>off-diagonal plots
>
>On 2/7/2015 9:50 AM, meng wrote:
>> If there are many character variables,and I want to get the mosaic plot of
>> every pair of each variable,how to do then?
>>
>>
>> If the variables are numeric, I can use pairs to get paired scatter
--
QQ: 1733768559
At 2015-02-07 17:04:26,"Jim Lemon" wrote:
>Hi meng,
>It's not too hard to get a mosaic plot of two character variables:
>
>x<-sample(LETTERS[1:3],20,TRUE)
>y<-sample(LETTERS[24:26],20,TRUE)
>mosaicplot(table(x,y))
>
>If you coul
If both x and y are all character, paired scatter plot is a little bit strange
I think.
--
QQ: 1733768559
At 2015-02-06 23:52:34,"Duncan Murdoch" wrote:
>On 06/02/2015 6:46 AM, meng wrote:
>> Hi all:
>> If there are two numeric variable:x,y, and I can g
Hi all:
If there are two numeric variable:x,y, and I can get paired scatter plot by
function "pairs".But if x and y are character, and I want to get paired mosaic
plot,which function should be used then?
Many thanks!
My best.
--
QQ: 1733768559
[[alternative HTML version deleted]
Very sorry that I ignore the question type.Yest,it's not an R question.
Many thanks for your answer.
--
QQ: 1733768559
At 2015-02-05 18:26:01,"peter dalgaard" wrote:
>
>On 05 Feb 2015, at 09:20 , meng wrote:
>
>> Hi all:
>> If I want to test
Hi all:
If I want to test whether the mean of a set of normal distributed data is
different from a value(e.g. 0), I can use one sample t test . But if the data
is not normal distributed,what kind of method should be used?
Many thanks!
My best.
[[alternative HTML version delete
(650) 467-7374
>>
>> "Data is not information. Information is not knowledge. And knowledge
>> is certainly not wisdom."
>> H. Gilbert Welch
>>
>>
>>
>>
>> On Sun, May 11, 2014 at 6:42 AM, Uwe Ligges
>> wrote:
>>>
>>
boxplot(count ~ spray, data = InsectSprays)
The order of the plot is: A,B,C,D,E,F.
If I want the order of the plot is: F,B,E,A,D,C, how to do then?
Many thanks!
My best
--
QQ: 1733768559
[[alternative HTML version deleted]]
__
R
Thanks.
--
QQ: 1733768559
At 2014-04-15 17:56:01,"Jim Lemon" wrote:
>On 04/15/2014 07:51 PM, meng wrote:
>> Yes,it works !
>> What's the reason for it?
>>
>Hi meng,
>As I understand it, the lattice graphics system produces plot
>objects, not
Yes,it works !
What's the reason for it?
Many thanks!
--
QQ: 1733768559
At 2014-04-15 14:36:41,"Jim Lemon" wrote:
>On 04/15/2014 11:46 AM, meng wrote:
>> Hi all:
>> I met a question about the output of plot.
>> I want to output 3 plots.
>&
Hi all:
I met a question about the output of plot.
I want to output 3 plots.
Method1: by function histogram{lattice}
Method2: by function hist{graphics}
But method1 failed(the output is empty),and only method 2 works.
I can't find out the reason,and many thanks for your help.
#Method1---failed
Hi all:
I met a question about plotting the predict values of time series.
The data is treering which belongs to R.
I want to predict the next 3 years(1980,1981,1982) result.
pred<-predict(ar(treering),n.ahead=3)
plot(treering)
lines(pred$pred,col="red",lwd=3)
I can only find a little bit red
Ok,many thanks for your detailed answer.
At 2013-09-18 19:20:26,"Duncan Murdoch" wrote:
>On 13-09-18 1:38 AM, meng wrote:
>> Oh,yes, I found out this according to your reply.Thanks.
>>
>> As to time series analysis, in order to show the effect of smoot
.
At 2013-09-18 08:49:51,"Duncan Murdoch" wrote:
>On 13-09-17 6:36 PM, meng wrote:
>> Thanks for your reply.
>>
>> Is "fitted(lm(...))" the same as "values" of lines(values)?
>>
>> If yes,then why the range of lines(values) is
Thanks for your reply.
