C   C   G   G
C  C  C   C   T   T   G   G   A   A   C   C   G   G   C   C


I would translate D as follow ( just for the first line)
C (V7)
C (V9)
T   G   A   C   G   C
C (V8)
C (V10)
T   G   A   C   G   C


Any help would be appreciated

Regards
M

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Re: [R] quantile function -> I need only the quantile value itself

2011-03-28 Thread Mohamed Lajnef
HI Laszlo,

q<-quantile(small_df,probs=0.95)
q[[1]]
[1] 12.85

Regrads





Le 28/03/11 11:37, Bodnar Laszlo EB_HU a écrit :
> Hi,
>
> I am using the quantile function currently and I have just bumped into a 
> little problem.
>
> I have a very small data frame something like this:
>
> small_df<- 
> c(7,3,4,7,1,10,12,1,12,4,4,8,6,11,9,10,4,13,3,9,6,5,2,10,7,14,2,7,10,10,7,8,2,11,3,10,11,3,11,14,12,7,6,11)
> small_df
>
> Now in the next step I would like to calculate the quantile value of the 
> previous data frame at the probability of 95% the following way:
> quantile(small_df,probs=0.95)
>
> The result:
>> 95%
>> 12.85
> Now I need to get only the number 12.85. I do not want to present the 
> probability "95%" here. Is there any possible way to "tell" the program to do 
> so?
>
> Regards,
> Laszlo
> 
> Ez az e-mail és az összes hozzá tartozó csatolt melléklet titkos 
> és/vagy jogilag, szakmailag vagy más módon védett információt 
> tartalmazhat. Amennyiben nem Ã--n a levél címzettje akkor a levél 
> tartalmának közlése, reprodukálása, másolása, vagy egyéb más úton 
> történÅ' terjesztése, felhasználása szigorúan tilos. Amennyiben 
> tévedésbÅ'l kapta meg ezt az üzenetet kérjük azonnal értesítse az 
> üzenet küldÅ'jét. Az Erste Bank Hungary Zrt. (EBH) nem vállal 
> felelÅ'sséget az információ teljes és pontos - címzett(ek)hez történÅ' 
> - eljuttatásáért, valamint semmilyen késésért, kapcsolat 
> megszakadásból eredÅ' hibáért, vagy az információ felhasználásából 
> vagy annak megbízhatatlanságából eredÅ' kárért.
>
> Az üzenetek EBH-n kívüli küldÅ'je vagy címzettje tudomásul veszi és 
> hozzájárul, hogy az üzenetekhez más banki alkalmazott is hozzáférhet az 
> EBH folytonos munkamenetének biztosítása érdekében.
>
>
> This e-mail and any attached files are confidential and/...{{dropped:19}}
>
>
>
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Re: [R] adding linear regression data to plot

2011-03-16 Thread Mohamed Lajnef
Hi Jan,

If you want to plot the fitted values  of lm function, you can use 
abline function
? abline ( te get help)

Regrads
M

Le 15/03/11 23:26, derek a écrit :
> Hello R,
>
> I would like to print regression data in graph. I mean the output from:
> k=lm(formula,data)
> summary(k)
>
> Or somehow extract and print only coefficients and R-squared.
>
>
>
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/adding-linear-regression-data-to-plot-tp3357946p3357946.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


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Hôpital CHENEVIER  #
40, rue Mesly  #
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Re: [R] create data set from selection of rows

2011-03-15 Thread Mohamed Lajnef
Hi,

Le 15/03/11 16:09, e-letter a écrit :
> Readers,
>
> For a data set:
>
> text1,23,text2,45
> text1,23,text3,78
> text1,23,text3,56
> text1,23,text2,45
>
> The following command was entered:
>
> datasubset<-data.frame(dataset[,3]=="text3")
datasubset<-subset(dataset,dataset[,3]=="text3")
> The result of
>
> datasubset
>
> is
>
> TRUE
> TRUE
>
> The required result is
>
> text1,23,text3,78
> text1,23,text3,56
>
> What is the correct command to use please?
>
> Thanks in advance.

> Regards
M

> __
> 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.
>


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Hôpital CHENEVIER  #
40, rue Mesly  #
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Re: [R] Plot confidence intervals

2011-02-23 Thread Mohamed Lajnef
Hi Iker,

matlines  function to plot 956 % confidence intervals after using 
predict function.
barplot function for errors bars.

hope this help
M
Regards


Le 23/02/11 16:00, Iker Vaquero Alba a écrit :
> Hello, list:
>
> I'm not sure about where to send this question. I have several 
> repeatability calculations, together with their 95% confidence intervals, and 
> I would like to plot them, in a way similar to error bars. I was wondering if 
> there is any specific function to do this, or any method I can apply in R, as 
> I haven't found anything like that in the R book.
>
> Thank you very much
>
> Iker Vaquero-Alba
> Centre for Ecology and Conservation
> Daphne du Maurier Building
> University of Exeter, Cornwall Campus
> Treliever Road
> TR10 9EZ Penryn
> U.K.
>
>
>
>
>   [[alternative HTML version deleted]]
>
>
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


-- 

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Hôpital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
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Re: [R] nrow()

2011-02-22 Thread Mohamed Lajnef
Hi Sandra,

What about ?is.na function ?

