[R] as.numeric question
Dear all, Is there a way of keeping the actual information in a factor when converting it to numeric? Example: What I want: as.numeric(factor(5:10)) 5, 6, 7, 8, 9, 10 How R works: as.numeric(factor(5:10)) #R example on 'as.numeric' [1] 1 2 3 4 5 6 Thank you very much. Paulo Paulo M. Brando Instituto de Pesquisa Ambiental da Amazonia (IPAM) Santarem, PA, Brasil. Av. Rui Barbosa, 136. Fone: + 55 93 3522 55 38 www.ipam.org.br E-mail: [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] RES: survival
28 30 interaction(parcel2, sp2)=1.5 15744 7142 33 32 34 Sorry, but I still cannot find what is going wrong. Regards, Paulo -Mensagem original- De: Thomas Lumley [mailto:[EMAIL PROTECTED] Enviada em: Wednesday, March 08, 2006 9:15 AM Para: Paulo Brando Cc: r-help@stat.math.ethz.ch Assunto: Re: [R] survival On Wed, 8 Mar 2006, Paulo Brando wrote: Dear R-helpers, We marked 6000 leaves from 5 SPECIES - 10 individuals/species - in two different TREATMENTs: a control and a dry-plot from which 50% of incoming precipitation was excluded. We followed those leaves for 42 months and noted the presence and absence at each visit. I then carried out a Cox Harzard model to see differences in leaf mortality between parcels and among species over time: leaves.cox - coxph(Surv(time, censo) ~ treatment + species, data= wsuv) When I plot 'survfitt(leaves.cox)', I come up with a survivor curve that starts at 1 ends at 0.4. The problem is that at time 42 almost all leaves are dead. I wander if surfit plot at time 42 should also be close to zero? Yes, it probably should. It is a bit hard to tell what is going wrong, though. If you do plot(survfit(Surv(time,censo)~1,data=wsuv) plot(survfit(Surv(time,censo)~species,data=wsuv) plot(survfit(Surv(time,censo)~treatment,data=wsuv) plot(survfit(Surv(time,censo)~interaction(treatment,species),data=wsuv) you will get Kaplan-Meier estimates of survival curves. Looking at these may tell you what is happening. -thomas I followed examples from Venables and Ripley' book. (These analysis are quite new for me). summary(leaves.cox) Call: coxph(formula = Surv(time, censo) ~ (treatment) + species, data = wsuv) n= 140840 coef exp(coef) se(coef) z p treatment -0.0209 0.98 0.00847 -2.47 0.014 species 0.0712 1.07 0.00296 24.07 0.000 exp(coef) exp(-coef) lower .95 upper .95 treatment0.98 1.021 0.963 0.996 species 1.07 0.931 1.068 1.080 Rsquare= 0.004 (max possible= 1 ) Likelihood ratio test= 590 on 2 df, p=0 Wald test= 587 on 2 df, p=0 Score (logrank) test = 588 on 2 df, p=0 My best regards and thanks in advance! Paulo Paulo M. Brando Instituto de Pesquisa Ambiental da Amazonia (IPAM) Santarem, PA, Brasil. Av. Rui Barbosa, 136. Fone: + 55 93 3522 55 38 www.ipam.org.br E-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] RES: How to read an excel data into R?