Yes, abline is much better,but it can only add straight line,and
lines(fitted(lm(y~x)) ~ x) can do the same thing but not constrain to straight
line.
At 2013-09-17 20:45:08,"Bretschneider (R)" wrote:
Dear
meng,
Re:
Hi all:
I met a question ab
Thanks for your reply.
Is "fitted(lm(...))" the same as "values" of lines(values)?
If yes,then why the range of lines(values) is different from
range(fitted(lm(...)))?
If no, what "values" refers to?
At 2013-09-17 20:56:04,"Duncan Murdoch" wrote:
Hi all:
I met a question about "lines".
attach(cars)
plot(dist ~ speed)
#add the regression line to the plot
lines(fitted(lm(dist~speed)) ~ speed)
plot(dist ~ speed)
#what kind of curve does the following command add to the plot?
lines(fitted(lm(dist~speed)))
My question is :
what kind of c
Hi all:
My data:
f1 f2 f3 f4 y
1 -4.45 3 6.44 3.721
2 25.66 121 39.03 3.55 0
3 -9.75 12 11.77 3.40 1
4 -12.09 17 13.75 3.33 0
5 -3.53 4 6.47 3.24 0
6 13.18 69 26.55 3.201
... ...
I want to plot f1 for y=0 and 1; f2 for y=0 and 1...and the same as
Hi all:
I met a question about lmer.
fm1 <- lmer(Reaction ~ Days + (Days|Subject), sleepstudy)
summary(fm1)
...
Fixed effects:
Estimate Std. Error t value
(Intercept) 251.405 6.825 36.84
Days 10.467 1.5466.77
...
My question:
Why p values of (Intercept) an
remaining variables ¨C and there are NO remaining
variables.
-
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77840-4352
From: meng [mailto:laomen...@163.com]
Sent: Monday, June 3, 2013 8:55 PM
To:dcarl...@t
xas A&M University
>College Station, TX 77840-4352
>
>-Original Message-
>From: r-help-boun...@r-project.org
>[mailto:r-help-boun...@r-project.org] On Behalf Of meng
>Sent: Saturday, June 1, 2013 10:03 AM
>To: R help
>Subject: [R] error about MCA
>
>Hi,all:
&g
Hi,all:
I want to perform multiple correspondance analysis via MCA{FactoMineR}.
The data is in the attachment.
My code:
dat<-read.delim("e:\\mydata.txt",header=T)
MCA(dat,quanti.sup=7,quali.sup=1:6)
Error in `[.data.frame`(tab, , i) : undefined columns selected
My question:
Why does the error ha
How to get p values from the result then?
At 2013-05-28 13:54:27,"David Winsemius" wrote:
>
>On May 27, 2013, at 7:59 PM, meng wrote:
>
>> Hi all:
>> As to the polr {MASS} function, how to find out p values of every
>> parameter?
>>
>
Hi all:
As to the polr {MASS} function, how to find out p values of every parameter?
>From the example of R help:
house.plr <- polr(Sat ~ Infl + Type + Cont, weights = Freq, data = housing)
summary(house.plr)
How to find out the p values of house.plr?
Many thanks.
Best.
[[alternat
Many thanks for your detailed reply.
I'll read your mail thoroughly. Thanks!
At 2013-05-23 21:56:29,"Greg Snow" <538...@gmail.com> wrote:
Meng,
This really comes down to what question you are trying to answer. Before
worrying about details of default contrasts and
It's not homework.
I met this question during my practical work via R.
The boss is an expert of biology,but he doesn't know statistics.So I must find
the right method to this work.
At 2013-05-22 17:30:34,"Uwe Ligges" wrote:
>
>
>On 22.05.2013 07:09, meng
Thanks.
As to the data " warpbreaks", if I want to analysis the impact of
tension(L,M,H) on breaks, should I order the tension or not?
Many thanks.
At 2013-05-21 20:55:18,"David Winsemius" wrote:
>
>On May 20, 2013, at 10:35 PM, meng wrote:
>
&g
Hi all:
If the explainary variables are ordinal,the result of regression is different
from
"unordered variables".But I can't understand the result of regression from
"ordered
variable".
The data is warpbreaks,which belongs to R.