Hope this help

Regards,
ML



Le 22/02/11 16:11, Sandra Stankowski a écrit :
> NROW(data[jan,16] != NaN) 


-- 
####
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Re: [R] barplot with errorbars

2011-02-17 Thread Mohamed Lajnef
Hi Maria,

Look at  barplot function

? barplot  for help

Regards
M
Le 17/02/11 17:00, Lathouri, Maria a écrit :
> Dear all
>
> I have six variables of the average metal concentrations
>
> Var1 4.77
> Var2 23.5
> Var3 5.2
> Var4 12.3
> Var5 42.1
> Var6 121.2
>
> I want to plot them as a barplot with error bars. Could you help me?
>
> Cheers
> Maria
>
>   [[alternative HTML version deleted]]
>
> __
> 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.
>


-- 

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Hôpital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
mohamed.laj...@inserm.fr   #
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Re: [R] darcs patch: Apply on data frame

2011-02-17 Thread Mohamed Lajnef
Hi chen,
avoid naming the function with specific expressions (:,;.) and try 
this code as follow

Scripranksum<- function(a){
   g1<- c(1, 2, 3)
   g2<- c(4, 5, 6, 7, 8, 9, 10, 11, 12)
   c<- wilcox.test(a[g1], a[g2])
   return (c$p.value)
}

Best
Mohamed

Le 17/02/11 08:15, Mr rong chen a écrit :
> Scrip:ranksum<- function(a){
>g1<- c(1, 2, 3)
>g2<- c(4, 5, 6, 7, 8, 9, 10, 11, 12)
>c<- wilcox.test(a[g1], a[g2])
>return c$p.value
> }


-- 
########
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Re: [R] removing lower and upper quantiles from an arry

2011-02-17 Thread Mohamed Lajnef
Thierry,

you forgot !

i[! quantile(i,.25) >= i & i <= quantile(i,.75)]

Best
M
Le 17/02/11 11:18, ONKELINX, Thierry a écrit :
> You need two logical test and then combine them with&  (AND) or | (OR)
>
> i[quantile(i,.25)>= i&  i<= quantile(i,.75)]
>
> Best regards,
>
> Thierry
>
> 
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek
> team Biometrie&  Kwaliteitszorg
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> Research Institute for Nature and Forest
> team Biometrics&  Quality Assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> tel. + 32 54/436 185
> thierry.onkel...@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more than 
> asking him to perform a post-mortem examination: he may be able to say what 
> the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not 
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
>
>> -Oorspronkelijk bericht-
>> Van: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] Namens Maas James Dr (MED)
>> Verzonden: donderdag 17 februari 2011 11:09
>> Aan: r-help@r-project.org
>> Onderwerp: [R] removing lower and upper quantiles from an arry
>>
>> I'm trying to work out the simplest way to remove the upper
>> and lower quantiles, in this case upper and lower 25% from an
>> array.  I can do it in two steps but when I try it in one, it
>> fails.  Is there something simple missing from my syntax or
>> are there other simple elegant way to accomplish this?
>>
>> Thanks
>>
>> J
>>
>>> i<-1:20
>>> i2<- i[i>> i3<- i[i>quantile(i,.25)]
>>> i4<- i[quantile(i,.25)<  i>  quantile(i,.75)]
>> Error: unexpected '>' in "i4<- i[quantile(i,.25)<  i>"
>>
>> ===
>> Dr. Jim Maas
>> University of East Anglia
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 

Mohamed Lajnef,IE INSERM U955 eq 15#
Pôle de Psychiatrie#
Hôpital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
mohamed.laj...@inserm.fr   #
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Sec : 01 49 81 32 90   #
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Re: [R] removing lower and upper quantiles from an arry

2011-02-17 Thread Mohamed Lajnef
Thierry,

you forgot !

i[! quantile(i,.25) >= i & i <= quantile(i,.75)]