Hi ling, save your file as 'csv'. I always use the following script: NAME - read.table(C:LOCATION/FILE NAME.csv, header = TRUE, sep = ,, na.string=.) Paulo M. Brando Instituto de Pesquisa Ambiental da Amazonia (IPAM) Santarem, PA, Brasil. Av. Rui Barbosa, 136. Fone: + 55 93 522 55 38 www.ipam.org.br E-mail: [EMAIL PROTECTED] -Mensagem original- De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Em nome de Ling Jin Enviada em: Wednesday, June 22, 2005 4:47 PM Para: r-help@stat.math.ethz.ch Assunto: [R] How to read an excel data into R? Hi all, Does anybody know the easiest way to import excel data into R? I copied and pasted the excel data into a txt file, and tried read.table, but R reported that Error in read.table(data_support.txt, sep = , header = T) : more columns than column names Thanks! Ling __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] RES: another aov results interpretation question
Dear All, I created a script to calculate averages - two groups: parcel and date - and, based on these averages, make a graph. The problem is that 'R' does not recognize the first column even if I try to insert one. A brief example Raw data: Data - sample(1:100, 30, replace = FALSE, prob = NULL) Date - rep(c(02/30/2004,03/15/2004, 04/16/2004, 06/14/2004, 07/08/2004), 6) Parcel - rep(c(DRY, CONTROL),15) Df - data.frame(Parcel, Date, Data) Res - tapply(Df$Data, list(Date = Df$Date, Parcel = Df$Parcel), mean) Res Parcel Date CONTROL DRY 02/30/2004 53.0 52.3 03/15/2004 54.0 67.7 04/16/2004 54.7 30.0 06/14/2004 27.7 20.0 07/08/2004 59.0 38.0 colnames(Res) [1] CONTROL DRY matplot(Res[,1], Res[,-1], type = l) ### It does not recognize the colunm Date. Why? Thanks in advance! Paulo Paulo M. Brando Instituto de Pesquisa Ambiental da Amazonia (IPAM) Santarem, PA, Brasil. Av. Rui Barbosa, 136. Fone: + 55 93 522 55 38 www.ipam.org.br E-mail: [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] RES: Learning, if and else
Thank you very much! All methods are very useful.
Paulo M. Brando
Instituto de Pesquisa Ambiental da Amazonia (IPAM)
Santarem, PA, Brasil.
Av. Rui Barbosa, 136.
Fone: + 55 93 522 55 38
www.ipam.org.br
E-mail: [EMAIL PROTECTED]
-Mensagem original-
De: Adaikalavan Ramasamy [mailto:[EMAIL PROTECTED]
Enviada em: Tuesday, June 14, 2005 3:17 AM
Para: Paulo Brando
Cc: R-help@stat.math.ethz.ch
Assunto: Re: [R] Learning, if and else
You can also use switch() instead of ifelse(), which makes the code a
bit easier to read. The downside to this is that switch does not take
vectorised input and thus you need a loop. For example
# data
dbh - c(30,29,15,14,30,29)
form - factor(c(tree, tree, liana, liana, palm, palm))
df - data.frame(dbh, form)
out - numeric( nrow(df) )
for(i in 1:nrow(df)){
x - as.numeric( df[i, 1] )
y - as.character( df[i, 2] )
out[i] - switch( y,
tree = exp( -4.898 + 4.512*log(x) - 0.319*(log(x))^2 ),
liana = 10^(0.07 + 2.17 * log10(x)),
NA
)
}
Or slightly more efficient solution is
out - apply( df, 1, function(z){
x - as.numeric(z[1]); y - as.character(z[2])
switch( y,
tree = exp( -4.898 + 4.512*log(x) - 0.319*(log(x))^2 ),
liana = 10^(0.07 + 2.17 * log10(x)),
NA
)
})
Regards, Adai
On Mon, 2005-06-13 at 15:32 -0700, Paulo Brando wrote:
Dear Rs,
I have tried to use conditional expressions to calculate biomass for
different life forms (trees, lianas, and palms).
Here is an example:
lifeform
dbh form
1 30 tree
2 29 tree
3 28 tree
4 27 tree
5 26 tree
6 25 tree
7 24 tree
8 23 tree
9 22 tree
10 21 tree
11 20 tree
12 15 liana
13 14 liana
14 13 liana
15 12 liana
16 11 liana
17 10 liana
18 9 liana
19 8 liana
20 7 liana
21 6 liana
22 5 liana
23 30 palm
24 29 palm
25 28 palm
26 27 palm
27 26 palm
28 25 palm
29 24 palm
30 23 palm
31 22 palm
32 21 palm
33 20 palm
### I want to include biomass
lifeform$biomass -
{
if(lifeform$form==tree)
exp(-4.898+4.512*log(dbh)-0.319*(log(dbh))^2)
else{
if (lifeform$form==liana)
10^(0.07 + 2.17 * log10 (dbh))
else (NA)
}
Warning message:
the condition has length 1 and only the first element will be used
in:
if (lifeform$form == tree) exp(-4.898 + 4.512 * log(dbh) -
### But I always get the message warning message above.
I looked for similar examples in R mail list archive, but they did not
help a lot.