If I use the "unordered variable"(tension):Levels: L M H
The res
value of drug2 is 0.0285 * ,which indicates that the total count of
drug2 is significantly different from drug1 at 0.05 level,which contradict to
the result of anova.
Can you give me some help about that ?
Many thanks.
Best
At 2013-05-13 18:56:01,"Achim Zeileis" wrot
Hi all:
I have a question about poisson regression.
My data:
drug result count
1 1 8
1 254
2 1 20
2 2 44
My model:
model<- glm(count ~ drug*result, family = poisson)
My result:
summary(model)
Coefficients:
Estimate Std. Error z
Hi all:
I have a question about one-way anova.
The data is ¡°sleep¡±which belongs to R.
code1:
summary(lm(extra~group))
Estimate of group2(1.58) is the difference between mean of group1 and
group2,and t value(1.861) and p value(0.0792) is the same as 2 sample t
test,and the code is ¡°
his "only difference"is significant.
But as to g1 and g3:
g1: a + d + c + a:d + a:c
g3: a + d + c + a:d + a:c + d:c + a:d:c
The only difference between g1 and g3 is "d:c + a:d:c".
What's "d:c + a:d:c" refers to?
Thanks.
At 2013-04-29 14:33:25,&
27;t find out the "only difference" between g1 and g3,so I don't know what
"anova(g1, g3, test = "Chisq")" tests for.
Also, what "/" sign following age in g1 refers to?
Many thanks and sorry for many quesions.
Best
At 2013-04-24 22:22:55,
Understand.
Many thanks for your detailed explaination.
At 2013-04-24 22:22:55,"Achim Zeileis" wrote:
>On Wed, 24 Apr 2013, meng wrote:
>
>> Hi,Achim:
>> Can all the analysis you mentioned via loglm be performed via
>> glm(...,family=poisson)?
>
Hi,Achim:
Can all the analysis you mentioned via loglm be performed via
glm(...,family=poisson)?
Many thanks.
At 2013-04-24 19:37:10,"Achim Zeileis" wrote:
>On Wed, 24 Apr 2013, meng wrote:
>
>> Hi all:
>> For stratified count data,how to perform regres
Hi all:
For stratified count data,how to perform regression analysis?
My data:
age case oc count
1 1 121
1 1 226
1 2 117
1 2 259
2 1 118
2 1 288
2 2 1 7
2 2 2 95
age:
1:<40y
2:>40y
ca
Many thanks for your reply.
At 2013-03-14 19:22:41,"Jim Lemon" wrote:
>On 03/14/2013 12:10 PM, meng wrote:
>> Thanks for your notice.
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> At 2013-03-14 00:18:48,"John Kane&quo
Hi,all:
I met a problem of nls.
My data:
xy
60 0.8
80 6.5
100 20.5
120 45.9
I want to fit exp curve of data.
My code:
> nls(y ~ exp(a + b*x)+d,start=list(a=0,b=0,d=1))
Error in nlsModel(formula, mf, start, wts) :
singular gradient matrix at initial parameter estimates
I can't find out
erence magnitude
>>
>> So strange to find the attachment is disappear.
>> Resent again.
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> At 2013-03-13 13:01:01,"Pascal Oettli" wrote:
>> >Hi,
>>>
>> &
Many thanks
At 2013-03-14 00:23:43,"Richard M. Heiberger" wrote:
Meng,
What seems to be going on is that the covariates are handled very differently
in TukeyHSD and in glht.
Please see the interaction_average and covariate_average arguments to glht.
I ran your exa
avid L Carlson
>Associate Professor of Anthropology
>Texas A&M 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 John Kane
>> Sent: Wedn
Hi all:
I have a question about multi-comparison.
The data is in the attachment.
My purpose:
Compare the predicted means of the 3 methods(a,b,c) pairwisely.
I have 3 ideas:
#idea1
result_aov<-aov(y~ method + x1 + x2)
TukeyHSD(result_aov)
difflwr upr p adj
b-a 0.845
So strange to find the attachment is disappear.
Resent again.
At 2013-03-13 13:01:01,"Pascal Oettli" wrote:
>Hi,
>
>The attachment has been deleted. Please be more specific.