Best
M
Le 17/02/11 11:18, ONKELINX, Thierry a écrit :
> You need two logical test and then combine them with&  (AND) or | (OR)
>
> i[quantile(i,.25)>= i&  i<= quantile(i,.75)]
>
> Best regards,
>
> Thierry
>
> 
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek
> team Biometrie&  Kwaliteitszorg
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> Research Institute for Nature and Forest
> team Biometrics&  Quality Assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> tel. + 32 54/436 185
> thierry.onkel...@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more than 
> asking him to perform a post-mortem examination: he may be able to say what 
> the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not 
> ensure that a reasonable answer can be extracted from a given body of data.
> ~ John Tukey
>
>
>> -Oorspronkelijk bericht-
>> Van: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] Namens Maas James Dr (MED)
>> Verzonden: donderdag 17 februari 2011 11:09
>> Aan: r-help@r-project.org
>> Onderwerp: [R] removing lower and upper quantiles from an arry
>>
>> I'm trying to work out the simplest way to remove the upper
>> and lower quantiles, in this case upper and lower 25% from an
>> array.  I can do it in two steps but when I try it in one, it
>> fails.  Is there something simple missing from my syntax or
>> are there other simple elegant way to accomplish this?
>>
>> Thanks
>>
>> J
>>
>>> i<-1:20
>>> i2<- i[i>> i3<- i[i>quantile(i,.25)]
>>> i4<- i[quantile(i,.25)<  i>  quantile(i,.75)]
>> Error: unexpected '>' in "i4<- i[quantile(i,.25)<  i>"
>>
>> ===
>> Dr. Jim Maas
>> University of East Anglia
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>


-- 

Mohamed Lajnef,IE INSERM U955 eq 15#
Pôle de Psychiatrie#
Hôpital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
mohamed.laj...@inserm.fr   #
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Sec : 01 49 81 32 90   #
fax : 01 49 81 30 99   #




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Re: [R] sampling

2011-02-17 Thread Mohamed Lajnef
Hi ,

what about split function ?
?split  divided x into 2 data.frame

a<-split(x,1:2)

a[[1]] first data frame
a[[2]] second data frame

regrads
M




Le 17/02/11 05:35, yf a écrit :
> I want to sample from the ID. For each ID, i want to have 2 set of data. I
> try the sample() function but it didn't work.
>
>> x<-data.frame(id=c(1,1,1,2,2,2,2,3,3,3,4,4), v1=c(1:12), V2=c(12:23))
>> x
> id v1 V2
> 1   1  1 12
> 2   1  2 13
> 3   1  3 14
> 4   2  4 15
> 5   2  5 16
> 6   2  6 17
> 7   2  7 18
> 8   3  8 19
> 9   3  9 20
> 10  3 10 21
> 11  4 11 22
> 12  4 12 23


-- 

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Hôpital CHENEVIER  #
40, rue Mesly  #
94010 CRETEIL Cedex FRANCE #
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Re: [R] Filter data

2010-12-02 Thread mohamed . lajnef

Hi Diogenas,

Try this,
data<-subset(poli, poli$year>1994)

Best,
M

Diogenas  a écrit :



Hello,
I understand that question is probably stupid, but ...

I have data (polity IV index)
"country","year","democ","autoc","polity","polity2"
"1","Afghanistan ",1800,1,7,-6,-6
"2","Afghanistan ",1801,1,7,-6,-6
"3","Afghanistan ",1802,1,7,-6,-6

I need to create new data sets which includes only cases after year 1995.
I've tried this command:
poli2<-poli[poli$year > 1994,] ,
 however it generated this:
Error in poli[poli$year > 1994, ] : incorrect number of dimensions.
Any ideas how i can overcome this?

Thank in advance for your help,





-
Diogenas
--
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[R] Path analysis with logistic regression

2010-10-27 Thread Mohamed Lajnef
Dear all,

I would like to make a path analysis after retrieving relevant variables 
(continuous or binary) after logistic regression.

So I try to model the dependent variable (0/1)  based on the significant 
predictors using path analysis

Could you tell me if there is a suitable package in R to do this?

Any help would be appreciated

Regards

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Re: [R] which function

2010-06-02 Thread mohamed . lajnef

Hi Arnaud
Try this

l[((which(100420==l))-1):which(100420==l)]

Regards
Mo


arnaud Gaboury  a écrit :


Dear group,

Here is a list of elements :


l <-
c("100415", "100416", "100419", "100420", "100421", "100422",
"100423", "100426", "100427", "100428", "100429", "100430", "100503",
"100504", "100505", "100506", "100507", "100510", "100511", "100512",
"100513", "100514", "100517", "100518", "100519", "100520", "100521",
"100524", "100525", "100526", "100527", "100528", "100531")



l[3:4]

[1] "100419" "100420"  #result is fine



l[(which(100420==l))-1:which(100420==l)]

[1] "100419" "100416" "100415"  #can't understand why I do not have same
result than the above line


Thank You for help

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Re: [R] R eat my data

2010-05-25 Thread Mohamed Lajnef

cheks the comments sent by  David!