I am quite new to 'R'. Any material that covers this theme?
Thank you very much!
Paulo
PS. Sorry about the last e-mail. I did not change the message title.
Paulo M. Brando
Instituto de Pesquisa Ambiental da Amazonia (IPAM)
Santarem, PA, Brasil.
Av. Rui Barbosa, 136.
Fone: + 55 93 522 55 38
www.ipam.org.br
E-mail: [EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] RES: install R 2.1.0 patched from source on FC3
Dear Rs,
I have tried to use conditional expressions to calculate biomass for
different life forms (trees, lianas, and palms).
Here is an example:
lifeform
dbh form
1 30 tree
2 29 tree
3 28 tree
4 27 tree
5 26 tree
6 25 tree
7 24 tree
8 23 tree
9 22 tree
10 21 tree
11 20 tree
12 15 liana
13 14 liana
14 13 liana
15 12 liana
16 11 liana
17 10 liana
18 9 liana
19 8 liana
20 7 liana
21 6 liana
22 5 liana
23 30 palm
24 29 palm
25 28 palm
26 27 palm
27 26 palm
28 25 palm
29 24 palm
30 23 palm
31 22 palm
32 21 palm
33 20 palm
### I want to include biomass
lifeform$biomass -
{
if(lifeform$form==tree)
exp(-4.898+4.512*log(dbh)-0.319*(log(dbh))^2)
else{
if (lifeform$form==liana)
10^(0.07 + 2.17 * log10 (dbh))
else (NA)
}
Warning message:
the condition has length 1 and only the first element will be used in:
if (lifeform$form == tree) exp(-4.898 + 4.512 * log(dbh) -
### But I always get the message warning message above.
I looked for similar examples in R mail list archive, but they did not
help a lot.
I am quite new to 'R'. Any material that covers this theme?
Thank you very much!
Paulo
Paulo M. Brando
Instituto de Pesquisa Ambiental da Amazonia (IPAM)
Santarem, PA, Brasil.
Av. Rui Barbosa, 136.
Fone: + 55 93 522 55 38
www.ipam.org.br
E-mail: [EMAIL PROTECTED]
-Mensagem original-
De: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] Em nome de Peter Dalgaard
Enviada em: Monday, June 13, 2005 9:04 AM
Para: Weiwei Shi
Cc: R-help@stat.math.ethz.ch
Assunto: Re: [R] install R 2.1.0 patched from source on FC3
Weiwei Shi [EMAIL PROTECTED] writes:
Hi,
I have some problem when I tried to install R from source:
work as root
cd /root/dls/R-patched # this is where I unzip the file:
R-patched_2005-06-08.tar.gz
make clean
# i need my R_HOME=/usr/lib/R and I need to create libR.so for
RSPython, so I did:
./configure --prefix=/usr/lib/R --enable-R-shlib
make
But I found the R is built in /root/dls/R-patched/bin instead of
/usr/lib/R/bin
Did I miss something here?
Yes.
make install
--
O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B
c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K
(*) \(*) -- University of Copenhagen Denmark Ph: (+45)
35327918
~~ - ([EMAIL PROTECTED]) FAX: (+45)
35327907
__
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PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Learning, if and else
Dear Rs,
I have tried to use conditional expressions to calculate biomass for
different life forms (trees, lianas, and palms).