>
>Regards,
>Pascal
>
>On 13/03/13 10:20, meng wrote:
>> Hi all:
>&g
Hi all:
Is there a plot tool to use different color indicates difference magnitude of
data?
The plot is in the attachment.
Many thanks.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Many thanks.
I understand from your reply.
Thanks.
At 2013-03-11 15:45:14,"PIKAL Petr" wrote:
>Hi
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of meng
>> Sent: Sunday,
if x1 and x2
are centered first).
On Sat, Mar 9, 2013 at 3:35 AM, meng wrote:
Hi all:
My data is in the attachment.
I want to analysis the mean difference of y between 2 sex.
My code:
result_lm<-lm(y~factor(sex) + x1 + x2)
summary(result_lm)
The result of "factor(sex)m" 136.83, is
On Sat, Mar 9, 2013 at 10:35 AM, meng wrote:
Hi all:
My data is in the attachment.
I want to analysis the mean difference of y between 2 sex.
My code:
result_lm<-lm(y~factor(sex) + x1 + x2)
summary(result_lm)
The result of "factor(sex)m" 136.83, is the mean difference of y between
Hi all:
My data is in the attachment.
I want to analysis the mean difference of y between 2 sex.
My code:
result_lm<-lm(y~factor(sex) + x1 + x2)
summary(result_lm)
The result of "factor(sex)m" 136.83, is the mean difference of y between 2
sex,and the corresponding p value is 0.07618.
My quest
Hi all:
I have a quesion about ANOVA: Is SS(Sum of Square) of a specific factor
constant with the number of factors changing?
dat1 includes one factor g1,and g1's SS is called SS_g1_dat1.
dat2 includes two factors g1,g2,and g1's SS is called SS_g1_dat2.
My quesion is: Is SS_g1_dat1 equals to S
Thanks.
At 2013-01-04 17:41:23,"Jorge I Velez" wrote:
Dear meng,
Check I understood correctly, the Ryacas package might do what you want. Check
its vignette at
http://cran.r-project.org/web/packages/Ryacas/vignettes/Ryacas.pdf
Best,
Jorge.-
On Fri, Jan 4, 2013 at 5:5
No,I have no x1 and x2 values.
At 2013-01-04 17:32:24,"PIKAL Petr" wrote:
>Hi
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of meng
>> Sent: Friday, January 04, 2013
Hi all:
I have a question about the computation of expression:
y = 1+(2x1+2)-3*(5x1-1)+(3x2+3)-2*(x2-1)
The result of y is:
y = 11-13x1+x2
How can I compute the result of y via R function?
Many thanks!
My best.
[[alternative HTML version deleted]]
___
Ok,I'll follow your suggestion.
Thanks!
At 2012-12-31 11:20:17,arun wrote:
>HI Meng,
>
>
>Just try:
>rep(z,rach=nrow(dat))
>#[1] 0.1 10.0 100.0
> rep(z,chair=nrow(dat))
>#[1] 0.1 10.0 100.0
>rep(z,times=nrow(dat))
>#[1] 0.1 10.0 100.0 0.1 1
Many thanks arun!
Your answer is the best one :)
At 2012-12-31 10:51:35,arun wrote:
>HI Meng,
>
>NO problem.
>#In fact,
>sweep(dat,2,z,"*") #will be data.frame
> # x1 x2 x3
>#1 2 120 2500
>#2 5 200 5000
>#3 8 300 6200
> str(sweep(dat,2
s 100, which is confusing.
> t(t(dat)*z)
>#x1 x2 x3
>#[1,] 2 120 2500
>#[2,] 5 200 5000
>#[3,] 8 300 6200
>
>A.K.