M



Changbin Du a écrit :

> length(count.fields("/home/cdu/operon/id_name_gh5.txt"))
[1] 1932

It is 1932 lines when count in R



On Tue, May 25, 2010 at 8:52 AM, Mohamed Lajnef 
mailto:mohamed.laj...@inserm.fr>> wrote:


Hi Changbin,

Try to use this code in R to count the lines of  your file without
open it

 length(count.fields("id_name_gh5.txt"))
Regards Mohamed




Changbin Du a écrit :

HI, Dear R community,

My original file has 1932 lines, but when I read into R, it
changed to 1068
lines, how comes?


c...@nuuk:~/operon$ wc -l id_name_gh5.txt
1932 id_name_gh5.txt


 


gene_name<-read.table("/home/cdu/operon/id_name_gh5.txt",
sep="\t",
   


skip=0, header=F, fill=T)
 


dim(gene_name)
   


    [1] 10683


 




-- 



Mohamed Lajnef,IE INSERM U955 eq 15
Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr <mailto:mohamed.laj...@inserm.fr>
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99




--
Sincerely,
Changbin
--

Changbin Du
DOE Joint Genome Institute
Bldg 400 Rm 457
2800 Mitchell Dr
Walnut Creet, CA 94598
Phone: 925-927-2856





--


Mohamed Lajnef,IE 
INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99 


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Re: [R] R eat my data

2010-05-25 Thread Mohamed Lajnef

Hi Changbin,

Try to use this code in R to count the lines of  your file without open it

 length(count.fields("id_name_gh5.txt")) 

Regards 
Mohamed





Changbin Du a écrit :

HI, Dear R community,

My original file has 1932 lines, but when I read into R, it changed to 1068
lines, how comes?


c...@nuuk:~/operon$ wc -l id_name_gh5.txt
1932 id_name_gh5.txt


  

gene_name<-read.table("/home/cdu/operon/id_name_gh5.txt", sep="\t",


skip=0, header=F, fill=T)
  

dim(gene_name)


[1] 10683


  



--


Mohamed Lajnef,IE 
INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99 


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Re: [R] sort a data.frame

2010-05-20 Thread Mohamed Lajnef

The solution given by Jim is more correct

dd[order(as.numeric(substr(dd$b,4,5))),]

regards
M
jim holtman a écrit :

dd


  b x y z
1  chr2 A 8 1
2  chr1 D 3 1
3 chr15 A 9 1
4 chr13 C 9 2
  

# add column with just numbers
dd$sort <- as.integer(gsub("\\D+", "", dd$b))
dd[order(dd$sort),]  # notice it is a numeric, not character order


  b x y z sort
2  chr1 D 3 11
1  chr2 A 8 12
4 chr13 C 9 2   13
3 chr15 A 9 1   15
  



On Thu, May 20, 2010 at 8:28 AM, Yuan Jian  wrote:
  

Hello,

I have a dataframe:
dd <- data.frame(b = c("chr2", "chr1", "chr15", "chr13"),
  x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
   z = c(1, 1, 1, 2))



dd
  

  b x y z
1  chr2 A 8 1
2  chr1 D 3 1
3 chr15 A 9 1
4 chr13 C 9 2

Now I want to sort them according column "b", but only its number is considered:
  b x y z
1  chr1 D 3 1
2 chr13 C 9 2
3 chr15 A 9 1
4  chr2 A 8 1

thanks
jian






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[R] t.test per line

2010-05-05 Thread Mohamed Lajnef

Dear All,

I apologize for this trivial question, I can not find the solution
I try to use t.test function per line  in the data.frame, But i 
dont'understand the error message


my program is as follows

group1<-gl(2,20)
fun<-function(x){
m<-data.frame(group1,x)
  #anova(aov(x ~ group1, m))
t.est(x ~ group1,m)
}

ttest<- apply(data, 1, fun)
Error in  t.test.default(x = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,  :
 data are essentially constant


Any help would be appreciated

ML
Best



--


Mohamed Lajnef,IE 
INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99 


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Re: [R] timing a function

2010-05-04 Thread mohamed . lajnef


Hi,

? proc.time() for more help

regards
Ml

pdb  a écrit :



Hi,
I want to time how long a function takes to execute. Any clues on what to
search for to achieve this?

Thanks in advance.
--
View this message in context:   
http://r.789695.n4.nabble.com/timing-a-function-tp2126319p2126319.html

Sent from the R help mailing list archive at Nabble.com.

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Re: [R] apply fun to df returning a matrix

2010-04-30 Thread Mohamed Lajnef

Hi Soeren

Apply or aggregate functions

best regards
M
soeren.vo...@eawag.ch a écrit :
Hello, a data.frame, df, holds the numerics, x, y, and z. A function, 
fun, should return some arbitrary statistics about the arguments, e.g. 
the sum or anything else. What I want to do is to apply this function 
to every pair of variables in df, and the return should be a matrix as 
found with cov. How can I achieve that? Thanks, Sören


df <- data.frame(x=1:10, y=11:20, z=21:30);
fun <- function(x){
  return(sum(x));
}
# and now???