Here is an example:
lifeform
dbh form
1 30 tree
2 29 tree
3 28 tree
4 27 tree
5 26 tree
6 25 tree
7 24 tree
8 23 tree
9 22 tree
10 21 tree
11 20 tree
12 15 liana
13 14 liana
14 13 liana
15 12 liana
16 11 liana
17 10 liana
18 9 liana
19 8 liana
20 7 liana
21 6 liana
22 5 liana
23 30 palm
24 29 palm
25 28 palm
26 27 palm
27 26 palm
28 25 palm
29 24 palm
30 23 palm
31 22 palm
32 21 palm
33 20 palm
### I want to include biomass
lifeform$biomass -
{
if(lifeform$form==tree)
exp(-4.898+4.512*log(dbh)-0.319*(log(dbh))^2)
else{
if (lifeform$form==liana)
10^(0.07 + 2.17 * log10 (dbh))
else (NA)
}
Warning message:
the condition has length 1 and only the first element will be used in:
if (lifeform$form == tree) exp(-4.898 + 4.512 * log(dbh) -
### But I always get the message warning message above.
I looked for similar examples in R mail list archive, but they did not
help a lot.
I am quite new to 'R'. Any material that covers this theme?
Thank you very much!
Paulo
PS. Sorry about the last e-mail. I did not change the message title.
Paulo M. Brando
Instituto de Pesquisa Ambiental da Amazonia (IPAM)
Santarem, PA, Brasil.
Av. Rui Barbosa, 136.
Fone: + 55 93 522 55 38
www.ipam.org.br
E-mail: [EMAIL PROTECTED]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] aggregate and stack
Dear All, I have tried to calculate tree mean growth but I think the structure I used below (growthresumo) is not the most elegant, even though it worked. The only problem I had in this first part was that I cannot use 'summary', just 'mean' (sorry but 'R' is pretty new for me). growthresumo - aggregate(growth[,c(16,19,23,27,31,35,39,43,47,52,56,60,64,68,72,76,81,85,89,93,97,101,105,109,113,117,121,125,129,133,137, 141,145,149,153,157,161,165,169,173,177,181,185,189,194,197,201,205,209,213,217,221,225,229,233,237,241)], by=(growth[,c(3,8)]),MEAN,na.rm=TRUE) #after growth is calculated, I want to stack the results in just one colunm. growthvertical - c(growthresumo[,3],...,growthresumo[,50]) # this is very time consuming though Parcel - c(C9,S8...C9,S8) # 50 items date c(DATE1DATE50) growthpermonth - data.frame(Parcel, Date, growthvertical) Thank you very much! Paulo Paulo Brando Inst. de Pesquisa Ambiental da Amazônia (IPAM) Rua Rui Barbosa,136. 68.005.080 Santarém, PA, Brasil. Fone/Fax ++ 55 93 522 5538 www.ipam.org.br [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Basic matematical functions with NAs
Dear All, I've tried to sum columns -- different species of flowers, fruits plus twigs -- with NAs to get litterfall/trap, and then after use litterfall to calculate production (litterfall (grams)/ hectare/ day. But R 'sees' litterfall/trap as a string. My question: How to use basic mathematical functions to deal with NAs in data management. Example (as you can note I have many missing values -- no fruit fell in the trap. area ponto date pseco psaco pliquido florg1 flor1 florg2 flor2 florg3 flor3 frutog1 fruto1 frutog2 fruto2 frutog3 fruto3 frutog4 fruto4 frutog5 fruto5 frutog6 fruto6 frutog7 fruto7 frutog8 fruto8 twigs A A 1 38233 17.7 1.6 7.1 0.266 1 A AA 1 38233 12.5 8.7 3.8 A AB 1 38233 13.9 1.7 3.2 0.421 3 A B 1 38233 12.1 1.6 1.5 0.248 2 0.435 7 0.16 1 A BORDA 1 38233 A C 1 38233 15.6 1.7 4.9 0.374 2 0.298 3 0.231 1 A F 1 38233 14 1.5 3.5 0.366 45 0.153 1 0.15 1 Paulo Brando Inst. de Pesquisa Ambiental da Amazônia (IPAM) Rua Rui Barbosa,136. 68.005.080 Santarém, PA, Brazil. Fone/Fax ++ 55 93 522 5538 www.ipam.org.br [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