>
>
>
>
>
>
>
>- Original Message -
>From: meng
>To: R help
>Cc:
>Sent: Sunday, December 30, 2012 5:26 AM
>
" wrote:
>
>On 30-12-2012, at 11:26, meng wrote:
>
>> hi all:
>> Here's a dataframe(dat) and a vector(z):
>>
>> dat:
>> x1 x2x3
>> 0.2 1.2 2.5
>> 0.5 2 5
>> 0.8 3 6.2
>>
>>> z
>> [1]
=11:13)
> as.matrix(dat)
> x1 x2
>[1,] 1 11
>[2,] 2 12
>[3,] 3 13
>t(as.matrix(dat)) * c(2, 3)
> [,1] [,2] [,3]
>x1246
>x2 33 36 39
>
>>
>> Andrius
>>
>>
>> 2012/12/30 Neal H. Walfield
>>
>> &
at I want is:
x1 x2 x3
1 0.1 40 700
2 0.2 50 800
3 0.3 60 900
At 2012-12-31 00:13:33,"Neal H. Walfield" wrote:
>At Sun, 30 Dec 2012 18:26:45 +0800 (CST),
>meng wrote:
>>
>> hi all:
>> Here's a dataframe(dat) and a vector(z):
>>
>>
"rach" mistakenly.
What's strange to me is :No error reply appears,but show me the result:
> dat*rep(z,rach=nrow(dat))
x1x2x3
1 0.1 0.4 0.7
2 20.0 50.0 80.0
3 300.0 600.0 900.0
Why does it happen,and what "rach" means?
Many thanks.
At 2
hi all:
Here's a dataframe(dat) and a vector(z):
dat:
x1 x2x3
0.2 1.2 2.5
0.5 2 5
0.8 3 6.2
> z
[1] 10 100 100
I wanna do the following:
10*x1,100*x2,1000*x3
My solution is using the loop for z and dat(since the length of z is the same
as ncol of dat),which is t
a.frame,lapply(lapply(1,function(y) dat),function(y)
> {y$f<-y$a/median(y$a);y$x_new<-y$f*y$x;y}))
>A.K.
>
>
>
>
>
>
>
>From: meng
>To: arun
>Cc: R help
>Sent: Monday, December 24, 2012 5:29 AM
>Subject: Re:Re: [
ks for you.
>set.seed(5)
>dat<-data.frame(x=sample(1:20,6,replace=TRUE),a=sample(20:40,6,replace=TRUE))
> dat1<-transform(dat,f=a/median(a),x_new=x*(a/median(a)))
> head(dat1,2)
># x af x_new
>#1 5 31 1.016393 5.081967
>#2 14 36 1.180328 16.524590
>A.K
A data set(dat),has 2 variables: x and a, and 100 rows.
I wanna add 2 variables,and call the new data set dat1:
var1:f = a/median(a)
var2:x_new = x*f
My solution:
dat1<-transform(dat,f = a/median(a),x_new = x*f)
But gets error reply which says that "f" is not exits since dat has no
variables c
as.numeric() is work
Settled!
2011/11/11 Lao Meng
> t2-t1 is ok,but only can get"Time difference of 786 days",but I wanna get
> the "786" only.How can I do then?
>
> Thanks!
>
>
> 2011/11/11 Lao Meng
>
>> time1 = 2008-03-09
>> time2
t2-t1 is ok,but only can get"Time difference of 786 days",but I wanna get
the "786" only.How can I do then?
Thanks!
2011/11/11 Lao Meng
> time1 = 2008-03-09
> time2 = 2010-9-10
>
> How to compute how many years between time1 and time2?
>
> Thanks!
&g
time1 = 2008-03-09
time2 = 2010-9-10
How to compute how many years between time1 and time2?
Thanks!
best
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PLEASE do read the
between the two ratios.
Thanks,
Meng
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide
ot the same test. The concept of linear regression applies here
> (and you might take any introductory at your hand to refresh that
> concept). The intercept is estimated from the whole sample not just
> group==1, dfs are 20-2, not sum(group==1)-1!
>
> best regards
>
> Am 17.08.2
Hi all:
If I have a dataframe of N columns.If I wanna get the min(or max,or
mean...etc)of the whole dataframe,how to do it quickly?
What I can do is only:
min(data[,1:ncol(data)])
Any other suggestion?
Thanks!
best
[[alternative HTML version deleted]]
__
st 0 (one sample t-test),
> so the t-statistic is (mean-0)/sd, having n-k (sample size - number of
> parameters) degrees of freedom.
>
> cc<-s1$coef["(Intercept)",1:2]
> 2*(1-pt(cc[1]/cc[2],df=18))
>
>
> hth.
>
> Am 16.08.2011 07:25, schrieb Lao Meng:
>
Yes, you remind me of this!