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Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
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Re: [R] Find solution for an error in the condition of if

2010-04-30 Thread Mohamed Lajnef

Dear Nuel,

could you send the error messages in English please. Adds & between the 
two conditions and try again


if ( (data[pa,k]== df[,j]) && (data[ch,k]==i))

Best
M



anderson nuel a écrit :

Dear r-help,

Could you help me to find a solution for this error:

Il y a eu 50 avis ou plus (utilisez warnings() pour voir les 50 premiers)

  

warnings()


Messages d'avis :
1: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... :
  la condition a une longueur > 1 et seul le premier élément est utilisé
2: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... :
  la condition a une longueur > 1 et seul le premier élément est utilisé


ch=3
pa=c(1,2)
r=2
t=4
nb=15
ni=array(0,c( r,t))

for ( i  in 1:r){

for (j in 1:t ){

for (k in 1:nb){

if ( (data[pa,k]== df[,j]) & (data[ch,k]==i)){

ni[i,j]=ni[i,j]+1
}
}
}
}

Best  Regards

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INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
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tel : 01 49 81 31 31 (poste 18470)
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Re: [R] median Q?

2010-04-21 Thread Mohamed Lajnef

Hi,

Try this median(v2[v1==1])

Mohamed
Regards

Vlatka Matkovic Puljic a écrit :

I have additional Q:

v1 is gender (M=1 and F=2)
v2 is age

I want R to calculate median only for M (1),
but my comand is not good :)

while(v1=1){median(v2,na.rm=TRUE)}
Error: unexpected '=' in "while(Q2="

2010/4/21 Vlatka Matkovic Puljic 

  

Thank you!

2010/4/21 Joshua Wiley 







Hello,


summary() removes NAs by default.  You can get the same results using

median(year, na.rm=TRUE)

see ?median

HTH,

Josh

On Wed, Apr 21, 2010 at 6:40 AM, Vlatka Matkovic Puljic
 wrote:
  

Hi,

I have dataset n1 and v1 (years).
when i ask
median(year)
[1] NA

but if i put summary of dataset n1:
summary(n1)
R produces median (together with min/max/mean)

why it is so?


--
**
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095/8618 171

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--
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http://www.joshuawiley.com/

  


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095/8618 171






  



--


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INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
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tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
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Re: [R] sum specific rows in a data frame

2010-04-14 Thread Mohamed Lajnef

Hi Arnaud,

Try  aggregate function

regards
M

arnaud Gaboury a écrit :

I have a data frame called "pose":


  DESCRIPTION QUANITY CLOSING.PRICE
1   WHEAT May/101467.75
2   WHEAT May/102467.75
3   WHEAT May/101467.75
4   WHEAT May/101467.75
5 COTTON NO.2 May/101 78.13
6 COTTON NO.2 May/103 78.13
7 COTTON NO.2 May/101 78.13

I would like to sum the quantity for each category (i.e WHEAT and
COTTON),but I have no idea how to write it in a simple manner. The number or
rows will change every day.
TY for any help.

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Re: [R] compare multiple values with vector and return vector

2010-04-02 Thread mohamed . lajnef

Hi Joris

Try Is.element function: is.element (x,y)

Regards
mohamed

Joris Meys  a écrit :


Dear all,

I have a vector, and for each element I want to check whether it is equal to
any element from another vector. I want a vector of logical values with the
length of the first one as return. In R this would be :


x <- 1:10
sapply(x,function(y){any(y==c("2","3","4"))})

[1] FALSE  TRUE  TRUE  TRUE FALSE FALSE FALSE FALSE FALSE FALSE

It works pretty smooth, but I have the feeling there's a less complicated
way of doing it. My code should be readable by programmers who are not
really familiar with R, but I hate to use for-loops as I have pretty huge
datasets. Anybody an idea?
thank you in advance.

Cheers
Joris

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[R] Empirical Pvalue

2010-02-17 Thread Mohamed Lajnef

Dear R users,

Can someone explain me the interpretation of  empirical p-value ( 
max(t))  used to test the genetic association with PLINK?


Any help will be appreciated
Regards

M


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INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99 


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Re: [R] Is the aggregate function the best way to do this?

2010-02-17 Thread Mohamed Lajnef

Hi Bert,

apply(data,1,max) gives the max value for each row in the data

M

Bert Jacobs a écrit :

Hi Mohamed,
Thx for the answer.
And what should I do if my fun function is not 'sum' but 'max'?
Bert
-Original Message-
From: Mohamed Lajnef [mailto:mohamed.laj...@inserm.fr] 
Sent: 17 February 2010 10:02

To: Bert Jacobs
Cc: r-help@r-project.org
Subject: Re: [R] Is the aggregate function the best way to do this?

Hi Bret,

Try to use rowSums function

Regrads
M

Bert Jacobs a écrit :
  

Hi,

 


I'm having a dataframe 'Subset1' with a number of factor variables and 160
numerical variables

Now I want to make sums for all rows that have the same values for the
different factor variables, except for the factor variables:


VAR1,VAR2,VAR3
  

who may have the same values.

With the formula given below this works great, but in a situation with


15000
  
rows and 13 factor variables the calculation takes more than 2 minutes. 


So my question is: Does anyone knows if there exists a faster alternative?



  
 


Subset1.AGG <-
as.data.frame(aggregate(Subset1[,(ncol(Subset1)-159):ncol(Subset1)],
list(VAR1 = Subset1$VAR1,VAR2=Subset1$VAR2,VAR3 = Subset1$VAR3), FUN=sum)


)
  
 


Thank you very much for helping me out,

Bert


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INSERM U955 eq 15

Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
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Sec : 01 49 81 32 90
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Re: [R] Is the aggregate function the best way to do this?

2010-02-17 Thread Mohamed Lajnef

Hi Bret,

Try to use rowSums function

Regrads
M

Bert Jacobs a écrit :

Hi,

 


I'm having a dataframe 'Subset1' with a number of factor variables and 160
numerical variables

Now I want to make sums for all rows that have the same values for the
different factor variables, except for the factor variables: VAR1,VAR2,VAR3
who may have the same values.

With the formula given below this works great, but in a situation with 15000
rows and 13 factor variables the calculation takes more than 2 minutes. 

So my question is: Does anyone knows if there exists a faster alternative? 

 


Subset1.AGG <-
as.data.frame(aggregate(Subset1[,(ncol(Subset1)-159):ncol(Subset1)],
list(VAR1 = Subset1$VAR1,VAR2=Subset1$VAR2,VAR3 = Subset1$VAR3), FUN=sum) )

 


Thank you very much for helping me out,

Bert


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Re: [R] Error of Stepwise Regression with number of rows in use has changed: remove missing values?

2010-02-16 Thread Mohamed Lajnef


Hi Kum,

If you look at the code step function  (  by typing  step in the R 
console), the condition (if (length(fit$residuals) != n) ) is not 
fulfilled, this explains the error!

i hope this can help

Regards
M


Kum-Hoe Hwang a écrit :

Howdy, R Grues

I have enjoyed R, but I cannot solve one problem easily. Please help my problem.
When I tried the R script, I got the following Error. This error
results from input data file exported through a Excel spreadsheet
software.

 Error in step(lm(pop.rate ~ as.numeric(year) + as.factor(policy) +
as.numeric(nation.grant) +  :
  number of rows in use has changed: remove missing values?

Could you direct me to solve the Error?
Thanks in advance,


  

### outputs from R console ###
pop <- step(


+ lm(pop.rate ~ as.numeric(year) + as.factor(policy) +
as.numeric(nation.grant)
++ as.numeric(do.grant) + as.numeric(city.grant) +
as.numeric(DMZ.dist) + as.numeric(Seoul.dist), data=borderI.data,
na.action = na.omit)
+ )
Start:  AIC=494.27
pop.rate ~ as.numeric(year) + as.factor(policy) + as.numeric(nation.grant) +
as.numeric(do.grant) + as.numeric(city.grant) + as.numeric(DMZ.dist) +
as.numeric(Seoul.dist)
   Df Sum of SqRSSAIC
- as.numeric(do.grant)  1  0.71 6622.9 492.28
- as.factor(policy) 1  1.21 6623.4 492.29
- as.numeric(DMZ.dist)  1  1.91 6624.1 492.30
- as.numeric(city.grant)1  5.07 6627.3 492.36
- as.numeric(nation.grant)  1 11.51 6633.7 492.47
- as.numeric(year)  1 29.58 6651.8 492.80
  6622.2 494.27
- as.numeric(Seoul.dist)1673.22 7295.4 503.79
Step:  AIC=492.28
pop.rate ~ as.numeric(year) + as.factor(policy) + as.numeric(nation.grant) +
as.numeric(city.grant) + as.numeric(DMZ.dist) + as.numeric(Seoul.dist)
   Df Sum of SqRSSAIC
- as.factor(policy) 1  1.99 6624.9 490.32
- as.numeric(DMZ.dist)  1  2.09 6625.0 490.32
- as.numeric(city.grant)1  7.18 6630.1 490.41
- as.numeric(nation.grant)  1 20.08 6643.0 490.64
- as.numeric(year)  1 28.89 6651.8 490.80
  6622.9 492.28
- as.numeric(Seoul.dist)1697.46 7320.4 502.20
Step:  AIC=490.32
pop.rate ~ as.numeric(year) + as.numeric(nation.grant) +
as.numeric(city.grant) +
as.numeric(DMZ.dist) + as.numeric(Seoul.dist)
   Df Sum of SqRSSAIC
- as.numeric(DMZ.dist)  1  2.08 6627.0 488.35
- as.numeric(city.grant)1 10.65 6635.6 488.51
- as.numeric(nation.grant)  1 31.30 6656.2 488.88
- as.numeric(year)  1 31.44 6656.4 488.88
  6624.9 490.32
- as.numeric(Seoul.dist)1732.88 7357.8 500.80
Step:  AIC=488.35
pop.rate ~ as.numeric(year) + as.numeric(nation.grant) +
as.numeric(city.grant) +
as.numeric(Seoul.dist)
   Df Sum of SqRSSAIC
- as.numeric(city.grant)1  9.86 6636.9 486.53
- as.numeric(year)  1 31.42 6658.4 486.92
- as.numeric(nation.grant)  1 33.33 6660.3 486.95
  6627.0 488.35
- as.numeric(Seoul.dist)1754.40 7381.4 499.18