Thanks!
2011/8/16 Eik Vettorazzi
> Hi Lao,
> you tried to reinvent the wheel. Have a look at ?tapply
>
> tapply(sleep$extra,sleep$group,mean)
>
> Cheers
>
> Am 16.08.2011 09:41, schrieb Lao Meng:
> > Hi all:
> > My data:data(slee
Sorry,I miss a word:"If I wanna calculate each group's mean extra"
2011/8/16 Lao Meng
> Hi all:
> My data:data(sleep)
>
> If I wanna calculate each group's extra,what I can do is:
> #method1
> attach(sleep)
> mean(extra[group==1])
> mean(extra[gro
Hi all:
My data:data(sleep)
If I wanna calculate each group's extra,what I can do is:
#method1
attach(sleep)
mean(extra[group==1])
mean(extra[group==1])
#method2
result<-matrix(,0,2)
g<-split(sleep,sleep$group)
for(i in 1:length(g))
{
result<-rbind(result,data.frame(unique(g[[i]]$group),mean(g[[
Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> ---
>
> Sent from my phone. Please excuse my brevity.
>
> Lao Meng wrote:
>>
>> The mean of group1 is 0.75,just the same as intercept.
>> Acturally,R trea
d Controllers) .OO#. .OO#. rocks...1k
> ---
>
> Sent from my phone. Please excuse my brevity.
>
> Lao Meng wrote:
>
>> Hi all:
>> I have a question about lm on t-test.
>>
>> data(sleep)
Hi all:
I have a question about lm on t-test.
data(sleep)
I wanna perform t-test to test the difference between the 2 groups:
I can use:
t.test(extra~group)
The t.test result shows that:t = -1.8608; mean1=0.75,mean2=2.33
But I still wanna use:
summary(lm(extra~group))
Intercept=0.75,which is
Here's another question:
What's kind of difference does glht tests?
In other words,
If I get 4 intercepts and 4 slopes for all levels of GROUP,then glht tests
for the difference of the 4 intercepts or 4 slopes,or something else?
Thanks.
2011/7/27 Lao Meng
> Yes.
> A
Yes.
According to your suggestion,I modified my code.It works well.
Thanks you very much.
My best.
2011/7/26 Peter Ehlers
> On 2011-07-26 00:16, Lao Meng wrote:
>
>> Hi all:
>> There's a question about glht function.
>>
>> My data:data_ori,which inclu
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP<-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result <- lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(result, linfct = mcp(f_GROUP="Tukey") )
Error
Hi all:
I have a question about linear mixed model.
my linear mixed model with randomized slope and intercept with interaction
of time and group(g1,g2,g3):
model<- glmmPQL(log10(CD4) ~ time + factor(group)+ time:factor(group),
random = ~time|id)
What I get is only the main and interaction of time
If I have 3 groups,and for each group,I get the ratio(e.g. incidence rate).
Now I wanna compare 3 ratio pairwise,and get the corresponding p values,i.e:
group1 vs group2 ,p value=?
group1 vs group3 ,p value=?
group2 vs group3 ,p value=?
Which statistical test should be used?
Thanks a lot for yo
ow),M(median),H(High).
2011/6/28 David Winsemius
>
> On Jun 27, 2011, at 10:02 PM, Lao Meng wrote:
>
> Hi all,I have some questions about the covariants of regression.
>>
>> My target: To explore the trend of CD4 level through a period of time.
>>
>> Respon
Hi all,I have some questions about the covariants of regression.
My target: To explore the trend of CD4 level through a period of time.
Response variable: CD4 count
Explanatory variable:time
Also, the demology information is available,such as gender,occupation,income
level...
Q1,Are these varia
You may try "xyplot"
2011/5/20 1Rnwb
> Hello gurus,
>
> I have a dataframe containing two groups viz., 'control' and 'case', each
> of
> these groups contains longitudinal data for 100 subjects. I have to plot
> all
> these subjects on a single chart and then put a regression line for each of
>
ormat
since the "huge" number of variables. One possible way is to project the
multidimensional space into 2-D platform, but I could not find any good way
to do that. Any suggestions or comments will be really helpful!