Error in step(lm(pop.rate ~ as.numeric(year) + as.factor(policy) +
as.numeric(nation.grant) +  :
---
  number of rows in use has changed: remove missing values?
--




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Email : phdhw...@gmail.com

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Re: [R] binary

2010-01-25 Thread Mohamed Lajnef

Hi val,


Val a écrit :

Hi all

Assume I have a data set xx;

Group: 1=group1  , 2=group2

IQ:  1= High, 0 =low

fit <- glm(IQ ~group, data = xx, family = binomial())

summary(fit)

Results

   Estimate Std. Error z value Pr(>|z|)

(Intercept) -2.554560.210 -12.273  < 5e-16 ***

 group  0.36180  0.076   3.952 5.24e-05 ***

the odd ratio = exp(0.36180 )= 1.435912

My question is that the log-odd  estimate 0.3618  is it for group1 or group2?
  
normally 1vs2, glm takes 2 as reference, in the group1 the IQ increase 
by 0.3618compared to group 2



What does the odd ratio 1.43359 is interpreted?
  

in the group1 the IQ score  increase by 1.43359  compared to group 2


Thanks in advance
  


Regards
ML

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Re: [R] Error in lm() function

2009-11-11 Thread Mohamed Lajnef

Hi Bogaso,

Try this
vecnames<-names(test[,2:11])
fmla <- as.formula(paste("test[,1] ~ ", paste(vecnames, collapse= "+")))
res<-lm(fmla)

Regards
M


Bogaso a écrit :

Hi all,

I wanted to have a seasonality study like whether a particular month has
significant effect as compared to others. Here is my data :

0.10499 0   0   1   0   0   0   0   0   0   
0   0
0.00259 0   0   0   1   0   0   0   0   0   
0   0
-0.060150   0   0   0   1   0   0   0   
0   0   0
0.10721 0   0   0   0   0   1   0   0   0   
0   0
0.03597 0   0   0   0   0   0   1   0   0   
0   0
0.10584 0   0   0   0   0   0   0   1   0   
0   0
0.02063 0   0   0   0   0   0   0   0   1   
0   0
-0.035090   0   0   0   0   0   0   0   
0   1   0
-0.034850   0   0   0   0   0   0   0   
0   0   1
0.01632 0   0   0   0   0   0   0   0   0   
0   0
0.06844 1   0   0   0   0   0   0   0   0   
0   0
-0.017660   1   0   0   0   0   0   0   
0   0   0
0.00989 0   0   1   0   0   0   0   0   0   
0   0
0.11673 0   0   0   1   0   0   0   0   0   
0   0
0.01789 0   0   0   0   1   0   0   0   0   
0   0
-0.003230   0   0   0   0   1   0   0   
0   0   0
0.06811 0   0   0   0   0   0   1   0   0   
0   0
-0.012920   0   0   0   0   0   0   1   
0   0   0
-0.122440   0   0   0   0   0   0   0   
1   0   0
-0.066450   0   0   0   0   0   0   0   
0   1   0
-0.033550   0   0   0   0   0   0   0   
0   0   1
0.02308 0   0   0   0   0   0   0   0   0   
0   0
-0.117111   0   0   0   0   0   0   0   
0   0   0
0.06116 0   1   0   0   0   0   0   0   0   
0   0
0.02832 0   0   1   0   0   0   0   0   0   
0   0
0.01441 0   0   0   1   0   0   0   0   0   
0   0
-0.044120   0   0   0   1   0   0   0   
0   0   0
0.05558 0   0   0   0   0   1   0   0   0   
0   0
0.08363 0   0   0   0   0   0   1   0   0   
0   0
-0.010630   0   0   0   0   0   0   1   
0   0   0

i.e. all explanatory variables here are dichotomous. However once I run lm()
i got following error :

  

lm(dat[,1]~dat[,-1])


Error in model.frame.default(formula = dat[, 1] ~ dat[, -1],
drop.unused.levels = TRUE) : 
  invalid type (list) for variable 'dat[, -1]'


Can anyone please tell me what to do?

Best,

  



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tel.: 01 49 81 31 31 (poste 18470)

Sec : 01 49 81 32 90
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Re: [R] how to suppress the output from stepAIC?

2009-11-10 Thread mohamed . lajnef

Hi Jack,

try stepAIC with trace parameter: stepAIC(...,trace=FALSE)

Regards,
M

Jack Luo  a écrit :


Hi,

I am now running a cross-validation using coxph coupled with stepAIC for
model selection, is there anyway to suppress the output? It's too much.

-Jack

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Re: [R] How to sum and group data by DATE in data frame

2009-09-09 Thread Mohamed Lajnef

Hi Clair,
try to use this code

aggregate(toto$Income,by=list((substr(toto$Date,1,7))),sum)

Regards
mohamed

clair.crossup...@googlemail.com a écrit :

Dear all,

Lets say I have a data frame as follows:


  

Date <- as.Date(c('2006-08-23', '2006-08-30', '2006-09-06', '2006-09-13', 
'2006-09-20'))
Income <- c(73.79, 72.46, 76.32, 72.43, 72.62)
data.frame(Date, Income)


Date Income
1 2006-08-23  73.79
2 2006-08-30  72.46
3 2006-09-06  76.32
4 2006-09-13  72.43
5 2006-09-20  72.62
  


is there a way to group the data by month (summing the values in each
month), i.e.

Date   Income
2006-08  146.25
2006-09  221.37

Thanks in advance,
C.C.

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Re: [R] Averaging rows if a condition is true.

2009-09-07 Thread Mohamed Lajnef

Hi,

Try to use aggregate function
RSiteSearch ("aggregate")   #for help


Regards
ML

A Ezhil a écrit :

Dear All,

I have matrix (5 X 60) of subjects and their responses to a set of 
questions. All responses are classified into categories (500). I would like to 
average all subject's responses for each category. I wrote a code using a for 
loop but is not working. Could please tell me what's wrong with the code? I 
guess, there is a elegant R way of doing the same thing.

Thanks in advance.

Kind regards,
Ezhil


j <- 1; n <- dim(dat)[1]; cat <- as.character(dat[,1]);
row <- matrix(nrow=nrow(dat), ncol=ncol(dat));
for(i in 1:n-1) {
  if(cat[i] != cat[i+1]) {row[j, ] <- dat[j, ]}
  else {
  start <- j;
  end <- i;
  }
  row[j, ] <- colMeans(dat[j:i, ]);
  j+1;
  }

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Re: [R] Select top three values from data frame

2009-08-26 Thread Mohamed Lajnef

Noah Silverman a écrit :
I only have a few values in my example, but the real data set might have 
20-100 rows with A="X".  So how do I pick just the three highest ones?


-N

  

Hi,

and now?

df.mydata$B[order(df.mydata[df.mydata$A=="X" AND df.mydata$C < 2, 
]$B)][length(df.mydata$B)-3:length(df.mydata$B)]


cheers,
ML






On 8/26/09 2:46 AM, Ottorino-Luca Pantani wrote:
  

df.mydata[df.mydata$A=="X" AND df.mydata$C < 2, ]
will do the job ?

8rino

Noah Silverman ha scritto:


Hi,

I'm trying to find an easy way to do this.

I want to select the top three values of a specific column in a 
subset of rows in a data.frame.  I'll demonstrate.


ABC
x21
x41
x32
y15
y26
y38


I want the top 3 values of B from the data.frame where A=X and C <2

I could extract all the rows where C<2, then sort by B, then take the 
first 3.  But that seems like the wrong way around, and it also will 
get messy with real data of over 100 columns.


Any suggestions?

  


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Re: [R] Select top three values from data frame

2009-08-26 Thread Mohamed Lajnef

Noah Silverman a écrit :

Hi,

I'm trying to find an easy way to do this.

I want to select the top three values of a specific column in a subset 
of rows in a data.frame.  I'll demonstrate.




Hi,
did you try this?

data[data$A=='x'& data$C<2,]$B # data = your data frame

ABC
x21
x41
x32
y15
y26
y38


I want the top 3 values of B from the data.frame where A=X and C <2

I could extract all the rows where C<2, then sort by B, then take the 
first 3.  But that seems like the wrong way around, and it also will 
get messy with real data of over 100 columns.


Any suggestions?


regards

ML


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Re: [R] Package read large file

2009-08-20 Thread Mohamed Lajnef

Dear Jim & Michael

the file contains 1000 lines and 1.200.000 columns, txt format and  i 
wish to read all the file with R if  possible, or  read a few columns?


I tried without success to turn bigmemory and colbycol packages, also 
scan and readLines  function but  It takes a long time without anything.


I have windows X64 bits 8 Go

In advance thanks
ML




jim holtman a écrit :

A little more detail would be appropriate.  How large is "large"?
What is the format of the file?  What do you wnat to do with the data?
 Do you have to have it all in memory at once?  Does it exist in a
data base already?  What type of system are you running on?  How much
memory do you have?  

On Wed, Aug 19, 2009 at 4:51 AM, Mohamed Lajnef wrote:
  

Dear R-Users,

I am looking for packages that could read large files in R?
any suggestions are welcome.
Regards,
ML

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[R] Package read large file

2009-08-19 Thread Mohamed Lajnef

Dear R-Users,

I am looking for packages that could read large files in R?
any suggestions are welcome.
Regards,
ML

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