Thanks,
Meng
[[alternative HTML
Hi all:
Which function or package can perform "competitive risk model"?
Thanks a lot!
My best
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
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PLEASE do read the posting gui
You may try the following to perform anova:
anova(lm(y~x))
or
summary(aov(y~x))
2011/4/13 Ubuntu Diego
> Hi all,
>I have a very easy questions (I hope). I had measure a property of
> plants, growing in three different substrates (A, B and C). The rest of the
> conditions remained con
hi:
Another question abour overdispersion test.
I wanna make sure that:
if p value<0.05,then the data is NOT overdispersion;
if p value>=0.05,then the data IS overdispersion.
I'm not sure whether it's true,just get the above conclusion from simulated
data.
Thanks for your help.
2011/4/2
> H
Hi all:
I have a question about the sample size calculation.
It's a pilot study,which includes 2 groups(low,high),3 time point(3,6,9
monthes).Each person has 3 results which according to the
3 time points.So it's a longitudinal study.
I want to calculate the minimum sample size from the pilot st
Hi all:
I have a question on sample size calculation of 2 groups of data. If 2
groups of data are all normal distribution, then I can use the function
"n.indep.t.test.eq" from samplesize package.But if 2 groups of data are all
skewed distribution, but not normal distribution,how can I calculate the
No,even there's only one observation,boxplot can still be drawn.
x<-1
boxplot(x)
or
x<-1:3
boxplot(x)
...
2011/3/9 Dennis Murphy
> Hi:
>
> A box plot is based on a five number summary, so you need at a minimum five
> observations (and preferably at least twice that) to make a box plot a
> v
You may try analysis of covariance.
But,as you say"There are no gender differences in age ",then why not combine
2 gender's age and ignore the gender?
2011/3/1 Jonathan DuBois
> Hi,
>
> I am new to R, so I am unsure of the formula to set up this analysis.
> I would like to run a linear model wit
Well,it's my daily work on duty,not homework:)
2011/2/18 Dennis Murphy
> This is a built-in dataset in R - see ?HairEyeColor and str() it. I smell
> homework...
>
> Dennis
>
> On Thu, Feb 17, 2011 at 11:50 PM, Lao Meng wrote:
>
>> The data is in the attac
Thanks!
It works well.
2011/2/18 Henrique Dallazuanna
> Try this:
>
> xtabs(Freq ~ Hair + Eye + Sex, Dat)
>
> Using Dimitri's Dat example.
>
> On Fri, Feb 18, 2011 at 5:50 AM, Lao Meng wrote:
>
>> The data is in the attachment.
>>
>> What
Thanks!
It works well.
2011/2/18 Dimitris Rizopoulos
> say, 'Dat' is your data frame, then one way to do it is:
>
> with(Dat, tapply(freq, list(hair, eye, sex), c))
>
>
> I hope it helps.
>
> Best,
> Dimitris
>
>
>
>
> On 2/18/2011 8:50 AM,
The data is in the attachment.
What I wanna get is:
, , Sex = Male
Eye
HairBrown Blue Hazel Green
Black32 1110 3
Brown53 502515
Red 10 10 7 7
Blond 3 30 5 8
, , Sex = Female
Eye
HairBrown Blue Hazel Green
Blac
Hi all:
I have 3 questions about the poisson regression of contingency table.
Q1¡¢How to understand the "independent poisson process"as many books or paper
mentioned?
For instance:
Table1
---
treat canernon-cancersum
-
Hi all:
I have 3 questions about the poisson regression of contingency table.
Q1¡¢How to understand the "independent poisson process"as many books or paper
mentioned?
For instance:
Table1
---
treat canernon-cancersum
---
-- Forwarded message --
From: Lao Meng
Date: 2011/1/19
Subject: Re: [R] question about result of loglinear analysis
To: David Winsemius
My command and result are :
> result_sat<-summary(glm_sat)
> result_coe<-result_sat$coefficients
> result_coe
Hi all:
Here's a question about result of loglinear analysis.
There're 2 factors:area and nation.The raw data is in the attachment.
I fit the saturated model of loglinear with the command:
glm_sat<-glm(fre~area*nation, family=poisson, data=data_Analysis)
After that,I extract the coefficients:
res
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